## Mastering Physics Solutions: Impulse and Change in Velocity

Impulse and Change in Velocity

Part A = You will need a graph that passes through the following points: (0,0), (2.9,28), (3.1,28), and (6,0). Read more to see the solution.
Part B = The change in velocity of the superball is greater than the change in velocity of the clay.
Part C = The force exerted by the scale on the superball is greater than the force exerted by the scale on the clay.

Solutions Below:

A glob of very soft clay is dropped from above onto a digital scale. The clay sticks to the scale on impact. A graph of the clay’s velocity vs. time, Vclay(t), is given, with the upward direction defined as positive.

The experiment is then repeated, but instead of using the clay glob, a superball with identical mass is dropped from the same height onto the scale.

Both the clay and the superball hit the scale 2.9s after they are dropped. Assume that the duration of the collision is the same in both cases and the force exerted by the scale on the clay and the force exerted by the scale on the superball are constant.

Part A

Sketch the graph of the superball’s velocity vs. time, vball(t) , from the instant it is dropped (t = 0s) until it bounces to its maximum height (t = 6s). Assume that the superball undergoes an elastic collision with the scale, and that the scale’s recoil velocity is negligible. The light colored graph already present in the answer window is vclay(t).

You will need a graph that passes through the following points: (0,0), (2.9,28), (3.1,28), and (6,0). It starts out like this:

So draw your graph over top, like this:

Basically, the first 2 points on your graph will be over the same points as on the original. But the 3rd point will bounce up to positive 28m/s (or whatever the velocity in your version of the problem was). The time it takes to get there (the x-coordinate) will be however long it took the velocity on the original graph to go from -28m/s to zero. In the problem shown here, that’s 0.2s. So the 3rd point on this graph is (3.1,28). The 3.1 is just 2.9 + 0.2 (the 2.9 is the x coordinate from the second point on the original graph). In other words, the x coordinate for your 3rd point will be the same as the original graph’s – only the y-coordinate changes. The 4th point on the graph will be (6,0), since velocity will be zero at 6 seconds.

So to recap:

You will need a graph that passes through the following points: (0,0), (2.9,28), (3.1,28), and (6,0).

Part B

Based on your graph, is the change in velocity of the superball during its collision with the scale greater than, less than, or equal to the change in velocity of the clay during its collision with the scale?

A. The change in velocity of the superball is greater than the change in velocity of the clay.
B. The change in velocity of the superball is less than the change in velocity of the clay.
C. The change in velocity of the superball is equal to the change in velocity of the clay.

Since the superball is going from -28m/s to +28m/s, this is greater than the clay just going from -28m/s to 0m/s:

A. The change in velocity of the superball is greater than the change in velocity of the clay.

Part C

Is the force exerted by the scale on the superball greater than, less than, or equal to the force exerted by the scale on the clay?

A. The force exerted by the scale on the superball is greater than the force exerted by the scale on the clay.
B. The force exerted by the scale on the superball is less than the force exerted by the scale on the clay.
C. The force exerted by the scale on the superball is equal to the force exerted by the scale on the clay.

The acceleration of the superball is greater (again, going from -28m/s to +28m/s instead of just -28m/s to 0m/s), and since F = ma, the force has to be greater as well.

A. The force exerted by the scale on the superball is greater than the force exerted by the scale on the clay.

Tagged with: