Momentum and Collisions - Mission MC3 Detailed Help

A sad and a happy ball are dropped from a height of 1 meter. The sad ball collides with the ground and stops with very little bounce. The happy ball collides with the ground and bounces off with a significant enough speed to reach a height of 0.8 meters. Compared to the sad ball, the happy ball has ____ velocity change, ____ momentum change, and ____ impact force. Assume that the collision times are the same for each ball and that each ball has the same mass.


 
The momentum change of an object can be calculated from knowledge of the object's mass (m) and velocity change (∆v) using the formula:

Momentum Change = m • ∆v


 
Momentum Change - Impulse Theorem:
When a force is exerted upon an object in a collision, the object is said to have encountered an impulse. The impulse is simply the mathematical product of the force exerted on the object and the amount of time over which it was exerted. The impulse changes the object's momentum and is equal to the amount of momentum change.
 
Impulse = Momentum Change
or

F•t = m•∆v


 
The essential difference between the two balls is that the happy ball bounces and the sad ball hits the ground and stops. For two balls such as this with the same pre-collision speed, the ball that bounces is the ball that has the greatest velocity change. After all, if a sad ball changes its velocity from -10 m/s to 0 m/s, then there is a change in velocity of 10 m/s. But if a happy ball changes its velocity from -10 m/s to +6 m/s, then there is a change in velocity of 16 m/s. The ball with the greatest ∆v will also have the greatest m•∆v. And if impulse is equal to the momentum change, then the ball with the greatest m•∆v also encounters the greatest impulse.