Question

L10.7

A 0.250-kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce. What is the magnitude of the impulse J imparted to the clay by the floor during the impact? The force exerted by the floor on the clay is plotted as a function of time in the figure to the right. What must have been the maximum force F-max exerted by the floor on the clay?

Answer 1

The impulse is equal to the decrease of the momentum of the lump of clay. As the clay falls 1.45 meters, its velocity increases at the rate of 9.8 m/s each second. Let’s use the following equation to determine its velocity when it hits the floor.

vf^2 = vi^2 + 2 * a * d, vi = 0
vf^2 = 2 * 9.8 * 1.45
vf = √28.42
This is approximately 5.33 m/s.

Momentum = 0.250 * 5.33

This is approximately 1.33 kg * m/s. As the floor exerts an upward force on the lump of clay, its velocity will decrease to 0 m/s. My point is the impulse is equal to the momentum when it hit floor.

Impulse = Force * time
9.5 ms is the total time the force is applied. If we use this time, we will get the average force. The maximum force will be two times the average force.
t = 0.0095 seconds.

F * 0.0095 = 0.250 * 5.33
F = 0.250 * 5.33/0.0095 = 140.3
This is approximately140 N.
Maximum force = 280 N