1. SPH4U: Lecture 15 Today’s Agenda
Elastic collisions in one dimension
Analytic solution (this is algebra, not rocket science!)
Center of mass reference frame
Colliding carts problem
Two dimensional collision problems (scattering)
Solving elastic collision problems using COM and inertial reference frame transformations
Some interesting properties of elastic collisions
Center of mass energy and energy of relative motion

2. Momentum Conservation: Review
The concept of momentum conservation is one of the most fundamental principles in physics.
This is a component (vector) equation.
We can apply it to any direction in which there is no external force applied.
You will see that we often have momentum conservation even when kinetic energy is not conserved.

3. Comment on Energy Conservation
We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved.
Mechanical Energy is lost: (remember what this is??)
Heat (bomb)
Bending of metal (crashing cars)
Kinetic energy is not conserved since dissipative work is done during an inelastic collision! (here, KE equals mechanical energy)
Total momentum, PT, along a certain direction is conserved when there are no external forces acting in this direction.
F = ma = dPT/dt says this has to be true!! (Newton’s Laws)
In general, momentum conservation is easier to satisfy than mechanical energy conservation.
Remember: in the absence of external forces, total energy (including heat…) of a system is always conserved even when mechanical energy is not conserved.
How much do two objects that inelastically collide heat up?

4. Lecture 15, Act 1 Collisions
A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest.
What is the ratio of initial to final kinetic energy of the system?
(a) 1
(b)
(c) 2

5. Lecture 15, Act 1 Solution
No external forces in the x direction, so PX is constant. v m m m m
v / 2 x

6. Lecture 15, Act 1 Solution
Compute kinetic energies: v m m m m
v / 2

7. Lecture 15, Act 1 Another solution
We can write
P is the same before and after the collision.
The mass of the moving object has doubled, hence the kinetic energy must be half. m m m m

8. Lecture 15, Act 1 Another Question:
Is it possible for two blocks to collide inelastically in such a way that the kinetic energy after the collision is zero?

9. Lecture 15, Act 1 Another Question
Is it possible for two blocks to collide inelastically in such a way that the kinetic energy after the collision is zero?
YES: If the CM is not moving! CM CM

10. Elastic Collisions
Elastic means that kinetic energy is conserved as well as momentum.
This gives us more constraints
We can solve more complicated problems!!
Billiards (2-D collision)
The colliding objects have separate motions after the collision as well as before.
all 3D collision problems can be solved in 2 dimensions by using center of mass inertial reference frame
Start with a simpler 1-D problem
Final
Initial

11. Elastic Collision in 1-D what has to happen
Why is this elastic? m2 m1
initial
v1,i
v2,i
Maybe, it depends… x
v1,f
v2,f
final m1 m2
Kinetic energy  potential energy  kinetic energy
The spring is conservative

12. Elastic Collision in 1-D the spring is conservative
v1,i
v2,i
before m1 m2
Conserve PX: (no external forces!)
m1v1,i + m2v2,i = m1v1,f + m2v2,f x
v1,f
v2,f
after
Conserve Kinetic Energy: (it’s elastic!)
1/2 m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f
Suppose we know v1,i and v2,i
We need to solve for v1,f and v2,f
Should be no problem equations & 2 unknowns!

13. Elastic Collision in 1-D
However, solving this can sometimes get a little bit tedious since it involves a quadratic equation!!
A simpler approach is to introduce the Center of Mass Reference Frame
First, describe the solution to the problem using algebra.
Useful analysis and useful formulae
m1v1,i + m2v2,i = m1v1,f + m2v2,f
1/2 m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f
momentum
energy

14. Elastic Collision in 1-D special case: equal masses
If the masses of the two objects are equal the algebra is not too bad. Let’s see what we get…
Divide through by m = m1 = m2
m1v1,i + m2v2,i = m1v1,f + m2v2,f
1/2 m1v21,i + 1/2 m2v22,i = 1/2 m1v21,f + 1/2 m2v22,f
momentum
energy
v1,i + v2,i = v1,f + v2,f
v21,i + v22,i = v21,f + v22,f

15. Elastic Collision in 1-D special case: equal masses
Now just rearrange equations to bring v1,i and v1,f to left hand side, and v2,i and v2,f to rhs
Divide through energy equation by momentum equation which gives
Particles just trade velocities in 1-D elastic collision of equal mass objects (let equation talk to you…)
v1,i - v1,f = v2,f - v2,i
momentum
energy
v21,i - v21,f = v22,f - v22,i
(v1,i - v1,f )(v1,i + v1,f ) = (v2,f - v2,i )(v2,f + v2,i )
v1,i + v1,f = v2,f + v2,i
v2,f = v1,i
v1,f = v2,i
v1,i - v1,f = v2,f - v2,i

16. Elastic Collision in 1-D general case: unequal masses
Conserve linear momentum and mechanical energy, but now the masses are different:
Divide through energy equation by momentum equation which gives
Now solving these two linear equations is only a bit more complicated
m1(v1,i - v1,f ) = m2(v2,f - v2,I )
momentum
energy
m1(v21,i - v21,f ) = m2(v22,f - v22,I )
m1(v1,i - v1,f )(v1,i + v1,f ) =m2 (v2,f - v2,i )(v2,f + v2,i )
v1,i + v1,f = v2,f + v2,i
m1(v1,i - v1,f ) = m2(v2,f - v2,I )

17. Elastic Collision in 1-D general case: unequal masses
Algebra just gave us the following equations based on conservation of momentum and mechanical energy:
Now just solve for final velocities, v1,f and v2,f in terms of v1,i and v2,i
v1,i + v1,f = v2,f + v2,i
m1(v1,i - v1,f ) = m2(v2,f - v2,I )
v2,f = v1,i
v1,f = v2,i
When m1 = m2

18. Another way to solve elastic collision problems: CM Reference Frame
We have shown that the total momentum of a system of particles is the velocity of the CM times the total mass:
PNET = MVCM.
We have also discussed reference frames that are related by a constant velocity vector (i.e.they’re in relative motion).
Now consider putting yourself in a reference frame in which the CM is at rest. We call this the CM reference frame.
In the CM reference frame, VCM = 0 (by definition) and therefore PNET = 0.
This is a cool mathematical tool that makes the algebra solving this much simpler
(it doesn’t change the physical situation)

19. Lecture 15, Act 2 Force and Momentum
Two astronauts, one with more mass than the other, are floating at the same location holding on to a massive block. The astronauts are at rest with respect to a space station.
The two astronauts throw the massive block away from them at the same speed.
Which astronaut is moving faster with respect to the space station?
(a) More massive (b) less massive (c) same

20. Lecture 15, Act 2 Conceptual Solution
The external force in the x direction is zero (frictionless):
The CM of the systems can’t move!
Aha! this is the key!! X X X X CM x CM

21. Lecture 15, Act 2 Conceptual Solution
The external force in the x direction is zero (frictionless):
The CM of the systems can’t move!
The blocks will reach the end of their destination at the same time, but lighter astronaut will be further from the CM at this time.
His motion doesn’t count as much, since he is less massive
The lighter astronaut moves faster with respect to the space station! X X X X CM CM

22. Example 1: Using CM Reference Frame
A glider of mass m1 = 0.2 kg slides on a frictionless track with initial velocity v1,i = 1.5 m/s. It hits a stationary glider of mass m2 = 0.8 kg. A spring attached to the first glider compresses and relaxes during the collision, but there is no friction (i.e. energy is conserved). What are the final velocities? m1 m2
v1,i
v2,i = 0
VCM
+ = CM m1 m2 x m1
v1,f
v2,f m2

23. Step 1 Example 1... (for v2,i = 0 only) Four step procedure
First figure out the velocity of the CM, VCM.
VCM = (m1v1,i + m2v2,i), but v2,i = 0 in this case so
VCM = v1,i
So VCM = 1/5 (1.5 m/s) = 0.3 m/s
Step 1
(for v2,i = 0 only)

24. Don’t forget that each objects’ v will have a plus or minus sign depending upon the direction
Example 1...
If the velocity of the CM in the “lab” reference frame is VCM, and the velocity of some particle in the “lab” reference frame is v, then the velocity of the particle in the CM reference frame is v* where:
v* = v - VCM (where v*, v, VCM are vectors)
This is the “lab” frame velocity
VCM v v*
This is the CM frame velocity
If you were traveling along with the CM, you would see the velocity of the mass to be less than in the lab frame in this case

25. Step 2 Example 1... v*1,i = v1,i - VCM = 1.5 m/s - 0.3 m/s = 1.2 m/s
Calculate the initial velocities in the CM reference frame (all velocities are in the x direction):
Step 2
v*1,i = v1,i - VCM = 1.5 m/s m/s = 1.2 m/s
v*2,i = v2,i - VCM = 0 m/s m/s = -0.3 m/s
v*1,i = 1.2 m/s
v*2,i = -0.3 m/s

26. Example 1 continued... Movie
Now consider the collision viewed from a frame moving with the CM velocity VCM. ( jargon: “in the CM frame”)
v*1,i m2
v*2,i m1 m1 m2 x m2
v*2,f
v*1,f m1

27. Energy in Elastic Collisions:
Use energy conservation to relate initial and final velocities.
The total kinetic energy in the CM frame before and after the collision is the same, it’s elastic!! (look how we write this…)
But the total momentum is zero, both initial and final:
So:
Likewise for final v’s
(and the same for particle 2)
Therefore, in 1-D: v*1,f = -v* 1,i v*2,f = -v*2,i

28. Step 3 Example 1... Calculate the final velocities in the CM frame:
v*1,f = -v* 1,i v*2,f = -v*2,i
v*1,i m2
v*2,i m1 m1 m2 x
v*1,f = - v*1,i = -1.2m/s
v*2,f = - v*2,i =.3 m/s m1 m2

29. Step 4 Example 1... v* = v - VCM v*f = -v*,i v = v* + VCM
Now we can calculate the final velocities in the lab reference frame, using:
v = v* + VCM
v1,f = v*1,f + VCM = -1.2 m/s m/s = -0.9 m/s
v2,f = v*2,f + VCM = 0.3 m/s m/s = 0.6 m/s
v1,f = -0.9 m/s
v2,f = 0.6 m/s
Four easy steps! No need to solve a quadratic equation!!
Especially important in 2D

30. Lecture 15, Act 3 Moving Between Reference Frames
Two identical cars approach each other on a straight road. The red car has a velocity of 40 mi/hr to the left and the green car has a velocity of 80 mi/hr to the right.
What are the velocities of the cars in the CM reference frame?
(a) VRED = - 20 mi/hr (b) VRED = - 20 mi/hr (c) VRED = - 60 mi/hr
VGREEN = + 20 mi/hr VGREEN = +100 mi/hr VGREEN = + 60 mi/hr

31. Lecture 15, Act 3 Moving Between Reference Frames
The velocity of the CM is:
= 20 mi / hr
20mi/hr CM
So VGREEN,CM = 80 mi/hr - 20 mi/hr = 60 mi/hr
So VRED,CM = - 40 mi/hr - 20 mi/hr = - 60 mi/hr
The CM velocities are equal and opposite since PNET = 0 !!
80mi/hr
- 40mi/hr x

32. Summary: Using CM Reference Frame
(m1v1,i + m2v2,i)
: Determine velocity of CM
: Calculate initial velocities in CM reference frame
: Determine final velocities in CM reference frame
: Calculate final velocities in lab reference frame
Step 1
VCM =
Step 2
v* = v - VCM
Step 3
v*f = -v*i
Step 4
v = v* + VCM

33. Lecture 15, Act 3 Aside 20mi/hr CM
As a safety innovation, Volvo designs a car with a spring attached to the front so that a head on collision will be elastic. If the two cars have this safety innovation, what will their final velocities in the lab reference frame be after they collide?
80mi/hr
- 40mi/hr x

34. Lecture 15, Act 3 Aside Solution
v*GREEN,i = 60 mi/hr
v*RED,i = -60 mi/hr
v*GREEN,f = -v* GREEN,i v*RED,f = -v*RED,i
v*GREEN,f = -60 mi/hr v*RED,f = 60 mi/hr
v´ = v* + VCM
v´GREEN,f = -60 mi/hr + 20 mi/hr = - 40 mi/hr
v´RED,f = 60 mi/hr + 20 mi/hr = 80 mi/hr

35. Interesting Fact v*1,i v*2,i v*1,f = -v*1,i v*2,f = -v*2,i
We just showed that in the CM reference frame the speed of an object is the same before and after the collision, although the direction changes.
The relative speed of the blocks is therefore equal and opposite before and after the collision. (v*1,i - v*2,i) = - (v*1,f - v*2,f)
But since the measurement of a difference of speeds does not depend on the reference frame, we can say that the relative speed of the blocks is therefore equal and opposite before and after the collision, in any reference frame.
Rate of approach = rate of recession
v*1,f = -v*1,i
v*2,f = -v*2,i
This is really cool and useful too!

36. Example
A ball with mass 2.5x10-2 kg moving at 2.3 m/s collides with a stationary 2.0x10-2 kg ball. If the collision is elastic, determine the velocity of each ball after the collision.

37. Solution
A ball with mass 2.5x10-2 kg moving at 2.3 m/s collides with a stationary 2.0x10-2 kg ball. If the collision is elastic, determine the velocity of each ball after the collision.
K.E.
Momentum
Given
Re-arranging the Momentum equation with v2i=0, we obtain:

38. Substitute this into the K.E. equation:
Substitute in:
Note: Both balls end up moving in the same direction.

39. Solution by CM
A ball with mass 2.5x10-2 kg moving at 2.3 m/s collides with a stationary 2.0x10-2 kg ball. If the collision is elastic, determine the velocity of each ball after the collision.
Given

40. Example 5
A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m. The collision is elastic.
a. When the pendulum reaches the vertical position, calculate the speed of the bob just before it strikes the box.
b. Calculate the speed of the bob and the box just after they collide elastically.
c. Determine the impulse acting on the box during the collision.
d. Determine how far away from the bottom edge of the table, Δx, the box will the box strike the floor.

41. Example 5
A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m.
a. When the pendulum reaches the vertical position, calculate the speed of the bob just before it strikes the box.

42. Example 5
A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m.
b. Calculate the speed of the bob and the box just after they collide elastically.

43. Example 5
A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m.
c. Determine the impulse acting on the box during the collision.

44. Example 5
A pendulum of length L = 1.0 meter and bob with mass m = 1.0 kg is released from rest at an angle θ = 30° from the vertical. When the pendulum reaches the vertical position, the bob strikes a mass M = 3.0 kg that is resting on a frictionless table that has a height h = 0.85m.
d. Determine how far away from the bottom edge of the table, Δx, the box will the box strike the floor.

45. Example 5
At the location where the box would have struck the floor, now a small cart of mass M = 5.0 kg and negligible height is placed. The box lands in the cart and sticks to the cart in a perfectly inelastic collision.
Calculate the horizontal velocity of the cart just after the box lands in it.

46. QUESTION
The planet Saturn is moving in the negative x-direction at its orbital speed (with respect to the Sun) of 9.6 km/s. The mass of Saturn is 5.69x1026 kg. A kg spacecraft approaches Saturn, moving initially in the +x-direction at 10.4 km/s. The gravitational attraction of Saturn (a conservative force) causes the spacecraft to swing around it and head of into the opposite direction.
Determine the speed of the spacecraft after it is far enough away to be free of Saturn’s gravitational pull.

47. Solution
The planet Saturn is moving in the negative x-direction at its orbital speed (with respect to the Sun) of 9.6 km/s. The mass of Saturn is 5.69x1026 kg. A kg spacecraft approaches Saturn, moving initially in the +x-direction at 10.4 km/s. The gravitational attraction of Saturn (a conservative force) causes the spacecraft to swing around it and head of into the opposite direction.
Determine the speed of the spacecraft after it is far enough away to be free of Saturn’s gravitational pull.
Here the “collision” is not an impact but rather a gravitational interaction. So we can treat it as a one-dimensional elastic collision. And thus we will solve it accordingly.

48. Solution
The reason the spacecraft acquires so much additional speed is that Saturn is moving in its orbit. It does slow down, but its mass is so much greater than the spacecraft, it is not noticeable.