1. a) More on Elastic Collisions
Physics 211 Lecture 13
Today’s Concepts:
a) More on Elastic Collisions
b) Average Force during Collisions

2. In the last slides of the prelecture, it talked about KE in terms of different reference frames and in one of the reference frames the KE = 0 why?
Demo: throw
baton

3. More on Elastic Collisions
In CM frame, the speed of an object before an elastic collision is the same as the speed of the object after. m1 m2
v*1,i
v*2,i m2
v*1, f
v*2, f m1
|v*1,i - v*2,i|
= |-v*1, f + v*2, f |
= |-(v*1, f - v*2, f)|
= |v*1, f - v*2, f |
So the magnitude of the difference of the two velocities is the same before and after the collision.
But the difference of two vectors is the same in any reference frame.
Just Remember This
Rate of approach before an elastic collision is the same as the rate of separation afterward, in any reference frame!

4. ACT
Consider the two elastic collisions shown below. In 1, a golf ball moving with speed V hits a stationary bowling ball head on. In 2, a bowling ball moving with the same speed V hits a stationary golf ball.
In which case does the golf ball have the greater speed after the collision?
A) B) C) same
Demo: 283
2 ball pendula V V 1 2

5. ACT
A small ball is placed above a much bigger ball, and both are dropped together from a height H above the floor. Assume all collisions are elastic.
What height do the balls bounce back to? H
Before A
After B C

6. Explanation
For an elastic collision, the rate of approach before is the same as the rate of separation afterward: 3v v m v v m M v v M
Rate of approach = 2v
Rate of separation = 2v
Demo: ball drop

7. CheckPoint
A block slides to the right with speed V on a frictionless floor and collides with a bigger block which is initially at rest. After the collision the speed of both blocks is V/3 in opposite directions. Is the collision elastic?
A) Yes B) No V
V/3
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. 7

8. Comments But they don’t have the same speed. Tell me more.
But that’s not nearly enough.
But that’s not enough. No
But that’s not enough.
No, the speeds are not the same.
But that’s not enough.
! ! !
Good

9. CheckPoint Response Is the collision elastic? A) Yes B) No V V/3 M m
Before Collision
After Collision m M
A) Yes, the blocks do not stick together.
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.
B) No because the relative speed before the collision is V and after it's 2V/3 and since those two do not equal each other the collision is not elastic.
C) The masses of the blocks need to be known. 9 9

10. Forces during Collisionst ti tf t
So far we have ignored what happens during the actual collision, in other words, during time when the objects are actually touching. We have done this for the simple reason that the total momentum of a system of objects can not be changed by the forces that act between the objects during the collision. These forces can still be interesting, however, so its worth investigating them in a bit more detail before leaving this topic.
We start with the differential form of Newton's second law which relates the total force on an object to the rate of change of the objects momentum. We can rearrange this slightly tell us that the change in an objects momentum during a small time dt is just the total force acting on the object multiplied by the time interval. We can integrate this to find that the total change in the objects momentum during the time interval between t1 and t2 is the integral of the force acting on the object during this time.
We use this result to define the average force acting on the object during the collision and we see that this is just the change in momentum divided by the time interval. This should not be a surprise since it has the same form as the differential form of Newton's second law that we started with. We can use this new equation to relate the average force that acts over a finite time interval to the momentum change of the object during this time.
As a side note, integral of the force on some object during a collision is sometimes referred to as Impulse.
Fave
Impulse

11. ACT
Two blocks, B having twice the mass of A, are initially at rest on frictionless air tracks. You now apply the same constant force to both blocks for exactly one second. B F A F
air track
air track
The change in momentum of block B is:
A) Twice the change in momentum of block A
B) The same as the change in momentum of block A
C) Half the change in momentum of block A 11 11

12. ACT
Two boxes, one having twice the mass of the other, are initially at rest on a horizontal frictionless surface. A force F acts on the lighter box and a force 2F acts on the heavier box. Both forces act for exactly one second. Which box ends up with the bigger momentum?
A) Bigger box B) Smaller box C) same F 2F M 2M

13. CheckPoint
A constant force acts for a time Dt on a block that is initially at rest on a frictionless surface, resulting in a final velocity V. Suppose the experiment is repeated on a block with twice the mass using a force that’s half as big. For how long would the force have to act to result in the same final velocity? F
A) Four times as long.
B) Twice as long.
C) The same length.
D) Half as long.
E) A quarter as long.
Lets try it again ! 13 13

14. The experiment is repeated on a block with twice the mass using a force that’s half as big. For how long would the force have to act to result in the same final velocity? F
A) Four times as long.
B) Twice as long.
C) The same length.
A) FDt=mv. If the velocity is to remain constant between case 1 and case 2, then halving the force and doubling the mass would require 4 times as much time to balance the equation.
B) the momentum is the same, and the force is halved, so it would take twice the given time to get the same velocity
C) mv=Ft, so if you solve for t in each case using the given values then t will be the same. 14 14

15. The experiment is repeated on a block with twice the mass using a force that’s half as big. For how long would the force have to act to result in the same final velocity? F
A) Four times as long.
B) Twice as long.
C) The same length.
For the same final velocity we need to change the momentum by a factor of two since the mass is twice as large, p = m v.
If the momentum change is twice as large, the Impulse must be twice as large.
With only half the first force, an Impulse that is twice as large requires a time that is four times as long.
Impulse = (Force) (time ) = F Dt
A) Four times as long. 15 15

16. Comments But you guessed wrong.
We do have to be concerned with mass since p = mv.
WHAT is squared?
But you guessed wrong also.
Is it any clearer now?
You began correctly but the force is only half as much and we need to account for that.
Good; what does that mean about the time?
Good.
How much more?
What about the time?
But you guessed wrong also.
Good.

17. CheckPoint
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1 the ball bounces off a cement floor and in Case 2 the ball bounces off a piece of stretchy rubber. In which case is the average force acting on the ball during the collision bigger?
A) Case B) Case 2
Let’s try it again !
Case 1
Case 2

18. CheckPoint Responses Case 1 Case 2
In which case is the average force acting on the ball during the collision bigger?
A) Case B) Case 2
Case 1
Case 2
A) Because the same change in momentum happens in a shorter time.
B) It’s in contact with the ball for longer
Demo: sheet + raw eggs

19. The impulse must be the same but what about the force; Impulse = force x time .
Tell me more.
? ? ?
And you even guessed correctly!
But your answer was 2) and that’s wrong.

20. HW Problem with demo

21. pi DP = pf -pi pf -pi pf Another way to look at it: mvq
Another way to look at it: pf q
DP = pf -pi
-pi pf mv
|DPX | = 2mv cos(q)cosq
|DPY | = 0

22. |Fave | = |DP | /Dt = 2mv cos(q) /Dt

23. “Inelastic” here means “not totally elastic”.
This is not the totally inelastic collision where the ball sticks to the wall.
We tend to look at totally elastic collisions or totally inelastic collisions. This one is neither of our “usual” situations.

24. |DP| = |pf –pi| = m|vf –vi|
Another way to look at it: pf pf
-pi
DP = pf -pi
Fave = DP/Dt
|DP| = |pf –pi| = m|vf –vi|
Dt = DP/Fave

25. Demo – similar concept H time between bounces h = reading on scale vivf
Scale
Demo

26. Parting Thoughts/Questions
That always happens for a totally elastic collision.
Has today’s lecture helped?
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