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
A. Momentum = m*V.
Case A: Change = 0 - mV = -mV.
Case B: Change = -mV - mV = -2mV.
Answer: Case B.
B. Case A: a1 = -mV/T.
Case B: a2 = -2mV/T.
a2/a1 = (-2mV/T)/-mV/T = 2
a2 = 2a1.
Answer: Case B.
jmn
A.) In case B, the change in momentum is the largest. Why? Because in case A, the ball sticks to the wall and does not bounce, so its final momentum is zero. In case B, the ball bounces back with the same speed it came with, which means it has the opposite momentum. Therefore, the change in momentum is the largest in case B.
B.) In case A, the larger average force is involved. Why? Because when the ball sticks to the wall, it requires a greater force to bring it to rest. In case B, the ball bounces back, so it's not fully stopped by the wall, and thus requires a smaller average force. Therefore, the larger average force is involved in case A.
A.) In which of the two cases is the change in momentum the largest?
To determine the change in momentum in each case, we need to consider the final momentum subtracted by the initial momentum.
1. Case A: In this case, the ball sticks to the wall and does not bounce. When the ball sticks to the wall, it comes to a complete stop. Therefore, the final momentum is zero, and the change in momentum is equal to the initial momentum in the opposite direction.
2. Case B: In this case, the ball bounces back with the same speed that it came with. When the ball bounces back, its momentum changes direction. The final momentum is the negative of the initial momentum.
Comparing the two cases, we can see that the magnitude of the change in momentum in Case B is larger since it involves a reversal in direction. Therefore, the correct answer is:
2. Case B - The change in momentum is largest in Case B.
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?
The average force can be calculated by dividing the change in momentum by the time it takes for the momentum change to occur. In this case, we are assuming that the time taken is approximately the same for both cases.
1. Case A: In this case, the ball comes to a complete stop, resulting in a large change in momentum. However, since the ball sticks to the wall, the time for the momentum change to occur is short.
2. Case B: In this case, the ball bounces back, resulting in a smaller change in momentum compared to Case A. However, since the ball bounces back and has the same speed, the time for the momentum change to occur is longer.
Comparing the two cases, we can deduce that the larger average force is involved in Case A since the time for the momentum change is shorter but the change in momentum is still large.
Therefore, the correct answer is:
1. Case A - The larger average force is involved in Case A.