modul xi sem 1 2009-2010

Physics Modul/xi/sem-1/2009

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Chapter 1 : Linear Motion Basic Competency : 1.1.1 Analyzation of Linier motion, anguler motion and parabolic motion uses vector Contents :

a. Position vector b. Displacement vector c. Average velocity d. Vector velocity e. Magnitude of velocity f. Direction of velocity g. Instantaneous velocity h. Average acceleration i. Direction of acceleration j. Instantaneous acceleration

Position vector Position of point object is state with unit vector r = xi + yj for two dimention, and r = xi + yj + zk for three dimention examples : j A B i Position vector for point A is rA = 3i + 2j ; Position vector for point B is rB = 2i + 3j + 2 k Displacement A r B A particle moves from point A to point B : Displacement vector : r = rB - rA = (xB – xA) i + (yB – yA)j Displacement Magnitude : |r| = (xB – xA)2 + (yB – yA)2 (yB – yA) Direction of displacement : tg =


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(xB – xA) r x y Average velocity : v = = i + j t t t Vector of velocity : v = vxi + vyj y Direction of velocity : tg = x Instantaneous Velocity : x r t t Instantaneous velocity for an object that moves in area : r dr v = lim t~0 = t dt Vector of velocity: dr d dx dy v = = (xi + yj) = i + j = vx i + vyj dt dt dt dt Component of instantaneous velocity : x(t + t) – x(t) vx = lim t~0 t y(t + t) – y(t) vy = lim t~0 t Magnitude of velocity : |v| = vx

2 + vy2


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vy

Direction of velocity : tg = vx Determine the position of an object from the velocity equation : x = xo + vx dt dan y = yo + vy dt xo and yo is initial position ( at t = 0) Distance and Displacement t1 t2 t3 t t3 Displacement = v(t) dt t1 t2 t3 Distance = + v(t) dt + v(t) dt t1 t2

+ and – is based on the position of curva in a graph (-) If curva is under the time function line and (+) if curva is above the

time function line Acceleration Average Acceleration : v v2 – v1 a = = t t2 – t1 Average acceleration in area : v dv a = lim t~0 = t dt


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Acceleration vector : dv d dvx dvy a = = (vi + vj) = i + j = ax i + ayj dt dt dt dt Acceleration magnitude: |a| = ax

2 + ay2

ay

Acceleration direction tg = ax Determine velocity equation from the acceleration function : v = vo + a dt vo : is initial velocity t = 0 Additional Exercises Topic : linear motion

1. Particle P moves in straight line with position in time is stated with x = 3t2 – 4t + 36, t in second and x in metre. Determine :

a. Average velocity between t = 1s and t =3s b. Initial velocity c. velocity in t = 2s

2. Position of particle q in time is stated with r = 3t2 i + t3 j, Determine :

a. Average velocity between t = 1 s dan t = 3 s b. Initial velocity c. Velocity in t = 2 s

3. Particle P moves in a straight line and its position from O is stated with x = 3t2 – 24t +

36. Determine : a. Initial velocity of P b. P velocity in t = 2s c. Maximum distance P from the initial position (O)

4. A golf ball is hit from the tee near the top of the cliff. Coordinat x and y in time is stated

with equation : x = (18 m/s) t dan y = (4 m/s) t – (4,9 m/s3)t2.

a. write the position vector equation r in time uses unit vector i and j with defferential mothode, determine : b. velocity vector in time c. acceleration vector d. velocity in t = 3s e. acceleration in t = 3s

5. Particle P moves with its velocity is stated v = (4 + 2t2) i + (-10 -4t) j, t in second and v

in m/s. Determine velocity vector and position vector after its moves 3 second, assume P is start from initial position of O.


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6. A Particle moves with a acceleration a. Determine the velocity vector and position vector in time t if :

a. a = 6t, first P is in initial position and moves with j velocity b. a = 2i + j first P at rest at position 41 + j c. a = 16t2 i + 9t j. First P is in initial position and moves with velocity of i + 2j


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Anguler Motion Competency Standart : 1. Analysis universe phenomena and the its rule in term point clasical mechanical

Basic Competency 1.1 To analysis linear motin, anguler motion and parabolic motion uses vector Content :

1. Position an object that moves in anguler motion 2. Average anguler velocity 3. Instantenous anguler velocity 4. Average anguler acceleration 5. Instantenous anguler acceleration

Position an object that moves in anguler motion Position of an object that moves in anguler motion is stated by symbol of , angle that is covered in time t. Anguler velocity (Is an angle that is covered per unit of time

Average anguler velocity t

In graph -t : anguler velocity can be determined uses tangen of angle of slope (rad) t t (sekon)

ttg

and

Instantenous anguler velocity is : dtd

Determining of position from velocity equations


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t

o dt0

.

Anguler Acceleration (

Average anguler acceleration can be computed by t

Instantenous angular acceleration is stated by 2

2

dtrd

dtd

Determining of angular velocity equations from acceleration equation

t

o dt0

.

Exercisses

1. Angel position of a point on the wheel is stated as = (5 + 10t + 2t2) rad, where t in second . Determine :

a. angle position at t = 0 and t = 3s b. average anguler velocity from t = 0 until t = 3 s

2. A cd plate moves a round an axis z follows the equation (t) = 4.2 rad – (2,9 rad/s)t + (0.31 rad/s3)t3. Determine :

a. anguler velocity in time b. anguler acceleration in time c. initial anguler velocity d. anguler velocity at t = 5s

3. A CD plate is rotation round an axis z with equation of anguler acceleration = (0.24 rad/s3) t – 0.89 rad/s2

a. If o = 3.1 rad/s, determine the equation for (t) b. If o = 2.7 rad, determine (t)

4. Rotation of a roller machine is expressed by equation of (t) = 2.50t2 – 0.400t, with t in second and q in radian.

a. Determine anguler velocity vector in time t b. Calculate the anguler velocity at :

i. t = 0 ii. t = 1 second and

iii. t = 2


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c. When the roller is stop

5. A Bicycle wheel with radius of 0.33 m is rotation with anguler acceleration (t) = (1.40 rad/s2) – (0.20 rad/s3) t. a wheel is in a rest at t = 0

a. Determine the anguler velocity in time t b. Calculate positive maximum anguler velocity and positive maximum

angle displacement

Combination Motion Competency Standart : 2. Analysis universe phenomena and the its rule in term point clasical mechanical

Basic Competency 1.1 To analysis linear motin, anguler motion and parabolic motion uses vector Content : An object can do two motion in difference direction in the same time. An example is a boat that perpendicular across the river. Two kind of combination motion :

1. Linear motion with linear motion 2. Linear motion with non linear motion (prjectile motion)

1. Linear motion with linear motion

Position A : x = vx . t A y = vy . t y Distance A : s = x2 + y2 vy v vx x Total velocity : v = vx

2 + vy2

vy Direction velocity ( is determined by tg = ---- vx 2. Linear motion with non linear motion (prjectile motion) Projectile motion is a 2-dimensional curved motion of a particle subjected to constant acceleration. An example is a ball thrown obliquely into air. Velocity along every point on the trajectory changes all the time. The vertical motion is affect by the gravitational acceleration g directed toward the centre of the earth. Gravitational acceleration q has no effect on the horizontal motion of projectile. Hence horizontal component remains constant.


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vy v x vx H vo voy y vox R Component Horizontal : Horizontal velocity : vox = vo cos vx = vox = vo cos (linear motion) coordinat x : x = vx . t (linear motion) = vo cos t Component Vertical : Vertical velocity : voy = vo sin Vy = voy + gt (non linear motion) = vo sin - gt coordinat y : y = vo . t + ½ gt2 (non linear motion) = vo sin . t – ½ gt2 Position in time t is stated by (x,y) : x = vo cos a . t y = vo sin a . t – ½ gt2 Velocity in time t is stated by v : v = vx

2 + vy2

Direction of velocity (a) is determined by : vy tg = --- vx Special cases : 1. Maximum height (H) At maximum height the vertical component of the velocity is zero vy = 0, its causes : vo sin - gt = 0, so v0 sin tH = ---------- g

tH is time that projectile at maximum height


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Maximum height : y = vo sin . t – ½ gt2 = vo sin . tH – ½ gtH

2 2 vo sin a vo sin a = vo sin 1/2 g ---------- g g vo

2 sin2 yH = ------------

2g The horizontal distance when projectile at maximum height (xH) : xH = vo cos tH vo cos vo sin xH = ------------------------- g vo

2 sin 2 xH = ------------ 2g 2. Maximum Distance (R) It can be happen when y = 0 Time of maximum distance is twice times of maximum height

tR = 2tH 2 v0 sin tR = ------------ g xR = vo cos . t vo cos a . 2 v0 sin = ------------------------ g vo

2 . 2 sin . cos = ----------------------- g


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v0

2 sin 2 xR = -------------

g Exercises :

1. The pilot of an aeroplane flying on a straight course know from his instruments that his airspeed is 300 mph. He also knows that a 60 mph gale is blowing at an angle of 60o to his course. How can he calculate his velocity relative to the ground ?

2. A motor boat can moves with a maximum speed of 10 m/s relative to the water.

A river 400 m wide flowing at 5 m/s must be crossed in the shortest possible time to reach a point on the other bank directly opposite the starting point. In which direction must the boat be pointed and how long will it take to cross ?

3. A boy leaning over a railway bridge 49 ft hight sees a train approaching with

uniform speed and attempts to drop a stone down the tunnel. He releases the stone when the engine is 80 ft away from the bridge and sees the stone hit the ground 3 ft in front of the engine. What is the speed of the train ?

4. A ball is projected horizontally with a velocity vo of 8 ft/sec. Find its position

and velocity after ¼ sec (sees the figure) y v0 x 0 y vx = vo x vy= -gt v

5. A ball is thrown with an initial velocity vo of 10 m/sec directed at an angle q of 53o with the ground.

a. Find the x and y component of vo b. Find the position of ball and the magnitude and direction of its velocity

when t = 2 sec c. At the highest point of the ball’s path. What is the ball’s altitude (h) and

how much time has elapsed ? d. What is the ball’s range d ?

6. A catapull projects a stone with an initial velocity of 15 m/s and inclined at an

angle of 60o to the horizontal ground. Assume that air resistance is negligible. a. Calculate the initial vertical and horizontal component of the velocity


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b. State the vetical and horizontal component of the velocity when the stone is at the maximum height of its trajectory.

c. State the final velocity and horizontal component of the velocity just before it hits the horizontal ground again.

d. Calculate the stone’s maximum vertical displacement e. Find the total time that the stone is air-bone. f. What is the range of the stone ? g. What is the angle of projection that will give the stone a maximum range

? 7. As shown in figure below projectile is fired horizontally with a speed of 30 m/s

from the top of a cliff 80 m height. a. How long will it take the level ground at the base of the cliff ? b. How far from of the foot of the cliff will it strike ? c. With what velocity will it strike ? Vi = 30 m/s 20 m/s 40o 80 m 50m

8. As shown in fig. b. above a ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s with angle of 40o above horizontal. How far above or below its original level will the ball strike the opposite wall ?

9. A workman sitting on the top of the roof of house drops his hammer. The roof is

smooth and slopes at an angle of 30o to the horizontal. It is 10 m long and its lowest point is 10 m from the ground. How far from the house wall is the hammer when it hits the ground ?

N vx v vy mg sin mg cos mg y x


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Chapter 2. Elasticity and Harmonic Motion Young's Modulus

For the description of the elastic properties of linear objects like wires, rods, columns which are either stretched or compressed, a convenient parameter is the ratio of the stress to the strain, a parameter called the Young's modulus of the material. Young's modulus can be used to predict the elongation or compression of an object as long as the stress is less than the yield strength of the material.

Elastic Properties of Selected Engineering Materials

Material Density (kg/m3)

Young's Modulus 109 N/m2

Ultimate Strength Su 106 N/m2

Yield Strength Sy 106 N/m2

Steela 7860 200 400 250 Aluminum 2710 70 110 95

Glass 2190 65 50b ... Concretec 2320 30 40b ...

Woodd 525 13 50b ... Bone 1900 9b 170b ...

Polystyrene 1050 3 48 ... a Structural steel (ASTM-A36), b In compression, c High strength, d Douglas fir Data from Table 13-1, Halliday, Resnick, Walker, 5th Ed. Extended

Elasticity

Elasticity is the property of an object or material which causes it to be restored to its original shape after distortion. It is said to be more elastic if it restores itself more precisely to its original configuration. A rubber band is easy to stretch, and snaps back to near its original length when released, but it is not as elastic as a piece of piano wire. The piano wire is harder to stretch, but would be said to be more elastic than the rubber band because of the precision of its return to its original length. A real piano string can be struck hundreds of times without stretching enough to go noticeably out of tune. A spring is an example of an elastic object - when stretched, it exerts a restoring force which tends to bring it back to its original length. This restoring force is generally


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proportional to the amount of stretch, as described by Hooke's Law. For wires or columns, the elasticity is generally described in terms of the amount of deformation (strain) resulting from a given stress (Young's modulus). Bulk elastic properties of materials describe the response of the materials to changes in pressure.

Hooke's Law

In 1678, Robert Hooke announced the invention of the spring scale and the relationship for elastic materials that is now known as Hooke's Law. When an object is acted upon by a force, it can be compressed, stretched or bent. If when the force is removed, the object returns to its original shape, it is said to be elastic. Solids that do not return to their original configuration once they have been distorted are categorized as plastics. Hooke discovered that not only are certain materials (steel bars, rods, wire, springs, diving boards, and rubber bands) elastic, but the stretch they experience is directly proportional to the load that they

support. Elastic media will stretch until the reach their elastic limit, or yield point. After that point, they exhibit plastic deformation and will never return to their original shape. Ductile materials stretch thinner and thinner, while brittle materials break without any plastic deformation. Eventually all will rupture at their breaking point. To simplify our discussion, we are going to use springs as our example of an elastic medium. The formulas used to calculate the force required to stretch or compress an elastic medium with respect to its equilibrium position and its elasticity constant, k, are: Finternal = - kx force supplied by the spring to restore itself to equilibrium (Hooke's Law)Fexternal = kx force supplied by an external agent on the spring distorting it from equilibrium

This formula is only applicable to force acting on an ideal spring that has not surpassed its elastic limit. Note that the amount of force required by an external agent to stretch the spring depends on how far it has been displaced from its equilibrium position. That is, the force is not constant, it is variable.


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When two or more springs are combined in parallel (side by side) so that any applied force must stretch both springs simultaneously, the spring constant for the combination will be kparallel = k1 + k2 When two or springs are combined in series (one after another), an applied force may stretch one more than another. Recall a saying that a chain is only as "strong as its weakest link." The spring constant for the combination will be kseries = (1/k1 + 1/k2)-1

Elastic Potential Energy

Elastic potential energy is Potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring. It is equal to the work done to stretch the spring, which depends upon the spring constant k as well as the distance stretched. According to Hooke's law, the force required to stretch the spring will be directly proportional to the amount of stretch. Since the force has the form

F = -kx

then the work done to stretch the spring a distance x is

You may enter data in any of the boxes. Then click on the active text for the quantity you wish to calculate. The values will not be forced to be compatible until you click on a quantity to calculate.

Spring Potential Energy


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Since the change in Potential energy of an object between two positions is equal to the work that must be done to move the object from one point to the other, the calculation of potential energy is equivalent to calculating the work. Since the force required to stretch a spring changes with distance, the calculation of the work involves an integral.

The work can also be visualized as the area under the force curve:


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Periodic Motion

Periodic motion of some source object is necessary to produce a sustained musical sound (i.e., one with definite pitch and quality). For example, to produce a standard musical A (440 Hz), the source object must sustain periodic motion at 440 vibrations per second with a tolerance of less than 1 Hz -- the normal human ear can detect the difference between 440 Hz and 441 Hz. The conditions necessary for periodic motion are

1. elasticity - the capacity to return precisely to the original configuration after being distorted.

o a. a definite equilibrium configuration o b. a restoring force to bring the system back to equilibrium

2. A source of energy.

Fortunately, it is not hard to find vibrators which meet these conditions, hence the richness in variety of musical sound sources.

Terms for describing periodic motion. A mass on a spring is an example of periodic motion with a single frequency called Description of Periodic Motion Motion which repeats itself precisely can be described with the following terms:

Period: the time required to complete a full cycle, T in seconds/cycle Frequency: the number of cycles per second, f in 1/seconds or Hertz (Hz) Amplitude: the maximum displacement from equilibrium A

and if the periodic motion is in the form of a traveling wave, one needs also Velocity of propagation: v Wavelength: repeat distance of wave .

Period, Frequency and Amplitude

In a plot of periodic motion as a function of time, the period can be seen as the repeat time for the motion. The frequency is the reciprocal of the period.


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Simple harmonic motion Equilibrium condition

Simple Harmonic Motion

Simple harmonic motion is typified by the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law. The motion is sinusoidal in time and demonstrates a single resonant frequency.

Simple Harmonic Motion Equations

The motion equation for simple harmonic motion contains a complete description of the motion, and other parameters of the motion can be calculated from it.

The velocity and acceleration are given by


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The total energy for an undamped oscillator is the sum of its kinetic energy and potential energy, which is constant at

Simple Harmonic Motion Calculation

The motion equations for simple harmonic motion provide for calculating any parameter of the motion if the others are known.

If the period is T = s then the frequency is f = Hz and the angular frequency = rad/s.

The motion is described by :

Displacement = Amplitude x sin (angular frequency x time) Y = A sin t

Simple Pendulum

A simple pendulum is one which can be considered to be a point mass suspended from a string or rod of negligible mass. It is a resonant system with a single resonant frequency. For small amplitudes, the period of such a pendulum can be approximated by:


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Pendulum Motion

The motion of a simple pendulum is like simple harmonic motion in that the equation for the angular displacement is

which is the same form as the motion of a mass on a spring:

The anglular frequency of the motion is then given by

compared to for a mass on a spring

The frequency of the pendulum in Hz is given by and the period of

motion is then


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Velocity and Acceleration

Once you know the position of the oscillator for all times, you can work out the velocity and acceleration functions.

x(t) = A cos (t + )

The velocity is the time derivative of the position so:

v(t) = -A sin (t + ) The change from cos to sin means that the velocity is 90o out of phase with the displacement-when x = 0 the velocity is a maximum and when x is a maximum v = 0. This is seen nicely in the page on the connection between SHM and circular motion. The acceleration is the time derivative of the velocity so:

a(t) = -A cos (t + ) Notice also from the preceding that: a(t) = -2x The acceleration is exactly out of phase with the displacement. This can also be seen in the page on the connection between SHM and circular motion

Energy in Mass on Spring

The simple harmonic motion of a mass on a spring is an example of an energy transformation between potential energy and kinetic energy. In the example below, it is assumed that 2 joules of work has been done to set the mass in motion.


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Chapter 3 : Universal Gravitation Isaac Newton's theory of universal gravitation (part of classical mechanics) states the following:

Every single point mass attracts every other point mass in the universe by a force pointing along the line combining the two. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between the point masses:

where:

F is the magnitude of the gravitational force between the two point masses,

G is the gravitational constant, m1 is the mass of the first point mass, m2 is the mass of the second point mass, r is the distance between the two point masses.

Assuming SI units, F is measured in newtons (N), m1 and m2 in kilograms (kg), r in metres (m), and the constant G is approximately equal to 6.67 × 10−11 N m2 kg−2. G was first accurately measured in the Cavendish experiment by the British scientist Henry Cavendish in 1798, it was also the first test of theory of gravitation between masses in the laboratory. This was 111 years after the publication of "Philosophiae Naturalis Principia Mathematica" and 71 years after Newton's death, so all of Newton's calculations could not use the value of G; instead he could only calculate a force relative to another force.

Newton's law of gravitation resembles Coulomb's law of electrical forces. Newton's law is used to calculate the Gravitational force between two masses; similarly Coulomb's Law is used to calculate the magnitude of electrical force between two charged bodies. Coulomb's Law's equation has the product of two charges in place of the product of the masses which is in Newton's Law of Gravitation. Hence, according to Coulmb's Law, the electrical force is proportional to the product of the charged bodies divided by the distance between them.

Acceleration due to gravity

Let a1 be the acceleration experienced by the first point mass due to the gravitational force exerted on it by the second point mass. Newton's second law states that F = m1 a1, meaning that a1 = F / m1. Substituting F from the earlier equation gives:


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and similarly for a2.

Assuming SI units, gravitational acceleration (as acceleration in general) is measured in metres per second squared (m/s2 or m s-2). Non-SI units include galileos, gees (see later), and feet per second squared.

Laws of astrodynamics

The fundamental laws of astrodynamics are Newton's law of universal gravitation and Newton's laws of motion, while the fundamental mathematical tool is his differential calculus. Kepler's laws of planetary motion may be derived from these laws, when it is assumed that the orbiting body is subject only to the gravitational force of the central attractor. When an engine thrust or propulsive force is present, Newton's second law of motion applies, and Kepler's laws are temporarily invalidated.

The formula for escape velocity is easily derived as follows. The specific energy (energy per unit mass) of any space vehicle is composed of two components, the specific potential energy and the specific kinetic energy. The specific potential energy associated with a planet of mass M is given by

while the specific kinetic energy of an object is given by

Since energy is conserved, the total specific orbital energy

does not depend on the distance, r, from the center of the central body to the space vehicle in question. Therefore, the object can reach infinite r only if this quantity is nonnegative, which implies

The escape velocity from the Earth's surface is about 11 km/s, but that is insufficient to send the body an infinite distance because of the gravitational pull of the Sun. To escape the solar system from the vicinity of the Earth requires around 42 km/s velocity, but there will be "part credit" for the Earth's orbital velocity for spacecraft launched from Earth, if their further acceleration (due to the propulsion system) carries them in the same direction as Earth travels in its orbit.


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Chapter 4 : Momentum and Impuls The sports announcer says "Going into the all-star break, the Chicago White Sox have the momentum." The headlines declare "Chicago Bulls Gaining Momentum." The coach pumps up his team at half-time, saying "You have the momentum; the critical need is that you use that momentum and bury them in this third quarter."

Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop. Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team which is on the move has the momentum. If an object is in motion (on the move) then it has momentum.

Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum which an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object.

Momentum = mass • velocity

In physics, the symbol for the quantity momentum is the lower case "p". Thus, the above equation can be rewritten as

p = m • v

where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity.

The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg•m/s. While the kg•m/s is the standard metric unit of momentum, there are a variety of other units which are acceptable (though not conventional) units of momentum. Examples include kg•mi/hr, kg•km/hr, and g•cm/s. In each of these examples, a mass unit is multiplied by a velocity unit to provide a momentum unit. This is consistent with the equation for momentum.

Momentum is a vector quantity. As discussed in an earlier unit, a vector quantity is a quantity which is fully described by both magnitude and direction. To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg•m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. The


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direction of the momentum vector is the same as the direction of the velocity of the ball. In a previous unit, it was said that the direction of the velocity vector is the same as the direction which an object is moving. If the bowling ball is moving westward, then its momentum can be fully described by saying that it is 10 kg•m/s, westward. As a vector quantity, the momentum of an object is fully described by both magnitude and direction.

From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest. The momentum of any object which is at rest is 0. Objects at rest do not have momentum - they do not have any "mass in motion." Both variables - mass and velocity - are important in comparing the momentum of two objects.

The momentum equation can help us to think about how a change in one of the two variables might affect the momentum of an object. Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass of loaded cart is 1.0 kg and its momentum is 2.0 kg•m/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 2.0 kg and its momentum would be 4.0 kg•m/s. A doubling of the mass results in a doubling of the momentum.

Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in a quadrupling of the momentum. These two examples illustrate how the equation p = m•v serves as a "guide to thinking" and not merely a "plug-and-chug recipe for algebraic problem-solving."

Check Your Understanding Express your understanding of the concept and mathematics of momentum by answering the following questions. Click the button to view the answers. 1. Determine the momentum of a ... a. 60-kg halfback moving eastward at 9 m/s. b. 1000-kg car moving northward at 20 m/s. c. 40-kg freshman moving southward at 2 m/s. 2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ... a. its velocity were doubled. b. its velocity were tripled. c. its mass were doubled (by adding more passengers and a greater load) d. both its velocity were doubled and its mass were doubled.


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3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m = 120 kg) are running down the football field. Consider their ticker tape patterns below.

Compare the velocities of these three players. How many times greater is the velocity of the halfback and the velocity of the tight end than the velocity of the lineman? Which player has the greatest momentum? Explain

Momentum and Impulse Connection

As mentioned in the previous part of this lesson, momentum is a commonly used term in sports. When a sports announcer says that a team has the momentum they mean that the team is really on the move and is going to be hard to stop. The term momentum is a physics concept. Any object with momentum is going to be hard to stop. To stop such an object, it is necessary to apply a force against its motion for a given period of time. The more momentum which an object has, the harder that it is to stop. Thus, it would require a greater amount of force or a longer amount of time or both to bring such an object to a halt. As the force acts upon the object for a given amount of time, the object's velocity is changed; and hence, the object's momentum is changed.

The concepts in the above paragraph should not seem like abstract information to you. You have observed this a number of times if you have watched the sport of football. In football, the defensive players apply a force for a given amount of time to stop the momentum of the offensive player who has the ball. You have also experienced this a multitude of times while driving. As you bring your car to a halt when approaching a stop sign or stoplight, the brakes serve to apply a force to the car for a given amount of time to change the car's momentum. An object with momentum can be stopped if a force is applied against it for a given amount of time.

A force acting for a given amount of time will change an object's momentum. Put another way, an unbalanced force always accelerates an object - either speeding it up or slowing it down. If the force acts opposite the object's motion, it slows the object down. If a force acts in the same direction as the object's motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.

These concepts are merely an outgrowth of Newton's second law as discussed in an earlier unit. Newton's second law (Fnet = m • a) stated that the acceleration of an object is


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directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. When combined with the definition of acceleration (a = change in velocity / time), the following equalities result.

If both sides of the above equation are multiplied by the quantity t, a new equation results.

This equation represents one of two primary principles to be used in the analysis of collisions during this unit. To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force • time is known as impulse. And since the quantity m•v is the momentum, the quantity m• v must be the change in momentum. The equation really says that the

Impulse = Change in momentum One focus of this unit is to understand the physics of collisions. The physics of collisions are governed by the laws of momentum; and the first law which we discuss in this unit is expressed in the above equation. The equation is known as the impulse-momentum change equation. The law can be expressed this way: In a collision, an object experiences a force for a specific amount of time which results in a change in momentum. The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • v.

In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum. Consider a football halfback running down the football field and encountering a collision with a defensive back. The collision would change the halfback's speed and thus his momentum. If the motion was represented by a ticker tape diagram, it might appear as follows:

At approximately the tenth dot on the diagram, the collision occurs and lasts for a certain amount of time; in terms of dots, the collision lasts for a time equivalent to approximately nine dots. In the halfback-defensive back collision, the halfback experiences a force which lasts for a certain amount of time to change his momentum. Since the collision causes the rightward-moving halfback to slow down, the force on the


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halfback must have been directed leftward. If the halfback experienced a force of 800 N for 0.9 seconds, then we could say that the impulse was 720 N•s. This impulse would cause a momentum change of 720 kg•m/s. In a collision, the impulse experienced by an object is always equal to the momentum change.

Now consider a collision of a tennis ball with a wall. Depending on the physical properties of the ball and wall, the speed at which the ball rebounds from the wall upon colliding with it will vary. The diagrams below depict the changes in velocity of the same ball. For each representation (vector diagram, velocity-time graph, and ticker tape pattern), indicate which case (A or B) has the greatest change in velocity, greatest acceleration, greatest momentum change, and greatest impulse. Support each answer. Click the button to check your answer.

Vector Diagram

Greatest velocity change? Greatest acceleration? Greatest momentum change? Greatest Impulse?

Velocity-Time Graph


Ticker Tape Diagram


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Observe that each of the collisions above involve the rebound of a ball off a wall. Observe that the greater the rebound effect, the greater the acceleration, momentum change, and impulse. A rebound is a special type of collision involving a direction change in addition to a speed change. The result of the direction change is a large velocity change. On occasions in a rebound collision, an object will maintain the same

or nearly the same speed as it had before the collision. Collisions in which objects rebound with the same speed (and thus, the same momentum and kinetic energy) as they had prior to the collision are known as elastic collisions. In general, elastic collisions are characterized by a large

velocity change, a large momentum change, a large impulse, and a large force.

Use the impulse-momentum change principle to fill in the blanks in the following rows of the table. As you do, keep these three major truths in mind:

the impulse experienced by an object is the force•time the momentum change of an object is the mass•velocity change the impulse equals the momentum change

Click the button to view answers.

Force

(N)

time

(s) Impulse

(N*s)

Mom. Change

(kg*m/s)

Mass

(kg)

Vel. Change

(m/s)

1. 0.010 10 -4 2. 0.100 -40 10 3. 0.010 -200 50 4. -20 000 -200 -8 5. -200 1.0 50

There are a few observations which can be made in the above table which relate to the computational nature of the impulse-momentum change theorem. First, observe that the answers in the table above reveal that the third and fourth columns are always equal;


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that is, the impulse is always equal to the momentum change. Observe also that the if any two of the first three columns are known, then the remaining column can be computed. This is true because the impulse=force • time. Knowing two of these three quantities allows us to compute the third quantity. And finally, observe that knowing any two of the last three columns allows us to compute the remaining column. This is true since momentum change = mass • velocity change.

There are also a few observations which can be made which relate to the qualitative nature of the impulse-momentum theorem. An examination of rows 1 and 2 show that force and time are inversely proportional; for the same mass and velocity change, a tenfold increase in the time of impact corresponds to a tenfold decrease in the force of impact. An examination of rows 1 and 3 show that mass and force are directly proportional; for the same time and velocity change, a fivefold increase in the mass corresponds to a fivefold increase in the force required to stop that mass. Finally, an examination of rows 3 and 4 illustrate that mass and velocity change are inversely proportional; for the same force and time, a twofold decrease in the mass corresponds to a twofold increase in the velocity change.

Check Your Understanding

Express your understanding of the impulse-momentum change theorem by answering the following questions. Click the button to view the answers.

1. A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1 second; another 0.50 kg cart (#2) is pulled with a 2.0 N-force for 0.50 seconds. Which cart (#1 or #2) has the greatest acceleration? Explain.

Which cart (#1 or #2) has the greatest impulse? Explain. Which cart (#1 or #2) has the greatest change in momentum? Explain. 2. In a physics demonstration, two identical balloons (A and B) are propelled across the room on horizontal guide wires. The motion diagrams (depicting the relative position of the balloons at time intervals of 0.05 seconds) for these two balloons are shown below.

Which balloon (A or B) has the greatest acceleration? Explain. Which balloon (A or B) has the greatest final velocity? Explain. Which balloon (A or B) has the greatest momentum change? Explain. Which balloon (A or B) experiences the greatest impulse? Explain.

3. Two cars of equal mass are traveling down Lake Avenue with equal velocities. They both come to a stop over different lengths of time. The ticker tape patterns for each car are shown on the diagram below.


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At what approximate location on the diagram (in terms of dots) does each car begin to experience the impulse?

Which car (A or B) experiences the greatest acceleration? Explain. Which car (A or B) experiences the greatest change in momentum? Explain. Which car (A or B) experiences the greatest impulse? Explain.

4. The diagram to the right depicts the before- and after-collision speeds of a car which undergoes a head-on-collision with a wall. In Case A, the car bounces off the wall. In Case B, the car crumples up and sticks to the wall.

a. In which case (A or B) is the change in velocity the greatest? Explain. b. In which case (A or B) is the change in momentum the greatest? Explain. c. In which case (A or B) is the impulse the greatest? Explain. d. In which case (A or B) is the force which acts upon the car the greatest (assume contact times are the same in both cases)? Explain. 5. Jennifer, who has a mass of 50.0 kg, is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid hitting a deer crossing the road. She strikes the air bag, which brings her body to a stop in 0.500 s. What average force does the seat belt exert on her? If Jennifer had not been wearing her seat belt and not had an air bag, then the windshield would have stopped her head in 0.002 s. What average force would the windshield have exerted on her? 6. A hockey player applies an average force of 80.0 N to a 0.25 kg hockey puck for a time of 0.10 seconds. Determine the impulse experienced by the hockey puck. 7. If a 5-kg object experiences a 10-N force for a duration of 0.10-second, then what is the momentum change of the object?


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Real-World Applications

In a previous part of Lesson 1, it was said that

In a collision, an object experiences a force for a given amount of time which results in its mass undergoing a change in velocity (i.e., which results in a momentum change).

There are four physical quantities mentioned in the above statement - force, time, mass, and velocity change. The force multiplied by the time is known as the impulse and the mass multiplied by the velocity change is known as the change in momentum. The impulse experienced by an object is always equal to the change in its momentum. In terms of equations, this was expressed as

This is known as the impulse-momentum change theorem.

In this part of Lesson 1, we will examine some real-world applications of the impulse-momentum change theorem. We will examine some physics in action in the real world. In particular, we will focus upon

the affect of collision time upon the amount of force an object experiences, and the affect of rebounding upon the velocity change and hence the amount of force

an object experiences.

As an effort is made to apply the impulse-momentum change theorem to a variety of real-world situations, keep in mind that the goal is to use the equation as a guide to thinking about how an alteration in the value of one variable might affect the value of another variable.

The Affect of Collision Time upon the Force

First we will examine the importance of the collision time in affecting the amount of force which an object experiences during a collision. In a previous part of Lesson 1, it was mentioned that force and time are inversely proportional. An object with 100 units of momentum must experience 100 units of impulse in order to be brought to a stop. Any combination of force and time could be used to produce the 100 units of impulse necessary to stop an object with 100 units of momentum. This is depicted in the table below.


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Combinations of Force and Time Required to Produce 100 units of Impulse

Force Time Impulse 100 1 100 50 2 100 25 4 100 10 10 100 4 25 100 2 50 100 1 100 100

0.1 1000 100

Observe that the greater the time over which the collision occurs, the smaller the force acting upon the object. Thus, to minimize the affect of the force on an object involved in a collision, the time must be increased. And to maximize the affect of the force on an object involved in a collision, the time must be decreased.

There are several real-world applications of this phenomena. One example is the use of air bags in automobiles. Air bags are used in automobiles because they are able to minimize the affect of the force on an object involved in a collision. Air bags accomplish this by extending the time required to stop the momentum of the driver and passenger. When

encountering a car collision, the driver and passenger tend to keep moving in accord with Newton's first law. Their motion carries them towards a windshield which results in a large force exerted over a short time in order to stop their momentum. If instead of hitting the windshield, the driver and passenger hit an air bag, then the time duration of the impact is increased. When hitting an object with some give such as an air bag, the


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time duration might be increased by a factor of 100. Increasing the time by a factor of 100 will result in a decrease in force by a factor of 100. Now that's physics in action.

The same principle explains why dashboards are padded. If the air bags do not deploy (or are not installed in a car), then the driver and passengers run the risk of stopping their momentum by means of a collision with the windshield or the dashboard. If the driver or passenger should hit the dashboard, then the force and time required to stop their momentum is exerted by the dashboard. Padded dashboards provide some give in such a collision and serve to extend the time duration of the impact, thus minimizing the affect of the force. This same principle of padding a potential impact area can be observed in gymnasiums (underneath the basketball hoops), in pole-vaulting pits, in baseball gloves and goalie mitts, on the fist of a boxer, inside the helmet of a football player, and on gymnastic mats. Now that's physics in action.

Fans of boxing frequently observe this same principle of minimizing the affect of a force by extending the time of collision. When a boxer recognizes that he will be hit in the head by his opponent, the boxer often relaxes his neck and allows his head to move backwards upon impact. In the boxing world, this is known as riding the punch. A boxer rides the punch in order to extend the time of impact of the glove with their head. Extending the time results in decreasing the force and thus minimizing the affect of the force in the collision. Merely increasing the collision time by a factor of ten would result in a tenfold decrease in the force. Now that's physics in action.

Nylon ropes are used in the sport of rock-climbing for the same reason. Rock climbers attach themselves to the steep cliffs by means of nylon ropes. If a rock climber should lose her grip on the rock, she will begin to fall. In such a situation, her momentum will ultimately be halted by means of the rope, thus preventing a disastrous fall to the ground below. The ropes are made of nylon or similar material because of its ability to stretch. If the rope is capable of stretching upon being pulled taut by the falling climber's mass, then it will apply a force upon the climber over a longer time period. Extending the time over which the climber's momentum is broken results in reducing the force exerted on the falling climber. For certain, the rock climber can appreciate minimizing the affect of the force through the use of a longer time of impact. Now that's physics in action.

In racket and bat sports, hitters are often encouraged to follow-through when striking a ball. High speed films of the collisions between bats/rackets and balls have shown that the act of following through serves to increase the time over which a collision occurs. This increase in time must result in a change in some other variable in the impulse-momentum change theorem. Surprisingly, the variable which is dependent upon the time in such a situation is not the force. The force in hitting is dependent upon how hard the hitter swings the bat or racket, not the time of impact. Instead, the follow-through increases the time of collision and


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subsequently contributes to an increase in the velocity change of the ball. By following through, a hitter can hit the ball in such a way that it leaves the bat or racket with more velocity (i.e., the ball is moving faster). In tennis, baseball, racket ball, etc., giving the ball a high velocity often leads to greater success. Now that's physics in action.

You undoubtedly recall other illustrations of this principle. A common physics demonstration involves the catching of water balloons of varying speed and varying mass. A water balloon is thrown high into the air and successfully caught (i.e., caught without breaking). The key to the success of the demonstration is to contact the balloon with outstretched arms and carry the balloon for a meter or more before finally stopping its momentum. The effect of this strategy is to extend the time over which the collision occurred and so reduce the force. This same strategy is used by lacrosse players when catching the ball. The ball is "cradled" when caught; i.e., the lacrosse player reaches out for the ball and carries it inward toward her body as if she were cradling a baby. The effect of this strategy is to lengthen the time over which the collision occurs and so reduce the force on the lacrosse ball. Now that's physics in action.

Another common physics demonstration involves throwing an egg into a bed sheet. The bed sheet is typically held by two trustworthy students and a volunteer is used to toss the egg at full speed into the bed sheet. The collision between the egg and the bed sheet lasts over an extended period of time since the bed sheet has some give in it. By extending the time of the collision, the affect of the force is minimized. In all my years, the egg has never broken when hitting the bed sheet. On occasion the volunteer has a wayward toss and is not as accurate as expected. The egg misses the bed sheet and collides with the wall. In these unexpected cases, the collision between wall and egg lasts for a short period of time, thus maximizing the affect of the force on the egg. The egg brakes and leaves the wall and floor in a considerable mess. And that's no yolk!

The Effect of Rebounding

Occasionally when objects collide, they bounce off each other as opposed to sticking to each other and traveling with the same speed after the collision. Bouncing off each other is known as rebounding. Rebounding involves a change in the direction of an object; the before- and after-collision direction is different. Rebounding was pictured


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and discussed earlier in Lesson 1. At that time, it was said that rebounding situations are characterized by a large velocity change and a large momentum change.

From the impulse-momentum change theorem, we could deduce that a rebounding situation must also be accompanied by a large impulse. Since the impulse experienced by an object equals the momentum change of the object, a collision characterized by a large momentum change must also be characterized by a large impulse.

The importance of rebounding is critical to the outcome of automobile accidents. In an automobile accident, two cars can either collide and bounce off each other or collide, crumple up and travel together with the same speed after the collision. But which would be more damaging to the occupants of the automobiles - the rebounding of the cars or the crumpling up of the cars? Contrary to popular opinion, the crumpling up of cars is the safest type of automobile collision. As mentioned above, if cars rebound upon collision, the momentum change will be larger and so will the impulse. A greater impulse will typically be associated with a bigger force. Occupants of automobiles would certainly prefer small forces upon their bodies during collisions. In fact, automobile designers and safety engineers have found ways to reduce the harm done to occupants of automobiles by designing cars which crumple upon impact. Automobiles are made with crumple zones. Crumple zones are sections in cars which are designed to crumple up when the car encounters a collision. Crumple zones minimize the affect of the force in an automobile collision in two ways. By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse. Finally, the crumpling of the car lengthens the time over which the car's momentum is changed; by increasing the time of the collision, the force of the collision is greatly reduced.


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Momentum Conservation Principle

One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.

For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.

The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value. This law of momentum conservation will be the focus of the remainder of Lesson 2. To understand the basis of momentum conservation, let's begin with a short logical proof.

Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton's third law). This statement can be expressed in equation form as follows.

The forces act between the two objects for a given amount of time. In some cases, the time is long; in other cases the time is short. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as

Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that the impulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as


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But the impulse experienced by an object is equal to the change in momentum of that object (the impulse-momentum change theorem). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as

The above equation is one statement of the law of momentum conservation. In a collision, the momentum change of object 1 is equal to and opposite of the momentum change of object 2. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. In most collisions between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision. The total momentum of the system (the collection of two objects) is conserved.

A useful analogy for understanding momentum conservation involves a money transaction between two people. Let's refer to the two people as Jack and Jill. Suppose that we were to check the pockets of Jack and Jill before and after the money transaction in order to determine the amount of money which each possesses. Prior to the transaction, Jack possesses $100 and Jill possesses $100. The total amount of money of the two people before the transaction is $200. During the transaction, Jack pays Jill $50 for the given item being bought. There is a transfer of $50 from Jack's pocket to Jill's pocket. Jack has lost $50 and Jill has gained $50. The money lost by Jack is equal to the money gained by Jill. After the transaction, Jack now has $50 in his pocket and Jill has $150 in her pocket. Yet, the total amount of money of the two people after the transaction is $200. The total amount of money (Jack's money plus Jill's money) before the transaction is equal to the total amount of money after the transaction. It could be said that the total amount of money of the system (the collection of two people) is conserved. It is the same before as it is after the transaction.

A useful means of depicting the transfer and the conservation of money between Jack


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and Jill is by means of a table.

The table shows the amount of money possessed by the two individuals before and after the interaction. It also shows the total amount of money before and after the interaction. Note that the total amount of money ($200) is the same before and after the interaction - it is conserved. Finally, the table shows the change in the amount of money possessed by the two individuals. Note that the change in Jack's money account (-$50) is equal and opposite to the change in Jill's money account (+$50) .

For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. A common physics lab involves the dropping of a brick upon a cart in motion.

The dropped brick is at rest and begins with zero momentum. The loaded cart (a cart with a brick on it) is in motion with considerable momentum. The actual momentum of the loaded cart can be determined using the velocity (often determined by a ticker tape analysis) and the mass. The total amount of momentum is the sum of the dropped brick's momentum (0 units) and the loaded cart's momentum. After the collision, the momenta of the two separate objects (dropped brick and loaded cart) can be determined from their measured mass and their velocity (often found from a ticker tape analysis). If momentum is conserved during the collision, then the sum of the dropped brick's and loaded cart's momentum after the collision should be the same as before the collision. The momentum lost by the loaded cart should equal (or approximately equal) the momentum gained by the dropped brick. Momentum data for the interaction between the dropped brick and the loaded cart could be depicted in a table similar to the money table above.


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Before

Collision

Momentum

After

Collision

Momentum

Change in

Momentum

Dropped Brick 0 units 14 units +14 units

Loaded Cart 45 units 31 units -14 units

Total 45 units 45 units

Note that the loaded cart lost 14 units of momentum and the dropped brick gained 14 units of momentum. Note also that the total momentum of the system (45 units) was the same before the collision as it was after the collision.

Collisions commonly occur in contact sports (such as football) and racket and bat sports (such as baseball, golf, tennis, etc.). Consider a collision in football between a fullback and a linebacker during a goal-line stand. The fullback plunges across the goal line and collides in midair with the linebacker. The linebacker and fullback hold each other and travel together after the collision. The fullback possesses a momentum of 100 kg*m/s, East before the collision and the linebacker possesses a momentum of 120 kg*m/s, West before the collision. The total momentum of the system before the collision is 20 kg*m/s, West (review the section on adding vectors if necessary). Therefore, the total momentum of the system after the collision must also be 20 kg*m/s, West. The fullback and the linebacker move together as a single unit after the collision with a combined momentum of 20 kg*m/s. Momentum is conserved in the collision. A vector diagram can be used to represent this principle of momentum conservation; such a diagram uses an arrow to represent the magnitude and direction of the momentum vector for the individual objects before the collision and the combined momentum after the collision.

Now suppose that a medicine ball is thrown to a clown who is at rest upon the ice; the clown catches the medicine ball and glides together with the ball across the ice. The


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momentum of the medicine ball is 80 kg*m/s before the collision. The momentum of the clown is 0 m/s before the collision. The total momentum of the system before the collision is 80 kg*m/s. Therefore, the total momentum of the system after the collision must also be 80 kg*m/s. The clown and the medicine ball move together as a single unit after the collision with a combined momentum of 80 kg*m/s. Momentum is conserved in the collision.

Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis. Useful means of representing such analyses include a momentum table and a vector diagram. Later in Lesson 2, we will use the momentum conservation principle to solve problems in which the after-collision velocity of objects is predicted.


Express your understanding of the concept and mathematics of momentum by answering the following questions. Click on the button to view the answers.

1. When fighting fires, a firefighter must use great caution to hold a hose which emits large amounts of water at high speeds. Why would such a task be difficult?

2. A large truck and a Volkswagen have a head-on collision. a. Which vehicle experiences the greatest force of impact? b. Which vehicle experiences the greatest impulse? c. Which vehicle experiences the greatest momentum change? d. Which vehicle experiences the greatest acceleration? 3. Miles Tugo and Ben Travlun are riding in a bus at highway speed on a nice summer day when an unlucky bug splatters onto the windshield. Miles and Ben begin discussing the physics of the situation. Miles suggests that the momentum change of the bug is much greater than that of the bus. After all, argues Miles, there was no noticeable change in the speed of the bus compared to the obvious change in the speed of the bug. Ben disagrees entirely, arguing that that both bug and bus encounter the same force, momentum change, and impulse. Who do you agree with? Support your answer.


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4. If a ball is projected upward from the ground with ten units of momentum, what is the momentum of recoil of the Earth? ____________ Do we feel this? Explain.

5. If a 5-kg bowling ball is projected upward with a velocity of 2.0 m/s, then what is the recoil velocity of the Earth (mass = 6.0 x 1024 kg).

6. A 120 kg lineman moving west at 2 m/s tackles an 80 kg football fullback moving east at 8 m/s. After the collision, both players move east at 2 m/s. Draw a vector diagram in which the before- and after-collision momenta of each player is represented by a momentum vector. Label the magnitude of each momentum vector. See answer below.

7. In an effort to exact the most severe capital punishment upon a rather unpopular prisoner, the execution team at the Dark Ages Penitentiary search for a bullet which is ten times as massive as the rifle itself. What type of individual would want to fire a rifle which holds a bullet which is ten times more massive than the rifle? Explain.

8. A baseball player holds a bat loosely and bunts a ball. Express your understanding of momentum conservation by filling in the tables below.

9. A Tomahawk cruise missile is launched from the barrel of a mobile missile launcher. Neglect friction. Express your understanding of momentum conservation by filling in the tables below.

Answer to Question #6


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Momentum Conservation in Explosions

As discussed in a previous part of Lesson 2, total system momentum is conserved for collisions between objects in an isolated system. For collisions occurring in isolated systems, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions. In an explosion, an internal impulse acts in order to propel the parts of a system (often a single object) into a variety of directions. After the explosion, the individual parts of the system (which is often a collection of fragments from the original object) have momentum. If the vector sum of all individual parts of the system could be added together to determine the total momentum after the explosion, then it should be the same as the total momentum before the explosion. Just like in collisions, total system momentum is conserved.

Momentum conservation is often demonstrated in a Physics class with a homemade cannon demonstration. A homemade cannon is placed upon a cart and loaded with a tennis ball. The cannon is equipped with a reaction chamber into which a small amount of fuel is inserted. The fuel is ignited, setting off an explosion which propels the tennis ball through the muzzle of the cannon. The impulse of the explosion changes the momentum of the tennis ball as it exits the muzzle at high speed. The cannon experienced the same impulse, changing its momentum from zero to a final value as it recoils backwards. Due to the relatively larger mass of the cannon, its backwards recoil speed is considerably less than the forward speed of the tennis ball.


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In the exploding cannon demonstration, total system momentum is conserved. The system consists of two objects - a cannon and a tennis ball. Before the explosion, the total momentum of the system is zero since the cannon and the tennis ball located inside of it are both at rest. After the explosion, the total momentum of the system must still be zero. If the ball acquires 50 units of forward momentum, then the cannon acquires 50 units of backwards momentum. The vector sum of the individual momenta of the two objects is 0. Total system momentum is conserved.

As another demonstration of momentum conservation, consider two low-friction carts at rest on a track. The system consists of the two individual carts initially at rest. The total momentum of the system is zero before the explosion. One of the carts is equipped with a spring loaded plunger which can be released by tapping on a small pin. The spring is compressed and the carts are placed next to each other. The pin is tapped, the plunger is released, and an explosion-like impulse sets both carts in motion along the track in opposite directions. One cart acquires a rightward momentum while the other cart acquires a leftward momentum. If 20 units of forward momentum are acquired by the rightward-moving cart, then 20 units of backwards momentum is acquired by the leftward-moving cart. The vector sum of the momentum of the individual carts is 0 units. Total system momentum is conserved.

Equal and Opposite Momentum Changes

Just like in collisions, the two objects involved encounter the same force for the same amount of time directed in opposite directions. This results in impulses which are equal in magnitude and opposite in direction. And since an impulse causes and is equal to a change in momentum, both carts encounter momentum changes which are equal in magnitude and opposite in direction. If the exploding system includes two objects or two parts, this principle can be stated in the form of an equation as:


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If the masses of the two objects are equal, then their post-explosion velocity will be equal in magnitude (assuming the system is initially at rest). If the masses of the two objects are unequal, then they will be set in motion by the explosion with different speeds. Yet even if the masses of the two objects are different, the momentum change of the two objects (mass • velocity change) will be equal in magnitude.

The diagram below depicts a variety of situations involving explosion-like impulses acting between two carts on a low-friction track. The mass of the carts is different in each situation. In each situation, total system momentum is conserved as the momentum change of one cart is equal and opposite the momentum change of the other cart.

In each of the above situations, the impulse on the carts is the same - a value of 20 kg•cm/s (or cN•s). Since the same spring is used, the same impulse is delivered. Thus, each cart encounters the same momentum change in every situation - a value of 20 kg•cm/s. For the same momentum change, an object with twice the mass will encounter one-half the velocity change. And an object with four times the mass will encounter one-fourth the momentum change.


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Solving Explosion Momentum Problems

Since total system momentum is conserved in an explosion occurring in an isolated system, momentum principles can be used to make predictions about the resulting velocity of an object. Problem-solving for explosion situations is a common part of most high school physics experiences. Consider for instance the following problem:

A 56.2-gram tennis ball is loaded into a 1.27-kg homemade cannon. The cannon is at rest when it is ignited. Immediately after the impulse of the explosion, a photogate timer measures the cannon to recoil backwards a distance of 6.1 cm in 0.0218 seconds. Determine the post-explosion speed of the cannon and of the tennis ball.

Like any problem in physics, this one is best approached by listing the known information.

Given:

Cannon: m = 1.27 kg d = 6.1 cm t = 0.0218 s Ball: m = 56.2 g = 0.0562 kg

The strategy for solving for the speed of the cannon is to recognize that the cannon travels 6.1 cm at a constant speed in the 0.0218 seconds. The speed can be assumed constant since the problem states that it was measured after the impulse of the explosion when the acceleration had ceased. Since the cannon was moving at constant speed during this time, the distance/time ratio will provide a post-explosion speed value.

vcannon = d / t = (6.1 cm) / (0.0218 s) = 280 cm/s (rounded)

The strategy for solving for the post-explosion speed of the tennis ball involves using momentum conservation principles. If momentum is to be conserved, then the after-explosion momentum of the system must be zero (since the pre-explosion momentum was zero). For this to be true, then the post-explosion momentum of the tennis ball must be equal in magnitude (and opposite in direction) of that of the cannon. That is,

mball • vball = - mcannon • vcannon

The negative sign in the above equation serves the purpose of making the momenta of the two objects opposite in direction. Now values of mass and velocity can be substituted into the above equation to determine the post-explosion velocity of the tennis ball. (Note that a negative velocity has been inserted for the cannon's velocity.)

(0.0562 kg) • vball = - (1.27 kg) • (-280 cm/s)

vball = - (1.27 kg) • (-280 cm/s) / (0.0562 kg)


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vball = 6323.26 cm/s

vball = 63.2 m/s

Using momentum explosion, the ball is propelled forward with a speed of 63.2 m/s - that's 141 miles/hour!

It's worth noting that another method of solving for the ball's velocity would be to use a momentum table similar to the one used previously in Lesson 2 for collision problems. In the table, the pre- and post-explosion momentum of the cannon and the tennis ball. This is illustrated below.

Momentum

Before Explosion

Momentum

After Explosion

Cannon 0 (1.27 kg) • (-280 cm/s)

= -355 kg•cm/s

Tennis Ball 0 (0.0562 kg) • v

Total 0 0

The variable v is used for the post-explosion velocity of the tennis ball. Using the table, one would state that the sum of the cannon and the tennis ball's momentum after the explosion must sum to the total system momentum of 0 as listed in the last row of the table. Thus,

-355 kg•cm/s + (0.0562 kg) • v = 0

Solving for v yields 6323 cm/s or 63.2 m/s - consistent with the previous solution method.

Using the table means that you can use the same problem-solving strategy for both collisions and explosions. After all, it is the same momentum conservation principle which governs both situations. Whether it is a collision or an explosion, if it occurs in an isolated system, then each object involved encounters the same impulse to cause the same momentum change. The impulse and momentum change on each object are equal in magnitude and opposite in direction. Thus, the total system momentum is conserved.


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Check Your Understanding 1. Two pop cans are at rest on a stand. A firecracker is placed between the cans and lit. The firecracker explodes and exerts equal and opposite forces on the two cans. Assuming the system of two cans to be isolated, the post-explosion momentum of the system ____. a. is dependent upon the mass and velocities of the two cans b. is dependent upon the velocities of the two cans (but not their mass) c. is typically a very large value d. can be either a positive, negative or zero value e. is definitely zero 2. Students of varying mass are placed on large carts and deliver impulses to each other's carts, thus changing their momenta. In some cases, the carts are loaded with equal mass; in other cases they are unequal. In some cases, the students push off each other; in other cases, only one team does the pushing. For each situation, list the letter of the team which ends up with the greatest momentum. If they have the same momentum, then do not list a letter for that situation. Enter the four letters (or three or two or ...) in alphabetical order.

3. Two ice dancers are at rest on the ice, facing each other with their hands together. They push off on each other in order to set each other in motion. The subsequent momentum change (magnitude only) of the two skaters will be ____. a. greatest for the skater who is pushed upon with the greatest force b. greatest for the skater who pushes with the greatest force c. the same for each skater d. greatest for the skater with the most mass e. greatest for the skater with the least mass

4. A 62.1-kg male ice skater is facing a 42.8-kg female ice skater. They are at rest on the ice. They push off each other and move in opposite directions. The female skater moves backwards with a speed of 3.11 m/s. Determine the post-impulse speed of the male skater.

5. A 1.5-kg cannon is mounted on top of a 2.0-kg cart and loaded with a 52.7 gram ball. The cannon, cart, and ball are moving forward with a speed of 1.27 m/s. The cannon is ignited and launches a 52.7 gram ball forward with a speed of 75 m/s. Determine the post-explosion velocity of the cannon and cart.


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The Law of Momentum Conservation

The Law of Action-Reaction (Revisited)

A collision is an interaction between two objects which have made contact (usually) with each other. As in any interaction, a collision results in a force being applied to the two colliding objects. Such collisions are governed by Newton's laws of motion. In the second unit of The Physics Classroom,

Newton's third law of motion was introduced and discussed. It was said that.. in every interaction, there is a pair of forces acting on the two interacting objects. The size of the force on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

Newton's third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces which are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum). According to Newton's third law, the forces on the two objects are equal in magnitude. While the forces are equal in magnitude and opposite in direction, the acceleration of the objects are not necessarily equal in magnitude. In accord with Newton's second law of motion, the acceleration of an object is dependent upon both force and mass. Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force which results during the collision.

Consider the collision between the club head and the golf ball in the sport of golf. When the club head of a moving golf club collides with a golf ball at rest upon a tee, the force experienced by the club head is equal to the force experienced by the golf ball. Most observers of this collision have difficulty with this concept because they perceive the high speed given to the ball as the result of the collision. They are not observing unequal forces upon the ball and club head, but rather unequal accelerations.

Both club head and ball experience equal forces, yet the ball experiences a greater acceleration due to its smaller mass. In a collision, there is a force on both objects which causes an acceleration of both objects. The forces are equal in magnitude and opposite in direction, yet the least massive object receives the greatest acceleration.


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Consider the collision between a moving seven-ball and an eight-ball that is at rest in the sport of table pool. When the seven-ball collides with the eight-ball, each ball experiences an equal force directed in opposite directions. The rightward moving seven-ball experiences a leftward force which causes it to slow down; the eight-ball experiences a rightward force which causes it to speed up. Since the two balls have equal masses, they will also experience equal accelerations. In a collision, there is a force on both objects which causes an acceleration of both objects; the forces are equal in magnitude and opposite in direction. For collisions between equal-mass objects, each object experiences the same acceleration.

Consider the interaction between a male and female figure skater in pair figure skating. A woman (m = 45 kg) is kneeling on the shoulders of a man (m = 70 kg); the pair is moving along the ice at 1.5 m/s. The man gracefully tosses the woman forward through the air and onto the ice. The woman receives the forward force and the man receives a backward force. The force on the man is equal in magnitude and opposite in direction to the force on the woman. Yet the acceleration of the woman is greater than the acceleration of the man due to the smaller mass of the woman.

Many observers of this interaction have difficulty believing that the man experienced a backward force. "After all," they might argue, "the man did not move backward." Such observers are presuming that forces cause motion. In their minds, a backward force on the male skater would cause a backward motion. This is a common misconception that has been addressed elsewhere in The Physics Classroom. Forces cause acceleration, not motion. The male figure skater experiences a backwards force which causes his backwards acceleration. The male skater slows down while the woman skater speeds up. In every interaction (with no exception), there are forces acting upon the two interacting objects which are equal in magnitude and opposite in direction.

Collisions are governed by Newton's laws. The law of action-reaction (Newton's third law) explains the nature of the forces between the two interacting objects. According to the law, the force exerted by object 1 upon object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 upon object 1.


Express your understanding of Newton's third law by answering the following questions. Click the button to check your answers.

1. While driving down the road, a firefly strikes the windshield of a bus and makes a quite obvious mess in front of the face of the driver. This is a clear case of Newton's third law of motion. The firefly hit the bus and the bus hits the firefly. Which of the two forces is


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greater: the force on the firefly or the force on the bus?

2. For years, space travel was believed to be impossible because there was nothing which rockets could push off of in space in order to provide the propulsion necessary to accelerate. This inability of a rocket to provide propulsion in space is because ...

a. space is void of air so the rockets have nothing to push off of. b. gravity is absent in space. c. space is void of air and so there is no air resistance in space. d. ... nonsense! Rockets do accelerate in space and have been able to do so for a long time.. 3. Many people are familiar with the fact that a rifle recoils when fired. This recoil is the result of action-reaction force pairs. A gunpowder explosion creates hot gases which expand outward allowing the rifle to push forward on the bullet. Consistent with Newton's third law of motion, the bullet pushes backwards upon the rifle. The acceleration of the recoiling rifle is ... a. greater than the acceleration of the bullet. b. smaller than the acceleration of the bullet. c. the same size as the acceleration of the bullet. 4. Kent Swimm, who is taking Physics for the third year in a row (and not because he likes it), has rowed his boat within three feet of the dock. Kent decides to jump onto the dock and turn around and dock his boat. Explain to Kent why this docking strategy is not a good strategy.

5. A clown is on the ice rink with a large medicine ball. If the clown throws the ball forward, then he is set into backwards motion with the same momentum as the ball's forward momentum. What would happen to the clown if he goes through the motion of throwing the ball without actually letting go of it? Explain.

6. Chubby, Tubby and Flubby are astronauts on a spaceship. They each have the same mass and the same strength. Chubby and Tubby decide to play catch with Flubby, intending to throw her back and forth between them. Chubby throws Flubby to Tubby and the game begins. Describe the motion of Chubby, Tubby and Flubby as the game continues. If we assume that each throw involves the same amount of push, then how many throws will the game last?


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Using Equations as a Guide to Thinking

The three problems on the previous page illustrate how the law of momentum conservation can be used to solve problems in which the after-collision velocity of an object is predicted based on mass and velocity information. There are similar practice problems (with accompanying solutions) lower on this page which are worth the practice. However, let's first take a more qualitative approach to some collision problems. The questions which follow provide a real test of your conceptual understanding of momentum conservation in collisions.

Suppose that you have joined NASA and are enjoying your first space walk. You are outside the space shuttle when your fellow astronaut of approximately equal mass is moving towards you at 2 m/s (with respect to the shuttle). If she collides with you and holds onto you, then how fast (with respect to the shuttle do you both move after the collision?

This problem could be solved in the usual manner with a momentum table; the variable m could be used for the mass of the astronauts or any random number could be used for the mass of the astronauts (provided each astronaut had the same mass). In the process of solving the problem, the mass would cancel out of the momentum conservation equation and the post-collision velocities could be determined. However, there is a more conceptual means of solving this problem. In order for the momentum before the collision to be equal to the momentum after the collision, the after collision velocity must be smaller than the before collision velocity. How many times smaller must it be? By what factor must the velocity be decreased? Before the collision, the amount of mass in motion is m; after the collision, the amount of mass in motion is 2•m. The amount of mass in motion has doubled as the result of the collision. If the mass is increased by a factor of two, then the velocity must be decreased by a factor of 2. The before-collision velocity was 2 m/s so the after-collision velocity must be one-half this value: 1 m/s. Each astronaut is moving with a velocity of 1 m/s after the collision.


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The process of solving this problem involved using a conceptual understanding of the equation for momentum (p=m*v). This equation becomes a guide to thinking about how a change in one variable effects a change in another variable. The constant quantity in a collision is the momentum (momentum is conserved). For a constant momentum value, mass and velocity are inversely proportional. Thus, an increase in mass results in a decrease in velocity.

A twofold increase in mass, results in a twofold decrease in velocity (the velocity is one-half its original value); a threefold increase in mass results in a threefold decrease in velocity (the velocity is one-third its original value); etc. Of course, it is instructive to point out that this form of problem-solving is limited to situations in which one of the two objects is at rest before the collision and both objects move at the same speed after the collision. To further test your understanding of this type of quantitative reasoning, try the following two questions.

A large fish is in motion at 2 m/s when it encounters a smaller fish which is at rest. The large fish swallows the smaller fish and continues in motion at a reduced speed. If the large fish has three times the mass of the smaller fish, then what is the speed of the large fish (and the smaller fish) after the collision? Click the button to view answer.

A railroad diesel engine has five times the mass of a boxcar. A diesel coasts backwards along the track at 4 m/s and couples together with the boxcar (initially at rest). How fast do the two trains cars coast after they have coupled together? Click the button to view answer.


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Express your understanding of the concept and mathematics of momentum conservation by answering the following questions. Assume isolated systems and momentum conservation for each problem. Click the See Answer button to view answer. (If necessary, return to the instructional page on solving collision analysis problems.)

1. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 92.8 cm/s collides with a 1.50-kg cart (Cart B) moving leftward with a speed of 21.6 cm/s. The two carts stick together and move as a single object after the collision. Determine the post-collision speed of the two carts.

2. A 25.0-gram bullet enters a 2.35-kg watermelon and embeds itself in the melon. The melon is immediately set into motion with a speed of 3.82 m/s. The bullet remains lodged inside the melon. What was the entry speed of the bullet? (CAUTION: Be careful of the units on mass.)

3. A 25.0-gram bullet enters a 2.35-kg watermelon with a speed of 217 m/s and exits the opposite side with a speed of 109 m/s. If the melon was originally at rest, then what speed will it have as the bullet leaves its opposite side? (CAUTION: Be careful of the units on mass.)

4. In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (Cart B) which is initially at rest. The 0.500-kg cart rebounds with a speed of 45 cm/s in the opposite direction. Determine the post-collision speed of the 1.50-kg cart.

5. A 3000-kg truck moving rightward with a speed of 5 km/hr collides head-on with a 1000-kg car moving leftward with a speed of 10 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision


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velocity of the car and truck. (CAREFUL: Be cautious of the +/- sign on the velocity of the two vehicles.)

6. During a goal-line stand, a 75-kg fullback moving eastward with a speed of 8 m/s collides head-on with a 100-kg lineman moving westward with a speed of 4 m/s. The two players collide and stick together, moving at the same velocity after the collision. Determine the the post-collision velocity of the two players. (CAREFUL: Be cautious of the +/- sign on the velocity of the two players.)

Answers

1. The problem can be solved using a momentum table:




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Using Equations as a Recipe for Algebraic Problem-Solving

As discussed in a previous part of Lesson 2, total system momentum is conserved for collisions between objects in an isolated system. The momentum lost by one object is equal to the momentum gained by another object. For collisions occurring in an isolated system, there are no exceptions to this law. This law becomes a powerful tool in physics because it allows for predictions of the before- and after-collision velocities (or mass) of an object. In this portion of Lesson 2, the law of momentum conservation will be used to make such predictions. The law of momentum conservation will be combined with the use of a "momentum table" and some algebra skills to solve problems involving collisions occurring in isolated systems.

Consider the following problem:

A 15-kg medicine ball is thrown at a velocity of 20 km/hr to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.

Such a motion can be considered as a collision between a person and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity (v) across the ice.


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If it can be assumed that the affect of friction between the person and the ice is negligible, then the collision has occurred in an isolated system. Momentum should be conserved and the post-collision velocity (v) can be determined using a momentum table as shown below.

Before Collision After Collision

Person 0 (60 kg) • v

Medicine ball (15 kg) • (20 km/hr)

= 300 kg • km/hr (15 kg) • v

Total 300 kg • km/hr 300

Observe in the table above that the known information about the mass and velocity of the two objects was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expressions 60 kg • v and 15 kg • v were used for the after-collision momentum of the person and the medicine ball. To determine v (the velocity of both the objects after the collision), the sum of the individual momentum of the two objects can be set equal to the total system momentum. The following equation results:

60 • v + 15 • v = 300

75 • v = 300

v = 4 km/hr

Using algebra skills, it can be shown that v = 4 km/hr. Both the person and the medicine ball move across the ice with a velocity of 4 km/hr after the collision. (NOTE: The unit km/hr is the unit on the answer since the original velocity as stated in the question had units of km/hr.)

Now consider a similar problem involving momentum conservation.

A 0.150-kg baseball moving at a speed of 45.0 m/s crosses the plate and strikes the 0.250-kg catcher's mitt (originally at rest). The catcher's mitt immediately recoils backwards (at the same speed as the ball) before the catcher applies an external force to stop its momentum. If the catcher's hand is in a relaxed state at the time of the collision, it can be assumed that no net external force exists and the law of momentum conservation applies to the baseball-catcher's mitt collision. Determine the post-collision velocity of the mitt and ball.


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Before the collision, the ball has momentum and the catcher's mitt does not. The collision causes the ball to lose momentum and the catcher's mitt to gain momentum. After the collision, the ball and the mitt move with the same velocity (v) .

The collision between the ball and the catcher's mitt occurs in an isolated system, total system momentum is conserved. Thus, the total momentum before the collision (possessed solely by the baseball) equals the total momentum after the collision (shared by the baseball and the catcher's mitt). The table below depicts this principle of momentum conservation.


Ball 0.15 kg • 45 m/s = 6.75 kg•m/s (0.15 kg) • v

Catcher's Mitt 0 (0.25 kg) • v

Total 6.75 kg•m/s 6.75 kg•m/s

Observe in the table above that the known information about the mass and velocity of baseball and the catcher's mitt was used to determine the before-collision momenta of the individual objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. Finally, the expression 0.15 • v and 0.25 • v are used for the after-collision momentum of the baseball and catcher's mitt. To determine v (the velocity of both objects after the collision), the sum of the individual momentum of the two objects is set equal to the total system momentum. The following equation results: 0.15 kg • v + 0.25 kg • v = 6.75 kg•m/s 0.40 kg • v = 6.75 kg•m/s v = 16.9 m/s

Using algebra skills, it can be shown that v = 16.9 m/s. Both the baseball and the catcher's mitt move with a velocity of 16.9 m/s immediately after the collision and prior to the moment that the catcher begins to apply an external force.


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The two collisions above are examples of inelastic collisions. Technically, an inelastic collision is a collision in which the kinetic energy of the system of objects is not conserved. In an inelastic collision, the kinetic energy of the colliding objects is transformed into other non-mechanical forms of energy such as heat energy and sound energy. The subject of energy will be treated in a later unit of The Physics Classroom. To simplify matters, we will consider any collisions in which the two colliding objects stick together and move with the same post-collision speed to be an extreme example of an inelastic collision.

Now we will consider the analysis of a collision in which the two objects do not stick together. In this collision, the two objects will bounce off each other. While this is not technically an elastic collision, it is more elastic than the previous collisions in which the two objects stick together.

A 3000-kg truck moving with a velocity of 10 m/s hits a 1000-kg parked car. The impact causes the 1000-kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck immediately after the collision.

In this collision, the truck has a considerable amount of momentum before the collision and the car has no momentum (it is at rest). After the collision, the truck slows down (loses momentum) and the car speeds up (gains momentum).

The collision can be analyzed using a momentum table similar to the above situations.


Truck 3000 • 10 = 30 000 3000 • v

Car 0 1000 • 15 = 15 000

Total 30 000 30 000

Observe in the table above that the known information about the mass and velocity of the truck and car was used to determine the before-collision momenta of the individual


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objects and the total momentum of the system. Since momentum is conserved, the total momentum after the collision is equal to the total momentum before the collision. The after-collision velocity of the car is used (in conjunction with its mass) to determine its momentum after the collision. Finally, the expression 3000•v was used for the after-collision momentum of the truck (v is the velocity of the truck after the collision). To determine v (the velocity of the truck), the sum of the individual after-collision momentum of the two objects is set equal to the total momentum. The following equation results:

3000*v + 15 000 = 30 000

3000*v = 15 000

v = 5.0 m/s

Using algebra skills, it can be shown that v = 5.0 m/s. The truck's velocity immediately after the collision is 5.0 m/s. As predicted, the truck has lost momentum (slowed down) and the car has gained momentum.

The three problems above illustrate how the law of momentum conservation can be used to solve problems in which the after-collision velocity of an object is predicted based on mass-velocity information. There are additional practice problems (with accompanying solutions) later in this lesson which are worth the practice. However, be certain that you don't come to believe that physics is merely an applied mathematics course which is devoid of concepts. For certain, mathematics is applied in physics. However, physics is about concepts and the variety of means in which they are represented. Mathematical representations are just one of the many representations of physics concepts. Avoid merely treating these collision problems as mere mathematical exercises. Take the time to understand the concept of momentum conservation which provides the basis of their solution.

The next section of this lesson involves examples of problems which provide a real test of your conceptual understanding of momentum conservation in collisions. Before proceeding with the practice problems, be sure to try a few of the more conceptual questions which follow


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References

Francis Weston Sears, University Physics, Addison-Wesley Publishing Company, Inc,

London, 1964. Look Kwok Wai, Advance Level Physics, Pearson Education South Asia Pte Ltd,

Singapore, 2006. Lam Chok Sang, Topical Practice Execises Physics, Volume 1, Federal Publications,

Singapura, 2004. Bond Thomas, a-level Challenging drill questions for H1 – H2 Physics, Cosmic Service,

London, 2006 Bueche Frederick J. College Physics Tenth Editin, The McGraw-Hill Companies, United States of America, 2006


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Physics For XI Level

1st Semester

Compiled By : Ichwan Aryono

SMA NEGERI 3 YOGYAKARTA

2009


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Contents Preface ..........................................................................................................................ii 1. Chapter 1

Linear Motion ......................................................................................................... 1

2. Chapter 2 Elasticity and Harmonic Motion ............................................................................ 13

3. Chapter 3 Universal gravitation ............................................................................................ 22

4. Chapter 4 Momentum and Impuls ........................................................................................ 24

Reference .................................................................................................................... iii

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Documents

modul xi sem 1 2009-2010