Question

A particle of mass 5.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a horizontal frictionless surface with an amplitude of 2.00 m. A 7.00 kg object is dropped vertically on top of the 5.00 kg object as it passes through its equilibrium point. The two objects stick together.

(a) By how much does the amplitude of the vibrating system change as a result of collision?

(b) By how much does the period change?

(c) By how much does the energy change?

Answer #1

V at neutral point = Vo = Aω = 2x4.47 = 8.94 m/sec

x-direction momentum: This momentum will be the same before and
after dropping the 6 on the 4 →

6*0 + 5x4.47 = (5+7)*v → v = 1.86 m/s

The new KE = 0.5mv² = 0.5*12*(1.86)² = 20.8 Nm ; This will equal
the new spring energy:

20.8 Nm = .5kA² → A = √(41.6/100) = .6449 m, so the change in
amplitude is

a) 2-.6449 = 1.355 m

b) New period: T = 2π√(m/k) = 2π√(12/100) = 2.18 sec

....Old period: To = 2π√(5/100) = 1.41 sec

....Difference = 0.77 sec greater

c) KE1 = 0.5*5*(8.94)² = 199.81 Nm. From (above) KE2 = 20.8
Nm

....so, energy change is -179 Nm

d) The energy loss is released as heat between the (momentarily)
sliding surfaces of the 2 objects when they contact.

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