A particle of mass 5.00 kg is attached to a spring with a force constant of 200

Question

A particle of mass 5.00 kg is attached to a spring with a force constant of 200 N/m. It is oscillating on a horizontal frictionless surface with an amplitude of 3.00 m. A 8.00 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) Does the amplitude of the vibrating system increase or decrease as a result of the collision?

increases? or decreases? or no change?

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

(b) How does the period change?

no change? or increases ? or decreases?

By how much does the period change?
? s

(c) How does the mechanical energy of the system change?

no change? or increases? or decreases?

By how much does the energy change?
? J
(d) Account for the change in energy.

Explanation / Answer

Vibration: = (k/m) = (200/5) = 6.324 rad/sec

V at neutral point = Vo = A = 3*6.324 = 18.9737 m/sec

x-direction momentum: This momentum will be the same before and after dropping the 8 on the 5
8*0 + 6.324*5 = (8+5)*v v = 31.6228/13 = 2.432 m/s

The new KE = .5mv² = .5*13*2.432² = 38.461 Nm ; This will equal the new spring energy:

38.462Nm = .5kA² A = (76.9224/200) = .6202 m, so the change in amplitude is
a) 3-.6202 = 2.38 m

b) New period: T = 2(m/k) = 2(13/200) = 1.602 sec
....Old period: To = 2(5/200) = 0.993 sec
....Difference = 0.608 sec greater

c) KE1 = 0.5*5*18.9737² = 900 Nm. From (above) KE2 = 38.461 Nm
....so, energy change is -861.54 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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