A 1.90kg , horizontal, uniform tray is attached to a vertical ideal spring of fo

Question

A 1.90kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 190N/m and a 300g 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 16.9cm 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?

Explanation / Answer

a)

We have d = mg/k = 1.9*9.8/190 = 9.8 cm

Above A, the point is 9.8+16.9 cm = 26.7 cm

b)

omega = sqrt (k/m) = sqrt(190/1.9) = 10 rad/s

x = A cos omega*t

Omega*t = cos inverse (x/A) = cos inverse ( -9.8/16.9) = 2.18 rad

t = 2.18/10 = 0.218 seconds

c)

velocity, v = sqrt (k/m * (A^2-x^2)

v = sqrt ( 190/1.9 * (0.169^2-0.098^2))

v = 1.37 m/s

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