Interactive Lecture Demonstration 11 – Bouncing Clay and Basketball

In this demonstration, we will compare two collisions. In one collision, a piece of clay will be dropped onto a scale and will stick to the scale. In the second collision, a basketball (that has the same mass as the clay) will be dropped from the same height as the clay onto the scale, and will bounce off the scale. The scale is interfaced to a computer. The scale measures the amount of force exerted on it by whatever comes into contact with it. The computer will plot a graph of that force vs. time.

The purpose of this activity is to practice using Newton's Second Law in Impulse-Momentum form (sometimes called the Impulse-Momentum Theorem):

Impulse = F_net Dt = Dp.

The momentum of an object is defined to be its mass times its velocity: p = mv. Dt is the time during which the force acts on the object.

1. Suppose an object (the clay or basketball) of mass m = 625 grams is dropped from a height of 50 cm above the scale. What will be it momentum just before it hits the scale ? Do this as a DVAT problem.

How much time did the object take to fall the 50 cm ? Do not do this as a DVAT problem, but instead use Newton's Second Law in momentum form. You will need a force diagram for the object while it was falling.

2. Now consider the clay, which will stick to the scale after hitting it (situations in which two objects stick together after colliding are called completely inelastic collisions). What will be the change in momentum of the clay during the collision ? Choose your positive direction to be down. Draw the usual "before" and "after" pictures, and include a force diagram to show the forces acting during the collision.

3. Now consider the basketball, which will bounce off the probe after hitting it. If the basketball leaves the probe at the same speed with which it hit, we call this a perfectly elastic collision. In that case, how will the change in momentum of the basketball during the collision compare with the change in momentum that the clay had ? Prove your answer by calculating the basketball's change in momentum, assuming that it makes a perfectly elastic collision with the scale. Draw the usual "before" and "after" pictures, and include a force diagram to show the forces acting during the collision.

4. We hope you realized that the basketball's change in momentum would be twice as large as the change in momentum of the clay, since the clay was only stopped by the force of the probe, whereas the basketball had its momentum completely reversed. If the basketball's change in momentum is twice as large as the clay's, how must the impulse of the scale on the basketball compare with the impulse of the scale on the clay ?
5. We will now do the experiments already described. The computer will show a graph of force vs. time. The force represented will be the force the clay or basketball exerts on the scale (or equivalently, the force the scale exerts on the clay or basketball). Before you see the graphs, sketch below the general shape of the graph you expect to see. You don't need numbers on the axes, just the shape of the graph.

6. Now examine the actual graph produced when the clay collides with the scale. Is it similar to what you predicted ? According to the graph, about how long did the collision last ?

You also predicted what the change in momentum of the clay would be. According to the Impulse-Momentum Theorem, the clays change in momentum must be equal to the impulse of the scale's force on the clay. On the force vs. time graph, how can you see what the impulse of the force was ?

7. We hope you realize that the impulse is equal to the area on the graph between the force curve and the time axis. This can be seen from the definition of Impulse: I = FDt. When we find the area of a shape, we are simply multiplying its width (in this case, Dt) by its height (in this case, F). Since the curve has an irregular shape, we should imagine breaking up the graph into many vertical slices of roughly rectangular shape and adding up all the areas of those slices to get the total area under the graph. (For those who know calculus, this means we are integrating the function F vs t from the time the collision began until it ended).

The computer will calculate the area underneath the graph. Compare it with your prediction for the impulse given to the clay. Were you close ?

8. How will the force vs. time graph for the collision of the basketball with the scale compare with the graph for the clay ? Make a prediction, then look at the actual graph and see if you were right. Compare the impulse of the probe on the basketball (found in the same way we found the impulse on the clay) with your predicted impulse for that situation.