1206 - Concepts in Physics

Monday, October 5th

Notes• Assignment #3 has been posted

• Today’s tutorial will have two parts first 15 min: problem solving strategies and than assignment #2 will be discussed in detail

• Assignment #2 will be available Wednesday for pick-up (after class)

Definition of Impulse:The impulse J of a force is the product of the average force Favg and the time interval

∆t during which the force acts:

J = Favg * ∆t

impulse J = area under the force curve.

The linear momentum p of an object is the product of the object’s mass m and velocity

p = m * v

momentum p is a vector quantity and the unit is kg m/s

Definition of Momentum:

Impuls = change in MomentumFavg ∆t = pf - pi

Note: this was how Newton formulated his second law

From last time ...

Conservation of linear momentum

The total linear momentum of an isolated system remains constant (is conserved). An isolated system is one for which the vector sum of the average external forces acting on the system

is zero.

Principle of conservation of linear momentum

This principle applies to a system containing any number of objects, regardless of the internal forces, provided the system is isolated. Whether the system is isolated depends on whether the vector sum

of the external forces is zero. Judging whether a force is internal or external depends on which objects are included in the system.

Earlier we learned about energy conservation. Conservation of Impuls is equally important and used in many area’s of physics.

Internal forces - Forces that the objects within the system exert on each other. External forces - Forces exerted on the objects by agents external to the system.

Deriving conservation of linear momentumexample: collision of two masses m1 and m2:

Before: t = t0

m1

m2

v01

v02F21

F12 vf2

vf1After: t = tfDuring: t = tc

During the collision (t = tc) the forces F12 (force exerted on object 1 by object 2) and F21 (force exerted by object 2 on object 1) are action - reaction forces and therefore equal in magnitude and

opposite in direction (Newton’s third law). They are also internal forces, since they are forces that the two object exert on each other inside the system. Since the objects have wights (W1 and W2) due to the force of gravity, we also have external forces acting. Other external force would be air resistance

and friction - we will ignore them for simplicity reasons in this example.

We apply the impuls - momentum theorem on every object and obtain:

Object 1: (W1 + F12) Δt = m1 vf1 - m1 v01 Object 2: (W2 + F21) Δt = m2 vf2 - m2 v02

We add these equation to produce a single result for the system as a whole and obtain:

(W1 + W2 + F12 + F21) Δt = (m1 vf1 + m2 vf2) - (m1 v01 + m2 v02) { { { {External forces

Internal forces

total final momentum pf

total initial momentum p0

(W1 + W2 + F12 + F21) Δt = (m1 vf1 + m2 vf2) - (m1 v01 + m2 v02) { { { {External forces

Internal forces

total final momentum pf

total initial momentum p0

We can write that in words:

(Sum of average external forces + Sum of average internal forces) * elapsed time = Difference between final momentum and initial momentum

We know that the sum of internal forces is zero, so we can leave them out and write:

(Sum of average external forces) Δt = pf - pi

Suppose the sum of external forces is zero, which is true for every isolated system, then:

0 = pf - pi or pf = pi

We obtained the principle of conservation of linear momentum!!

Example: Imagine to balls colliding on a billiard table that is friction free.Use the momentum conservation principle in answering the following questions. (a) Is the total momentum of the two-ball system the same before and after the collision? (b) Does this change if the system contains only one of the two balls?

(a) Let’s collect all external forces involved into this two-ball system: Weights W1 and W2 of the balls

Normal forces (upward) on each ball FN1 and FN2

Since the balls do not accelerate in the vertical direction, the normal forces must balance the weights, so that the vector sum of the four external forces is zero. We defined the table as friction free.

Therefore we don’t have a net external force to change the total momentum of the two-ball system.The total momentum of this two-ball system is conserved.

(b) YOUR TURN

Example: Starting from rest, two skaters “push off” against each other on a smooth level ice, where friction is negligible. One is a woman (m1 = 54 kg), and the other one is a man (m2 = 88 kg). The woman moves

away with a velocity of vf1 = +2.5 m/s. What is the final velocity of the man? (Also called recoil velocity)

For a system consisting of the two skaters on level ice, the sum of external forces is zero. Thus, we can use the principle of conservation of linear

momentum to determine the man’s recoil velocity.The man has a larger mass, therefore (according to Newton’s second law) the acceleration he experience will be smaller and thus a smaller recoil velocity.

The total momentum after the “push off” must be equal to before. When the skaters start, they are at rest, therefore their total momentum is zero.

m1 vf1 + m2 vf2 = 0 m1 vf1 = - m2 vf2

vf2 = -(m1/m2) vf1 = -(54 kg/88 kg) 2.5 m/s = -1.5 m/sAs expected the recoil velocity of the man is smaller.

Note! The total linear momentum is conserved, even when the kinetic energy of the individual parts of a system changes. Kinetic energy changes because work is done by the internal forces.

YOUR TURN: A freight train is being assembles in a switching yard. Let’s look at two boxcars. Car 1 has a mass of

m1 = 65 x 103 kg and moves at a velocity of v01 = 0.80 m/s. Car 2 has a mass of m2 = 92 x 103 kg and a velocity of v02 = 1.30 m/s. It is faster, overtakes car 1 and couples to it.

Neglecting friction, find the common velocity vf of the car after they become coupled.

v02 v01 vf

YOUR TURN: A freight train is being assembles in a switching yard. Let’s look at two boxcars. Car 1 has a mass of

m1 = 65 x 103 kg and moves at a velocity of v01 = 0.80 m/s. Car 2 has a mass of m2 = 92 x 103 kg and a velocity of v02 = 1.30 m/s. It is faster, overtakes car 1 and couples to it.

Neglecting friction, find the common velocity vf of the car after they become coupled.

v02 v01 vf

before after

The boxcars form a system. The sum of external forces acting on the system is zero. So, we can use the linear momentum conservation principle.

(m1 + m2) vf = m1 v01 + m2 v02

vf = (m1 v01 + m2 v02)/(m1 + m2)

vf = {(65 x 103 kg)(0.80 m/s) + (92 x 103 kg)(1.30 m/s)}/(157 x 103 kg) = 1.1 m/s

Note! The value for the final velocity is between the two given initial velocities

The diagram below 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.

YOUR TURN: from last time

The diagram below 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.Case A has the greatest velocity change. The velocity change is -9 m/s in case A and only -5 m/s in case B.Case A has the greatest momentum change. The momentum change is dependent upon the velocity change; the object with the greatest velocity change has the greatest momentum change.

The impulse is greatest for Car A. The impulse equals the momentum change. If the momentum change is greatest for Car A, then so must be the impulse.The impulse is greatest for Car A. The force is related to the impulse (I=F*t). The bigger impulse for Car A is attributed to the greater force upon Car A. Recall that the rebound effect is characterized by larger forces; car A is the car which rebounds.

Collisions in one dimensionWe know, that the total linear momentum is conserved when two object collide, provided they constitute an isolated system. When object are atoms or subatomic particles, the total kinetic energy of the system is often conserved as well. So, the kinetic energy gained by one particle is lost by the other (in case of a two object system). In contrast, when macroscopic objects collide (like cars), the total kinetic energy after the collision is generally less than that before the collision. The kinetic energy is mainly lost in two ways: (1) it can be converted into heat because of friction(2) it is spent in creating permanent distortion or damage (deformation of the objects) - very hard objects suffer less permanent distortion than softer objects. Therefore softer object loose more kinetic energy.

Collisions are often classified according to wether the total kinetic energy changes during the collision:

1.) Elastic collision - total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision

2.) Inelastic collision - total kinetic energy of the system after the collision is not the same than before the collision; if object stick together after colliding, the collision is called completely

inelastic

The boxcars were an example for a completely inelastic collision.

ExampleA ball of mass m1 = 0.250 kg and velocity v01 = 5.00 m/s collides head-on with a ball of mass

m2 = 0.800 kg that is initially at rest (v02 = 0 m/s). No external forces act on the balls. Assume the collision is elastic, what are the velocities of the balls after the collision?

No external forces, therefore the total linear momentum of the two-ball system is conserved. (This is true no matter if the collision is elastic or not)

m1 vf1 + m2 vf2 = m1 v01 + 0

Since the collision is elastic, we also know that the total kinetic energy before and after the collision is the same.

1/2 (m1 vf12 + m2 vf22) = 1/2 m1 v012 + 0 --> m1 vf12 + m2 vf22 - m1 v012 = 0

We have two equations and two unknowns, which means we can solve the problem.

Use the first equation and re-arrange to find vf2 vf2 = (m1 v01 - m1 vf1)/m2 = (v01 - vf1)*(m1/m2)

Substitute this in second equation: m1 vf12 + m2 [(v01 - vf1)*(m1/m2)]2 - m1 v012

after simplification (your home work) we get: vf1 = {(m1-m2)/(m1+m2)} v01

vf2 = {(2m1)/(m1+m2)} v01

And putting in numbers: vf1 = -2.62 m/s and vf2 = 2.38 m/s

Collisions in two dimensionsA “head-on” collision is one-dimensional, because the velocities of the objects all point along a single line before and after contact. However in reality collisions occur in two of three dimensions. We will take a look at a two-dimensional case. Assume a system of two balls on a billiard table - we don’t

have to deal with external forces (each weight is balance by a normal force, the sum is zero) and we neglect friction. Momentum is a vector quantity and as for forces we can treat the x and y

components separately. This means the components (x and y) of the total momentum are conserved separately. So we can write:

x Component: pfx = p0x

y Component: pfy = p0y

YOUR TURN:

50°

35°

Θ

y

x

m1 = 0.150 kgv01 = 0.900 m/s

m2 = 0.260 kgv02 = 0.540 m/s

vf1 = ?vf1 = ?

vf2 = 0.700 m/s

Use momentum conservation to determine the magnitude and direction of the final velocity of ball 1

after collision.

The magnitude and direction of the final velocity of ball 1 can be obtained one the components vf1x and vf1y are known. The momentum conservation principle can be

used. (linear momentum for each component)

m1 = 0.150 kgv01 = 0.900 m/s

m2 = 0.260 kgv02 = 0.540 m/s

vf2 = 0.700 m/s

50 degrees to vertical incoming (ball 1) 35 degrees before and down from horizontal for ball 2 after

x Component: pfx = p0x

y Component: pfy = p0y

p = mv

put in numbers: v1fx = 0.63 m/s and vf1y = 0.12 m/s

v1fx

v1fy Θvf1 = sqrt(vf1x2 + vf1y2) = 0.64 m/s

direction given by Θ = tan-1 ((vfly/vf1x)) = 11°

Basic geometryLet’s look at plane geometry - 2 dimensions

In this document lowercase letters a, b, c, ... denote the sides of a polygon and uppercase letters A, B, C denote the vertices of a polygon. A polygon here is any geometric form with sides and vertices that can

be made out of linear functions (for example: triangle, square, rectangular, ...)Lowercase greek letters α, β, γ, ... are uses for angles.

Sum of angles: α + β + γ = 180° = πPerimeter: a + b + c

Area: 1/2*h*a (h height to vertex A, note the right angle 90° between h

and a)

Right triangle 90° at C (or π/2 radians) between h and a)

Area: 1/2*a*b (b=h)Perimeter: a + b + c

Pythagorean theorem: c2 = a2 + b2

Two triangles are called similar, when the corresponding angles are equal.

Rectangle with sides a and b and Square, where a = b

Perimeter: 2a + 2b = 2(a+b)Area: a*b

Diagonal: d = √a2+b2

Perimeter: 2a + 2a = 4aArea: a*a = a2

Diagonal: d = √a2+a2 = √2 a

Note the diagonal divides the rectangle into two triangles, for each one the area is given by 1/2 * a * b (where b = h). To get the area of the diagonal, we have to add the area of the two

triangles: 2*(1/2*a*b) = a*b. We have just shown that the area of a right angle triangle and a rectangle are connected! As a

consequence knowing one of them also means knowing the other. This is the case with many relations in physics as well - sometime it is not quite as obvious, but it is always worth thinking about it and finding them. It will deepen your understanding and safe

you from learning multiple formulas for the “same thing”.

Sum of angles in any polygone with four sides is 360° or 2π radians.

Trapezoid with sides a, b, c and d and height h. Sides a and c are parallel

Perimeter: a + b + c + dArea: 1/2(a+c)h

h

a2a1

Let’s show how this formula can be derived from starting off with trianglars and a rectangle:

rectangle: c*htriangle left: 1/2*h*a1

triangle right: 1/2*h*a2

Now add them all up: 1/2*h(2c+a1+a2)At the end we don’t want to use a1 and a2, so express them through a and c:

a1 + a2 = a - c and substitute above:

Area: 1/2*h(2c+a-c) = 1/2(a+c)h

The sum of angles in a polygon with n sides is given by (n-2)π radians or (n-2)180 degrees, so 180° (or π) in a triangle, 360° (or 2π) in square, rectangle, parallelogram, trapezoid,

540° (or 3π) in a pentagon, etc.

Now circles and ellipse:

knowing the radius of a circle determines circumference and

area as well.

diameter: d = 2 rcircumference: 2πr

area: πr2

Shaded area between two circles r1 < r2

area: πr22 - πr12 = π(r22 - r12)

Ellipse with semi-axis a and b and centre C.

area: πab

Assume three lines, a and b are parallel and c intersects both of them

α = β and α = α’ and β = β’