1. Physics 211 Lecture 12 Today’s Concepts: a) Elastic Collisions
b) Center-of-Mass Reference Frame
I'm still a little unsure about the difference between elastic and inelastic collisions. How can you tell the difference?
Hang on, Here we go!

2. Hour Exam 1 High Score of 100% Charles Huh Andrew Yu
Mean score of 82.1%
(Yes, the KEY got an A - )
Change in my office hours
2:30 to 3:30 on M, Tu, W
10:30 to 11:30 on Thursday

3. iClicker and “bonus points”
The Course Description mentions “bonus points” for correct answers with your iClickers.
That feature is not implemented this summer.
I am using the iClickers for feedback. Do well. But there are no add’l points for getting answers correct with your iClickers.

4. Summary / Paraphrase What are we doing today?
I need someone to summarize or restate or paraphrase what you learned in today’s Prelecture.
This is important and the Prelecture seems very well done – to me. But it is somewhat different.
What did YOU hear? What did you learn?
Why are we doing this?
Help me out. I need some feedback.

5. Center of Mass & Collisions so far:
The CM behaves just like a point particle
If then momentum is conserved
If you are in a reference frame moving along with the CM then the total momentum you measure is 0.

6. Inelastic / Elastic Collisions
Inelastic collisions are those in which the objects stick together. Think of balls of clay or gummy marshmellows. ( S p l a t ! ) KE will not be conserved.
Elastic collisions are like billiard balls or air hockey pucks or air track gliders with springs or rubber bands. ( B o i n g ! ) KE will be conserved.
Inelastic and Elastic collisions may also be called
Splat – Boing problems.

7. CheckPoint/ACT
A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest. Compare the initial and final kinetic energies of the system.
initial
final
A) Kinitial > Kfinal
B) Kinitial = Kfinal
C) Kinitial < Kfinal

8. CheckPoint Response A) Kinitial > Kfinal B) Kinitial = Kfinal
C) Kinitial < Kfinal
A) The boxes stick so it is not an elastic colllision
B) Because there are no external forces acting, the energy will be conserved.
C) There are two boxes moving instead of 1, so there is more energy.

9. Your Comments
Because this is not an elastic collision KE is not conserved. Due to this when the mass increases the KE increases. If KE increases, you need to have a source for that in mind.
So the collision is not elastic and some K lost during the collision
if the boxes stick that means there is an inelastic collision happening, thus there is some energy lost due to the internal forces of the boxes.
energy is conserved in an elastic collision True; but this is an INELASTIC collision.
since they are identical boxes, their kinetic energies will be the same Hmmm, That sounds like you’re answering a different question. How do the initial and final KE’s compare?
The momentum will we the same before and after the boxes collide and stick together, however the mass of the object moving doubles after the collision, making the initial kinetic energy smaller than the final kinetic energy. Good observation.
kinetic energy is lost with inelastic collisions
thats how it works
lost energy due to object material ? ? ?
Should be equal Tell me more.
The forces from the initial is the reason why the final force is moving I don’t understand.
The collision is inelastic so kinetic energy is not conserved. Good
Be careful That is often my response.
Inelastic collision ! ! !

10. Relationship between Momentum & Kinetic Energy
since
This is often a handy way to figure out the kinetic energy before and after a collision since p is conserved.
same p
initial
final

11. CheckPoint/ACT
A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is initially at rest. After the collision the green block is at rest and the red block is moving to the right. How does M compare to m? A) m > M B) M = m C) M > m D) Need more information
90% correct. Must conserve Energy and Momenturm
Before Collision
After Collision m M
We will now move on to consider a very interesting class of problems – that of collisions – and we will start with a very specific example to guide our discussion. Consider a box of mass m1 sliding along a horizontal frictionless floor in the positive x direction with an initial velocity vi. Suppose that there is a second box of mass m2 which is initially at rest, and that the first box collides with the second one. After the collision the boxes stick together and move with the same final velocity vf. Our job is to find a relationship between the initial and final velocity.
In this problem the system we are interested in is made up of the two boxes. Since the floor is horizontal and frictionless, the total external force on the system in the x direction is zero which means that total momentum of the two boxes in the x direction will be the same before and after the collision. At this point you might be wondering about the forces that will act between the boxes during the actual collision itself – wont these forces change the momentum of the system? [show middle picture] The answer is no – they will not – and the reason is simple: The forces between the boxes are not from an external source since both boxes are part of the system. In other words, the force by box 1 on box 2 will change the momentum of box 2 and the force by box 2 on box 1 will change the momentum of box 1, but the total momentum of the two boxes combined will not change since these forces are equal and opposite by Newton's third law. For this reason we never actually have to worry about what happens during the instant when the boxes collide – we can just focus on the total momentum before and after [fade out the middle picture].
In this example the initial momentum of the system in the x direction is due entirely to box 1 [show Pinitial = m1vi]. The final momentum of the system is due to both boxes [show Px,final=(m1+m2)vf]. Since the initial and final of the system has to be the same, we can solve for the final velocity in terms of the initial.
We see that the final velocity of the combined boxes is lower than the initial velocity of box 1 by itself. This makes because the mass of the moving stuff is bigger after the collision so the velocity after the collision has to be smaller in order to keep the momentum the same. 11

12. CheckPoint/ACT
A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is initially at rest. After the collision the green block is at rest and the red block is moving to the right. How does M compare to m? A) m > M B) M = m C) M > m D) Need more information
Before Collision
After Collision m M
We will now move on to consider a very interesting class of problems – that of collisions – and we will start with a very specific example to guide our discussion. Consider a box of mass m1 sliding along a horizontal frictionless floor in the positive x direction with an initial velocity vi. Suppose that there is a second box of mass m2 which is initially at rest, and that the first box collides with the second one. After the collision the boxes stick together and move with the same final velocity vf. Our job is to find a relationship between the initial and final velocity.
In this problem the system we are interested in is made up of the two boxes. Since the floor is horizontal and frictionless, the total external force on the system in the x direction is zero which means that total momentum of the two boxes in the x direction will be the same before and after the collision. At this point you might be wondering about the forces that will act between the boxes during the actual collision itself – wont these forces change the momentum of the system? [show middle picture] The answer is no – they will not – and the reason is simple: The forces between the boxes are not from an external source since both boxes are part of the system. In other words, the force by box 1 on box 2 will change the momentum of box 2 and the force by box 2 on box 1 will change the momentum of box 1, but the total momentum of the two boxes combined will not change since these forces are equal and opposite by Newton's third law. For this reason we never actually have to worry about what happens during the instant when the boxes collide – we can just focus on the total momentum before and after [fade out the middle picture].
In this example the initial momentum of the system in the x direction is due entirely to box 1 [show Pinitial = m1vi]. The final momentum of the system is due to both boxes [show Px,final=(m1+m2)vf]. Since the initial and final of the system has to be the same, we can solve for the final velocity in terms of the initial.
We see that the final velocity of the combined boxes is lower than the initial velocity of box 1 by itself. This makes because the mass of the moving stuff is bigger after the collision so the velocity after the collision has to be smaller in order to keep the momentum the same. 12

13. Newton’s Cradle
Demos: Newton's cradle,
Air track elastic collisions

14. In the CM reference frame, VCM = 0
Center-of-Mass Frame
In the CM reference frame, VCM = 0
In the CM reference frame, PTOT = 0
Films: cmel, cminel

15. Center of Mass Frame & Elastic Collisions
In the prelecture when finding the velocities of the objects in the center of mass, they did 5-2=3 and 0-2= -2 but then switched the signs to be -3 and 2, why?
The speed on an object is the same before and after an elastic collision as viewed in the CM frame: m1 m2
v*1,i
v*2,i m1 m2 m2
v*1, f
v*2, f m1

16. Example: 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?
+ = CM m1 m2
v*1, f
v*2, f
VCM + x
v*1,i
v*2,i = 0

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

18. Example
Now consider the collision viewed from a frame moving with the CM velocity VCM. m1 m2
v*1,i
v*2,i m1 m2 m2
v*1, f
v*2, f m1

19. Example
Step 2:
Calculate the initial velocities in the CM reference frame (all velocities are in the x direction): v
v = v* +VCM
v* = v - VCM
VCM v*
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

20. Example
Step 3:
Use the fact that the speed of each block is the same before and after the collision in the CM frame.
v*1, f = -v* 1,i v*2, f = -v*2,i m1 m2
v*1,i
v*2,i m1 m2 x
v*1, f = - v*1,i = -1.2m/s
v*2, f = - v*2,i =.3 m/s m2
V*1, f
V*2, f m1

21. Example
Calculate the final velocities back in the lab reference frame:
Step 4: v
v = v* +VCM
VCM v*
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!

22. CheckPoint/ACT
Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other.
The velocity of the center of mass of this system before the collision is
A) Toward the left B) Toward the right C) zero m 2m 2v v
Before Collision

23. CheckPoint Response 2m m This is the CM frame 2v v
The velocity of the center of mass of this system before the collision is
A) Toward the left B) Toward the right C) zero
This is the CM frame m 2m 2v v
Before Collision
A) because it has twice the mass
B) the net velocity is towards the right.
C) The total momentum of the system is zero. This means that the velocity of the center of mass must be zero.

24. CheckPoint/ACT
Two blocks on a horizontal frictionless track head toward each other as shown. One block has twice the mass and half the velocity of the other.
Suppose the blocks collide elastically. Picking the positive direction to the right, what is the velocity of the bigger block after the collision takes place?
A) v B) - v C) 2v D) - 2v E) zero m 2m 2v v +
Before Collision
This is the CM frame

25. CheckPoint Response 2m m V 2V This is the CM frame
Suppose the blocks collide elastically. Picking the positive direction to the right, what is the velocity of the bigger block after the collision takes place?
A) v B) - v C) 2v D) - 2v E) zero +
This is the CM frame
Before Collision m 2m 2V V
A) Since the collision is elastic, and the velocity of the center of mass is zero then the blocks simply travel backwards at their same speeds, (or opposite velocities).
B) the magnitude of the velocity stays the same but the direction changes
E) in order to conserve momentum, the blocks will both stop.

26. Do a problem where we have to go between the enter of mass reference frame and the labs reference frame

27. Do same steps for car 2
vcm is the same = final speed
Compare 1/2 mv 2 before and after for both cases

28. Special Case
Consider m1 coming in with velocity v1i
and colliding with mass m2 at rest.
What will be their final velocities?

29. Special Case

30. Final Results for this Special Case

31. 2D collisions Now conserve Px, Py (& K if elastic).
Scattering angle depends on impact parameter (glancing blow vs head-on)
Demo: pool table