Two cases of a collision between a ball and box are occur. In case 1 a ball of m

Question

Two cases of a collision between a ball and box are occur. In case 1 a ball of mass m is thrown horizontally with speed v0 at a stationary box of mass M. The ball bounces off the box and after the collision the box is moving to the right with V1 and the ball is moving to the left with speed vf. In case 2 a ball of mass m is thrown horizontally with speed v0 at a stationary box of mass M. The ball sticks to the box and after the collision the box (with the ball stuck to it) is moving to the right with speed V2. In both cases the box slides without friction. Assume all motion is horizontal. Which of the following statements best describes V1 and V2? options: V1 = V2 V1 > V2 V1 < V2

Explanation / Answer

In case 1, the ball's change in momentum is:
m*((-vf) - v0)

or:
-m*(vf + v0)

The reason for the choice in signs, is that vf occurs in the direction to the left, and v0 is in a direction to the right. Defining right as +x, then left is -x. As you can see, the rate of change in momentum of the ball is based upon a change in velocity that is GREATER than the individual speeds comprising it.

In case 2, the ball's velocity is to the right in both cases. Subtracting one velocity from another, will give a change in velocity that is SMALLER than the initial velocity itself.

Change in momentum:
m*(v2 - v0)

It is a guarantee that v2 be less than v0, since what you've done is add more inertia to the same amount of momentum. Thus, this subtraction will result in a value less than m*v0 on its own.

By contrast, the case A change in momentum, -m*(vf + v0), will be greater in magnitude than the initial momentum of m*v0 on its own. The reason being, is that it ADDS up two speeds, instead of subtracts them.

Conclusion:
A direction reversal will have greater change in momentum, than a simple slow down.

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