Question

A ball of clay with a mass of 4.0 kg, starting from rest, falls from a height of 0.20 m above a block with mass 1.0 kg that rests on top of a spring with spring constant 90 N/m. The block then sticks to the block and the system undergoes simple harmonic motion. Please submit your answers to the following questions on a separate sheet of paper (or multiple sheets of paper).

i. Write the formulas for the position, velocity, and acceleration as functions of time, with t = 0 corresponding to the moment that the ball hits the spring. Sketch graphs for each quantity over the first three cycles of the oscillations. Pay careful attention to where in the cycle the system’s oscillation starts.

ii. Write formulas for the kinetic energy and the total mechanical energy of the system as functions of time. Sketch graphs for each quantity over the first three cycles of the oscillations.

Answer 1

i] Velocity of the ball of clay just before hitting the block is:

v = [2gh]^1/2 = [2 x 9.8 x 0.2]^1/2 = 1.98 m/s

now it is given that the ball sticks to the block. So, the collision is inelastic and so its momentum will be conserved before and after.

mv + 0 = [M+m]V

=> V = 1.584 m/s

this is the velocity of the system.

k = 90 N/m

so, the frequency of oscillation will be:

also, maximum kinetic energy = maximum elastic potential energy

(1/2)(m+M)V² = (1/2)kA²

A = 0.373 m

this is the amplitude of the oscillation.

therefore, the position of the system as a function of time will be:

where x=A is the position at t = 0

velocity will be:

and acceleration is:

ii]

The kinetic energy of the system = (1/2)[m+M]V²

and total mechanical energy will be constant for the system after the ball hits the block and its value will be:

T = 6.273 J.