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Physics 42S IBTopic 2 – Mechanics 3

Energy and Momentum

Physics 42S IBMiles Macdonell Collegiate

2.3 – Work, energy and power

Nature of science:

Theories: Many phenomena can be fundamentally understood through application of the theory of conservation of energy. Over time, scientists have utilized this theory both to explain natural phenomena and, more importantly, to predict the outcome of previously unknown interactions. The concept of energy has evolved as a result of recognition of the relationship between mass and energy. (2.2)

Understandings:

Kinetic energy Gravitational potential energy Elastic potential energy Work done as energy transfer Power as rate of energy transfer Principle of conservation of energy Efficiency

Applications and skills:

Discussing the conservation of total energy within energy transformations

Sketching and interpreting force–distance graphs

Determining work done including cases where a resistive force acts

Solving problems involving power Quantitatively describing efficiency in energy

transfers

Guidance:

Cases where the line of action of the force and the displacement are not parallel should be considered

Examples should include force–distance graphs for variable forces

Theory of knowledge:

To what extent is scientific knowledge based on fundamental concepts such as energy? What happens to scientific knowledge when our understanding of such fundamental concepts changes or evolves?

Utilization:

Energy is also covered in other group 4 subjects (for example, see: Biology topics 2, 4 and 8; Chemistry topics 5, 15, and C; Sports, exercise and health science topics 3, A.2, C.3 and D.3; Environmental systems and societies topics 1, 2, and 3)

Energy conversions are essential for electrical energy generation (see Physics topic 5 and sub-topic 8.1)

Energy changes occurring in simple harmonic motion (see Physics sub-topics 4.1 and 9.1)

Aims:

Aim 6: experiments could include (but are not limited to): relationship of kinetic and gravitational potential energy for a falling mass; power and efficiency of mechanical objects; comparison of different situations involving elastic potential energy

Aim 8: by linking this sub-topic with topic 8, students should be aware of the importance of efficiency and its impact of conserving the fuel used for energy production

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Topic 2 Energy and Momentum

Data booklet reference:

W=Fscosθ

EK=12mv2

Ep=12k x2

∆ EP=mg∆h

power=Fv

Efficiency= useful work out

totalwork∈¿=useful power outtotal power∈¿¿

¿

- 2 -


Work, Energy, and Power

Work

If a force is acting on an object as it moves a certain distance, work is being done on the object. We define the work done on an object as:

W=Fscosθ

W is the work, in Joules ( J ) F is the force, in Newtons (N ) sis the displacement, in meters (m )θ is the angle between

F and s

When work is done by a force on an object the energy that the object possesses will change. If the force and displacement are in the same direction, energy is added to the object. If the force and displacement are in the opposite directions, energy is removed from the object.

When mechanical energy is added to a mass, the speed of a mass typically increases. When mechanical energy is removed from a mass, the speed of the mass typically increases.

Joules

The work being done on an object is the product of force and displacement in the direction of the force.

The unit of work is the Joule. 1J is the work done by a force of 1N when it moves a body a distance of 1m in the direction of the force.

1J=1N ∙m

1J=1kgm s−2∙m

1J=1kgm2 s−2

Example 1

A force of constant magnitude of 25N is applied to a body that moves along a straight path. Find the work done after the mass moves a distance of50.m.

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Example 2

Find the work done by the tension of a string as a mass attached to the end performs one horizontal revolution of uniform circular motion (constant speed).

Example 3

A mass is being pulled along a level road by a rope attached at an angle of 40.0 ° with the horizontal. The force in the rope is20.0N . What is the work done by this force moving the mass a distance of 8.00m along the level road?

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Example 4

A50.0 kg crate is pulled40.0m along a horizontal floor by a constant force exerted by a person,F A=100N , at an angle of37.0 ° above the horizontal. The floor exerts a force of friction in the opposite direction ofF f =50.0N .

(a) Determine the work done on the crate by the person.

(b) Determine the work done on the crate by friction.

(c) Determine the net work done on the crate.

- 5 -


Force – Displacement Graphs

Force-displacement graph can be used to calculate the work done on a body where the force is not constant. The area under a force-displacement graph of an object gives the work being done to the object.

By analyzing the units, we can see that the area under the graph is equal to the work being done.

Area=base×height

Area=m×N

Area=J

Example 6

Find the total change in energy of the body from Figure 1.

Homework: Giancoli – Page 161, Questions 1-16

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Figure 1 – a force-displacement graph

d /m108642

1000

750

500

250

0

F /N


Kinetic Energy

When work is being done to an object, we can be adding or removing kinetic energy.

From our kinematics equations:

v2=u2+2as

Replacinga= Fm , we get:

v2=u2+2 Fm

s

and isolating for Fs:

F ∙ s=12m (v2−u2 )

F ∙ sis the work done on the mass. This work is the change in kinetic energy. For the kinetic energy of an object:

∆ Ek=12m (v2−u2 )

If the initial speed of the object was0:

Ek=12mv2

Ek is the kinetic energy, in Joules (J )

m is the mass, in kilograms (kg)

v is the speed, in meters per second (ms−1)

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Example 1

How much kinetic energy does a 5.00kg particle have if it is moving at2.50ms−1?

Example 2

How fast is a12.5kg object traveling if it has 100.J of kinetic energy?

- 8 -


Work and Energy

To change the kinetic energy of a mass, and therefore the speed, work must be done to the mass.

E1+W=E2

Work can also represent the change in kinetic energy of the mass:

∆ EK=E2−E1

W=E2−E1

Example 3

A mass of 5.00kg is moving with an initial velocity of 12.0ms−1 is brought to rest by a force over a distance of12.0m. What is the magnitude of the force?

Example 4

A12.0 kg mass is moving with an initial velocity of+15.0ms−1. A force of−45.0N is applied over a displacement of+25.0m. Using work and energy, calculate the final speed of the mass.

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Example 5

Calculate the work required to accelerate a1000kg crate from a speed of20ms−1 to a speed of30ms−1.

Example 6

A50.0 kg mass is moving with an initial speed of12.0ms−1. A force of−150.N is applied over a displacement of−13.5m. Using work and energy, calculate the final speed of the mass.


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Gravitational Potential Energy

Potential energy is the amount of work that could be done by a force on an

Near the surface of the Earth (or any planet) gravitational potential energy can be approximated by the formula:

∆ EP=mg∆h

Where:

∆ EP is the change in gravitational potential energy, in Joules( J )

m is the mass, in kilograms (kg)

g is the gravitational field, in Newtons per kilogram(N k g−1 ), or in meters per second squared(ms−2)

∆ h is the change in height, in meters (m)

The height of the object is always relative to a fixed point, usually the surface of the Earth.

This formula for gravitational potential energy can only be used in a uniform gravitational field. Generally, this equation works near the surface of a planet with small distances(0−10km). If the distance is greater, we must use a more complex equation:

EP=−GMm

r

But, more on this later… (see Topic 10 – Fields)

Work done by Gravity

If an object moves horizontally along a surface parallel to the surface of the Earth, then the work done on the object by gravity is zero, since in this case the angle formed between the force and the displacement is90 °.

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Figure 2 – work done by gravity on a horizontally moving body

displacement

mgmgmg


If a body is thrown vertically upward to a height ofh, then the work done on the object is−mgh. The force of gravity is anti-parallel to the displacement.

If a body falls a vertical distanceh, then the work done on the object is+mgh. The force of gravity is parallel to the displacement.

Example 7

A ball of mass m=0.250 kg is thrown from a height of1.50m with an initial speed of 15.0ms−1.

(a) Determine the kinetic energy of the ball.

(b) Determine the work done by gravity on the ball and the speed of the ball as it reaches a height of9.00m directly above where it was thrown.

(c) Determine the work done by gravity and the speed of the ball at the point where the ball has displaced 9.00m [40 °above thehorizontal ] from where it was thrown.

- 12 -

Figure 3 – work done by gravity on a vertically moving body

h

mg

mg

mg

h

mg

mg

mg


Conservation of Mechanical Energy

In a closed, ideal system, the total energy of an object at any one point in time should be the same as the total energy of the same object at any other point in time, i.e. the total energy is constant.

If an outside force is applied, then the work done by this force must be considered when calculating the total energy.

In the following scenario, a ball is thrown into the air. As the ball rises, it gains potential energy. To gain potential energy, it must lose kinetic energy. However, the total energy of the ball remains constant.

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Figure 4 – a body moving through a uniform gravitational field

v5

v4

v2

v3

v1


Example 1

A0.750kg ball is dropped from rest at a height of5.5m. Calculate the total energy and the speed of the ball at a height of3.2m and the speed as it hits the ground.

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Figure 5 – a body falling in a uniform gravitational field

h=0.0m

h=3.2m

h=5.5m


Example 2

A0.50kgball rolls off a1.0m high flat, level table with a speed of4.0ms−1, as in figure 6. Calculate the speed of the ball when it strikes the floor.

Example 3

A roller coaster with a mass of350.kg is travelling at a speed of12.0ms−1 at point A on the frictionless track in Figure 7. Calculate the height at point B and the speed at point C.

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Figure 7 – A roller coaster moving along a track

C

h=?

v=?

h=10.0m

v=5.00ms−1B

A

v

h=1.0m

Figure 6 – A ball falling off a table.

v=4.0ms−1


Example 4

A500.kg roller coaster is travelling at5.00ms−1 along a track where there is friction as in Figure 8.

(a) Given the following diagram, what is the work being done by friction?

(b) If the work being done on the roller coaster by friction occurs over a distance of25.0m, what is the average force of friction?

- 16 -

Figure 8

v=10.5ms−1h=7.50m

v=5.00ms−1


Example 5

Determine the minimum speed of the mass in Figure 9 at the initial point such that the mass makes it over the barrier of heighth=7.5m.

Example 6

A pendulum of length1.000m is released from rest with the string starting at an angle of10.0 ° to the vertical. Find the speed of the mass on the end of the pendulum when it passes through its lowest position.

- 17 -

Figure 9

v=?h=7.5m


Example 7

A body of mass2.00kg , initially at rest, slides down a curved path of total length22.0m, as in figure 10. The body starts from a vertical height of5.00m from the bottom. When it reaches the bottom, its speed is found to equal6.00ms−1.

(a) Show that there is a force resisting the motion.

(b) Assuming the force to have a constant magnitude, determine the magnitude of the force.

- 18 -

Figure 10

m=2.00kg

∆ h=5.00m


Example 8

A mass of5.00kg with an initial velocity of2.0ms−1 is acted upon by a force of55N in the direction of motion. The motion is opposed by a frictional force. After traveling a distance of12m the velocity of the body becomes15ms−1. Determine the magnitude of the force of friction.

- 19 -


Elastic Potential Energy

Work done by a spring

According to Hooke’s Law, a spring applies a force of tension as such:

T = k x

To find the work done by extending or compressing a spring we can find the area under a force time graph.

W=12Tx

W=12kx x

W=12k x2

EP=12k x2

We can also derive the equation for elastic potential energy if substitute the average force of tension in the work equation. We can leave out thecosθas the displacement and the force will always be in the same direction.

EP=12(T+T 0)s

EP=12Tx

EP=12kxx

EP=12k x2

- 20 -

Figure 11 – a force displacement graph

x / m

T / N


Example 1

A mass of8.40kg rests on top of a vertical spring whose base is attached to the floor. The spring compresses by5.20 cm.

(a) Calculate the spring constant of the spring.

(b) Determine the energy stored in the spring.

Example 2

A mass of0.250 kg is attached to a spring with a spring constant ofk=50.0N m−1. The mass on the spring is then stretched horizontally a distance of0.135m and then released. Assume that the mass and spring rest on a frictionless surface.

(a) What is the potential energy of the spring when it is extended0.135m?

(b) What is the speed of the mass when it returns to the natural length of the spring?

(c) What is the speed of the mass when the spring is compressed0.100m?

- 21 -


Example 3

A mass of15.0kg is dropped on to a curved surface at a height of10.0m and it slides down the surface. At the end of the curved surface, there is a spring with a spring constant of200 N m−1 as in Figure 12.

(a) What is the potential energy of the mass before it is dropped?

(b) What is the speed of the mass at the end of the curved surface?

(c) What is the distance that the spring will compress when the mass is brought to rest?

- 22 -

Figure 12

k=200N m−1

10.0m

15.0kg


Example 4

A body of mass4.20 kg with an initial speed5.60ms−1 begins to move up an incline as shown in Figure 13:

The body will momentarily be brought to rest after colliding with a spring of spring constant220N m−1. The body stops at a vertical distance of0.850m above its initial position.

(a) Determine the amount by which the spring has compressed. There are no frictional forces.

(b) Repeat the calculation but now with a constant force of friction opposing the motion along the uphill part of the path. The length of the uphill part of the path is1.20m and the frictional force is15N .

- 23 -

Figure 13

∆ hv=5.60ms−1


Example 5

Determine the spring constant of a spring that makes a maximum extension of 0.178mwhen a mass ofm=12.5 kgis fixed to the end and is left to oscillate.


- 24 -

x=0.178m


Power

When a quantity of work∆W is being performed within a time interval∆ t the ratio:

P=∆W∆t

is the power developed. The units for Power are Watts(W ) which are equivalent to Joules per second. 1W=1J s−1

If you consider the ratio of work per unit time, we can derive another equation to give us power.

P=∆W∆t

P= F ∆s∆ t

P=F ∆s∆ t

P=Fv

power=Fv

Example 1

What is the minimum power required to lift a mass of50.0kg up a vertical of12m in5.0 s?

- 25 -


Example 2

What is the minimum power output of an engine of a car that moves with a speed of25.0ms−1 with an overall force of friction of 1.50×103 N?

Example 3

A70.0 kg jogger runs up a long flight of stairs in4.00 s. The vertical height of the stairs is4.50m.

(a) Estimate the jogger’s power output in watts.

(b) How much energy did this require?

- 26 -


Example 4

A car of a1400 kg travels on the highway experiences an average retarding force (friction, air resistance, etc.) of700N . Calculate the power required under the following circumstances:

(a) the car climbs a10 ° hill at a steady speed of80kmh−1.

(b) the car accelerates along a level road from90km h−1 to100 kmh−1 in a period of6.0 s to pass another car.

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Efficiency

Suppose that a mass is being pulled up along a rough incline with a constant speed, as in figure 14. Let the mass be15.0 kg and the angle of inclination be45.0 °. The constant frictional force opposing the motion isf=42N .

The forces on the mass are:

F=mg sin θ+f

¿106 N+42N

¿148N

If the force applied raises the mass a distance of20.0m along the plane, the work done by the force is:

W=148N ∙20.0m=2.96×103 J

The actual change in energy is∆ EP through the change in height of14.1m which is:

∆ EP=mgh

∆ EP=15.0 kg ∙9.81N k g−1 ∙14.1m=2.07×103 J

The efficiency with which the force raised the mass is:

Efficiency= useful work out

totalwork∈¿=useful power outtotal power∈¿¿

¿

- 28 -

Figure 14

R

mg

f

F

θ=45.0 °


¿ 2.07×103J

2.96×103J

¿0.699

Example

A0.50kg battery operated toy train moves with constant velocity 0.30ms−1along a level class. The power of the motor in the train is 2.0W and the total force opposing the motion of the train is5.0 N .

(a) What is the efficiency of the train’s motor?

(b) Assuming the efficiency and the opposing force stay the same, calculate the speed of the train as it climbs an incline10 ° to the horizontal.

- 29 -


Homework: Giancoli – Page 164, Questions 57-68Tsokos – Page 96-97, Questions 55-71

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2.4 – Momentum and impulse

Nature of science:

The concept of momentum and the principle of momentum conservation can be used to analyse and predict the outcome of a wide range of physical interactions, from macroscopic motion to microscopic collisions. (1.9)

Understandings:

Newton’s second law expressed in terms of rate of change of momentum

Impulse and force–time graphs Conservation of linear momentum Elastic collisions, inelastic collisions and

explosions

Applications and skills:

Applying conservation of momentum in simple isolated systems including (but not limited to) collisions, explosions, or water jets

Using Newton’s second law quantitatively and qualitatively in cases where mass is not constant

Sketching and interpreting force–time graphs Determining impulse in various contexts

including (but not limited to) car safety and sports

Qualitatively and quantitatively comparing situations involving elastic collisions, inelastic collisions and explosions

Guidance: Students should be aware that F = ma is

equivalent ofF=∆ pt only when mass is

constant Solving simultaneous equations involving

conservation of momentum and energy in collisions will not be required

Calculations relating to collisions and explosions will be restricted to one-dimensional situations

A comparison between energy involved in inelastic collisions (in which kinetic energy is not conserved) and the conservation of (total) energy should be made

International-mindedness:

Automobile passive safety standards have been adopted across the globe based on research conducted in many countries

Theory of knowledge: Do conservation laws restrict or enable further

development in physics?

Utilization:

Jet engines and rockets Martial arts Particle theory and collisions (see Physics sub-

topic 3.1)

Aims:

Aim 3: conservation laws in science disciplines have played a major role in outlining the limits within which scientific theories are developed

Aim 6: experiments could include (but are not limited to): analysis of collisions with respect to energy transfer; impulse investigations to determine velocity, force, time, or mass; determination of amount of transformed energy in inelastic collisions

Aim 7: technology has allowed for more accurate and precise measurements of force and momentum, including video analysis of real-life collisions and modelling/simulations of molecular collisions


Data booklet reference:

p=mv

F=∆ pt

EK=p2

2m

Impulse=Ft=∆ p


Momentum and Impulse

Linear Momentum

Linear Momentum ( p) is defined as the product of mass and velocity.

Momentum=mass×velocity

p=m∙v

The S.I. units for momentum arekg ms−1. Alternative units of Ns can also be used. Since velocity is a vector quantity, momentum must also be a vector quantity. The change in momentum,∆ p, is called the impulse.

In terms of momentum, Newton’s second Law of mechanics can be stated as

F=∆ p∆ t

The average net force on a body equals the rate of change of a body’s momentum.

If a mass of a body is constant, this formula reduces to the familiarF=ma:

F=∆ p∆ t

F=mv f−mvi

∆ t

F=m(v¿¿ f−v i)

∆ t¿

F=m ∆ v∆ t

F=ma

The advantage of the first equation is that it can also be used in cases where an objects mass is changing (for example, in the motion of a rocket).

- 34 -


Example 1

A0.100 kg ball moving at5.0ms−1 bounces off a vertical wall without a change in its speed. If the collision with the wall lasted for0.10 s, what force was exerted on the wall?

Example 2

A0.50kg ball bounces vertically off a hard surface. A graph of the velocity versus time is shown in figure 15. Find the magnitude of the momentum change of the ball during the bounce. The ball stayed in contact with the floor for0.15 s. What average force did the ball exert on the floor?

- 35 -

Figure 15 – a velocity time graph and free body diagram for a ball hitting the floor

−2

0.90.60.3

−4

0

2

t /s

v /ms−1

W

FN


Example 3

Bullets of mass30.0 g are being fired from a gun with a speed of300ms−1 at a rate of20 bullets per second. What force is being exerted on the gun?

Example 4

A cart moves in a horizontal line at a constant speed3.5ms−1. Rain starts to fall and fills the cart with water at a rate ofσ=0.20 kg s−1. (This means that in one second,0.20 kg has fallen on the cart.) The cart must keep moving at a constant speed. Determine the magnitude force that must be applied to the cart.

- 36 -


Example 5

Gravel falls vertically on a conveyor belt at a rate ofσ=5.00kg s−1 as shown in figure 16:

(a) Determine:(i) the force that must be applied on the belt to keep it moving at constant speed

v=1.75ms−1.

(ii) the power that must be supplied by the motor to turn the belt.

(iii) the rate at which the kinetic energy of the gravel is changing.

(b) Explain why the answers to (a) (ii) and (a) (iii) are different.

- 37 -

Figure 16 – a conveyor belt moving gravel

beltgravel


Impulse

The change in momentum is proportional to the amount of force transferred from one object to another. Mathematically we can write this as:

F=∆ p∆ t

The impulse is the total change in momentum.

Impulse=∆ p=F ∆t

The force generated by impulse is not constant. The average force (F) is often sufficient for a force acting over a period of time.

Impulse can also be calculated by finding the area under a Force-time graph.

In the graph the area is about2.5N s. The maximum force that acted on the body was1000N . The average Force is:

F=∆ p∆ t

¿ 2.5N s6.0×10−3 s

≈470N

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Figure 17 – a force-time graph

t /ms108642

1000

750

500

250

0

Force /N


Example 1

A baseball (m=0.14 kg) has an initial velocity of u=−38ms−1 as it approaches the bat. The bat is swung and hits the ball. The baseball leaves the bat with a velocity ofv=+58ms−1.

(a) Determine the impulse applied to the ball by the bat.

(b) Determine the average force in the following example if the time of contact was∆ t=1.6×10−3 s.

- 39 -


Example 2

Consider the force-time graph in Figure 18.

The force acts on a body of mass3.0kg initially at rest. Calculate:

(a) the initial acceleration of the body.

(b) the speed att=4.0 s.

(c) the speed att=6.0 s.

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Figure 18

654321t /s

−10

−5

0

5

10

15F /N


Example 3

A ball of mass0.250kg moves on a frictionless horizontal surface and hits a vertical wall at a speed of5.0ms−1. The ball rebounds with a speed of4.0ms−1.

(a) If the ball was in contact with the wall for0.150 s, find the average force that acted on the ball.

(b) The force in the previous example is assumed to vary with time. Deduce the maximum force that acted on the ball if the force acted on the ball as in Figure 19.

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Figure 19

0.150 st /s

F /N


Example 4

A force of1000.0 N acts on a body of40.0 kg, that is initially at rest, for a time interval of0.0500 s. What is the velocity of the mass?

Example 5

A body of mass12.5 kg moves with a velocity of+5.70ms−1. A force of−250N acts on the body for a period of timet=0.450 s. Determine the final velocity of the body.

- 42 -


Example 6

The following graph shows the applied force to a cart that has a mass of15.0kg.

(a) Determine the impulse of the force on the cart.

(b) The cart starts from rest. Determine the final momentum and speed of the cart.

Example 7

A body with a mass of2.50kgis in motion with a velocity of+1.50ms−1. It experiences a non- uniform force shown in the following graph.

(a) Calculate the initial momentum of the body.

(b) Calculate the impulse acting on the body.

(c) Determine the final velocity of the body.

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Figure 20

t /ms108642

1000

750

500

250

0

F /N

Figure 21

t /ms108642

1000

750

500

250

0

F /N


Example 8

A body with a mass of45.0 kgis in motion with a velocity of +18.5ms−1. It experiences a non- uniform force shown in the following graph.

(a) Calculate the initial momentum of the body.

(b) Calculate the impulse acting on the body.

(c) Determine the final velocity of the body.

Consider 2 bodies with the same momentum. A body that is brought to rest with a greater average force requires less time stop. A body that is brought to rest with a smaller average force requires more time.

The graph in Figure 21 represents 2 bodies undergoing the same impulse.

Example 9

A 70.0 kgperson lands on firm ground after jumping from a height of5.00m.

(a) Calculate the impulse experienced by the person from the ground.

(b) Calculate the average force exerted on the legs if the person lands stiff legged, change in distance of1.0cm.

- 44 -

Figure 22

t /s

F /N

t /s54321

−400

−300

−200

−100

0Force /N


(c) Calculate the average force exerted on the legs if the legs bend so the distance moved is0.50m.

- 45 -


Conservation of Momentum

The law of conservation of linear momentum states that the total linear momentum of a system of interacting bodies or particles remains constant. This is true for a closed system provided that no resultant external force interacts with the system.

In effect, the total momentum before and after a collisions or impulse must be equal.

Example 1

Two masses of2.0kg and3.0kg move to the right with speeds of4.0ms−1 and5.0ms−1 respectively. What is the total momentum of the system?

Example 2

A system consists of 2 masses; one mass of2.0kg moves to the right with a speed of 10.0ms−1 and another mass of4.0 kg moves to the left with a speed of8.0ms−1. What is the total momentum of the system?

- 46 -


Example 3

A mass of3.0kg is at rest is hit by a second mass of5.0 kg and a velocity of+4.0ms−1. The smaller mass moves away with a velocity of+3.0ms−1. What happens to the larger mass?

Example 4

Two masses of 2.0kg and4.0 kg are held together with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a speed of6.0ms−1, what is the speed of the other mass?


- 47 -


Conservation of Momentum in 2 Dimensions

When a collision happens in 2 dimensions, the momentum must be conserved in all directions, i.e. the horizontal momentum must be conserved and the vertical momentum must be conserved.

Example 1

Two identical curling stones of mass19.5 kgcollide. The first stone hits the stationary second stone with a velocity of5.00ms−1[N ]. If the velocity of the first stone is3.20ms−1[N 30 °W ] after the collision, find the velocity of the second stone after the collision.

- 48 -


Example 2

A mass of22.5 kg traveling at a velocity of14.0ms−1[N ] collides with a mass of17.0 kg traveling at a velocity of10.5ms−1[S]. If the17.0 kg mass has a velocity of8.50ms−1 [ S40°W ]after the collision, what is the velocity of the22.5kg mass?

- 49 -


Example 3

Consider a stationary body of mass12.0kg that is hit by a4.0 kg mass moving at12ms−1. The collision is not head on and the bodies move away from each other at an angle to the original direction of motion as shown in Figure 21. How can we find the speeds of the two bodies after the collision?

- 50 -

Figure 21

30 °60 °


Elastic Collisions

Consider the following scenario where 2 masses collide:

Before the collision:

p=m∙v¿80kgm s−1

Ek=12mv2

¿400 J

Scenario 1: Totally inelastic collision

p=m∙v¿80kgm s−1

Ek=12mv2

¿160 J

The momentum is conserved however kinetic energy is lost. In collisions that are completely inelastic the speed of separation is0ms−1. The masses ‘stick’ together. In this scenario the maximum amount of kinetic energy is lost.

Scenario 2: Inelastic collision

p=m1 ∙ v1+m2 ∙ v2¿80kgm s−1

Ek=12m1 v1

2+ 12m2 v2

2

¿220 J

The momentum is conserved however kinetic energy is still lost. In collisions that are inelastic the speed of separation does not equal the speed of approach.

vapp=10ms−1

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Figure 22 – two bodies before a collision

12kg10ms−1

8.0kg

Figure 22 (a) – a totally inelastic collision

12kg4ms−1

8.0kg

Figure 22 (b) – an inelastic collision

1ms−1 12kg 6ms−18.0kg


vsep=5ms−1

Scenario 3: Elastic collision

p=m1 ∙ v1+m2 ∙ v2¿80kgm s−1

Ek=12m1 v1

2+ 12m2 v2

2

¿400 J

The momentum and kinetic energy are conserved. In collisions that are elastic the speed of separation equals the speed of approach.

vapp=10ms−1

vsep=10ms−1

Calculating velocities for elastic collisions

To determine the velocities of particles after an elastic collision we need to use relative velocities. Particle A has a mass of3.0kg and particle B has a mass of4.0 kg.

The total momentum of the system is+1.0kg ms−1 and the total energy of the system is21.5 J .

We will count particle B as stationary and consider the relative velocity of particle A by adding+2.00ms−1 to each velocity.

The total momentum of the system before and after the collision can be calculated as:

pA1+ pB1=pA2+ pB 2

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Figure 22 (c) – an elastic collision

2ms−112kg 8ms−1

8.0kg

Figure 23 – calculations for elastic collisions

BAv=2ms−1v=3ms−1

BA

Figure 23 (a) – relative velocities in B’s frame of reference

v=0ms−1v=5ms−1


We can isolate for the final momentum of particle B:

pA1=p A2+pB2

pA1−p A2=pB2

mA v A 1−mA v A 2=mB v B2

mA (vA 1−vA 2 )=mB vB2

The total kinetic energy would be calculated in a similar fashion:

EKA 1+EKB 1=EKA2+EKB 2

We can isolate the final kinetic energy of particle B

EKA 1=EKA2+EKB2

12mA v A1

2 =12mA v A2

2 + 12mBv B2

2

mAv A 12 =mA v A 2

2 +mB vB22

mA v A 12 −mA v A 2

2 =mB v B22

mA (v A12 −v A2

2 )=mBvB 22

We can then divide the energy equation by the momentum equation:

mA (v A12 −v A2

2 )=mB vB 22 ; mA (vA 1−vA 2 )=mB vB2

mA (v A 12 −v A 2

2 )mA (v A 1−v A 2)

=mBvB2

2

mBvB2

And simplify to:

(vA 12 −v A 2

2 )(vA 1−v A 2 )

=v B2

(v A 1+v A2 ) (v A 1−v A 2)(v A 1−v A 2)

=vB 2

vA 1+v A2=vB2

This equation shows that the velocity of separation must equal the velocity of approach:

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vA 1=vB2−v A2

Replacing the velocity for particle B into the original momentum equation would give us:

vA 1+v A2=vB2

pA1=p A2+pB2

mA v A 1=mA v A 2+mB vB2

mAv A 1−mA v A 2=mB (v A 1+v A2 )

And solve for the final velocity of particle A

mA v A 1−mA v A 2=mB v A1+mB v A2

mA v A 1−mBv A 1=mA v A2+mB v A2

vA 1 (mA−mB )=vA 2(mA+mB)

v A1 (mA−mB )mA+mB

=v A2

For this example:

5ms−1 (3.0kg−4.0kg )3.0kg+4.0kg

=v A2

−0.71ms−1=v A 2

And the final velocity of particle B:

vA 1+v A2=vB2

5.0ms−1+(−0.71ms−1 )=v B2

+4.3ms−1=vB2

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The relative velocities of the particles are as follows:

We can revert the velocities back from their relative velocity to their actual velocity by subtracting+2.00ms−1. The actual velocities are:

The total momentum of the system is+1.0kg ms−1 and the total energy of the system is21.5 J . As both momentum and kinetic energy are conserved the collision is perfectly elastic.

The speed of separation is equal to the speed of approach.

vapp=5.0ms−1

vsep=5.0ms−1

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BA

Figure 23 (b) – relative velocities after the collision from B’s initial frame of reference

v2=4.3ms−1v2=0.71ms−1

Figure 23 (c) –velocities after the collision

BAv2=2.3ms−1v2=2.71ms−1


Example 1

A 0.300 kg toy train and 0.600 kg toy train are involved in an elastic collision on a straight section of a model rail. The 0.300 kg train, traveling at +3.50ms−1strikes the 0.600kg train that has a velocity of −2.00ms−1. Determine the velocities of the trains after the collision.

Example 2

Two balls are involved in an elastic head-on collision. If the first ball has a massm1=2.5kg was traveling at+4.0ms−1 and the second ball has a massm2=4.0kg was traveling at−2.0ms−1, determine the velocity of each ball after collision.

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Example 3

The velocity-time graph in Figure 23 shows a collision between 2 carts on a rail track. The first, a mass of400. kg moving at30.0ms−1 collides with a600. kgcart that is at rest.

(a) What evidence is there that the collision is elastic?

(b) Calculate the impulse applied to each carriage.

(c) Calculate the average force applied to each carriage.

Homework: Giancoli – Page 188, Questions 21-25Tsokos – Page 109-110, Questions 72-83

- 57 -

Figure 24

Mass 2

Mass 1

−4t /s

1.00.80.60.40.2

8

4

0

28

24

20

16

12

v /ms−1

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