A horizontal, uniform tray of mass mt = 1.50 kg is aBached to a vertical ideal s

Question

A horizontal, uniform tray of mass mt = 1.50 kg is aBached to a vertical ideal spring of force
constant k = 200 N/m , and a metal ball of mass mb = 250 g is in the tray. The spring is below the
tray, so it can oscillate up-­--and-­--down. The tray is then pushed down a distance D = 12.5 cm
below its equilibrium point (call this equilibrium point y = 0 ) and released from rest.
(a) At what height h (relative to the equilibrium point) 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 and the ball leaving the
tray?
(c) How fast is the ball moving just as it leaves the tray?

Explanation / Answer

a) The mass will leave the plate when the force of gravity and spring force balance each other. So, k*(.125-h) = m*g .125 -h = 1.5*9.8/200 = 0.0735 h = .125 - 0.0735 = 0.0515 m = 5.15 cm b)Time period of the ball = 2*pi*sqrt(m/k) = 2*3.14*sqrt(1.5/200) = 0.543863 sec angular frequency = 2*pi/0.543863=11.55288 rad/sec So, the equation is x(t) = -.125*cos(11.55288*t) 0.0515 = -.125*cos(11.55288*t) cos(11.55288*t) = - 0.0515/.125 = -0.412 11.55288*t = 1.99544425 t = 1.99544425/11.55288 = 0.17272 seconds c)Velocity of the ball = v Equating energy .5*200*(.125^2) = .5*1.5*v^2 + 1.5*9.8*0.0515 v^2 = ((.5*200*(.125^2)) - (1.5*9.8*0.0515 ))/(.5*1.5) = 1.073 v = 1.036307 m/s

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