Question

A 1.30 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 200 N/mand a 275 g metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 13.3 cm below its equilibrium point (call this point A) and released from rest.

A) How high above point A will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.) B)How much time elapses between releasing the system at point A and the ball leaving the tray? c)How fast is the ball moving just as it leaves the tray?

Answer 1

a) The ball and the tray part ways when the velocity of the ball exceeds the velocity of the tray.

As long as the spring is pushing up on the tray, the acceleration experienced by the ball and the tray are the same. Therefore their velocities are the same.

However, when the spring starts pulling on the tray instead of pushing, the tray experiences an additional deceleration that the ball does not.

Therefore the two part company when the spring reaches its uncompressed length and the spring starts pulling on the tray.

This is not the equilibrium point because at the equilibrium point the spring is providing just enough force to balance the weight of the tray and the ball.

Given this weight and the spring constant, you can compute the deflection of the spring at the equilibrium point

b) the easiest approach is to take advantage of the equation for simple harmonic motion:
the equation is d(t) = X cos (wt + p), where:

d(t) is the displacement at time t

X is the magnitude of oscillation about the equilibrium point (in this case 13.3 cm)

w is the angular frequency, for a mass on a spring this is
sqrt(k/M) where k is the spring constant and M is the mass

p is the phase, which depends on the initial conditions.

In this case, you know d(t) from part a), so you can find t.

c) you can differentiate the equation in part b) to get the velocity, or you can use energy considerations.

Total energy is constant and equals the sum of kinetic, potential, and spring energy.

The spring is at its full length at the time of release so the spring energy is 0 at that point. You know it height above A so you can compute its potential energy, so if you know the total energy, you can compute the kinetic energy and then the velocity.

At the point where the tray is released from rest, the kinetic energy is 0; you know the distance to A so you can compute the potential energy; and you know the spring constant and the amount the spring has compressed so you know the spring energy. The sum of the three is the total energy.