Consider two cases in which the same ball is thrown against a wall with the same initial velocity. In case A, the ball sticks to the wall and does not bounce.In Case B, the ball bounces back with the same speed that it came with.

a.) In which of the two cases is the change in momentum the largest?

1. Case A
2. Case B
3. Same for both cases

b.) Assuming that the time during which the momentum change takes place is approximately the same for these two cases, in which case is the larger average force involved?

1. Case A
2. Case B
3. Same for both cases

Let initial velocity = vi

final velocity = vf (negative if opposite in direction)
Change in momentum, ΔM
= m(vf)-m(vi)=m(vf-vi)
a)
Case A: vf=0, ΔM=-m(vi)
Case B: vf=-vi, ΔM=-2m(vi)
The absolute value of the change is greater in case B.
b)
Force = ΔM/Δt
Will let you decide which case has a larger average force.

a.) 2

b.) 2

Correct for both.

a) To determine the change in momentum, we can use the formula:

Change in momentum (Δp) = Final momentum (p_f) - Initial momentum (p_i)

In both cases, the initial momentum of the ball is the same because it has the same initial velocity. However, in Case A, the ball sticks to the wall, so the final momentum is zero. In Case B, the ball bounces back with the same speed, so the final momentum is equal to the initial momentum but in the opposite direction.

Therefore:
- Case A: Δp = 0 - p_i = -p_i
- Case B: Δp = -p_i - p_i = -2p_i

Comparing the magnitudes of the change in momentum, we see that |Δp| (absolute value of Δp) is greater in Case B. So the correct answer is 2. Case B.

b) The average force can be estimated using the impulse-momentum theorem, which states:

Impulse (J) = Change in momentum (Δp) = Average force (F) × Time (Δt)

Since we are assuming that the time during which the momentum change takes place is approximately the same for both cases, we can conclude that Δt is the same.

In Case A, since the ball sticks to the wall, the change in momentum (Δp) happens in a very short amount of time. This means that the average force (F) involved will be very large to produce the change in momentum in such a short time.

In Case B, the ball bounces back, and the change in momentum (Δp) takes place over a longer time period. This implies that the average force (F) involved will be smaller compared to Case A since the momentum change occurs over a greater duration.

Therefore, the correct answer is 1. Case A, where the larger average force is involved.