A particle of mass 3.00 kg is attached to a spring with a force constant of 150

Question

A particle of mass 3.00 kg is attached to a spring with a force constant of 150 N/m. It is oscillating on a frictionless, horizontal surface with an amplitude of 5.00 m. A 7.00-kg object is dropped vertically on top of the 3.00-kg object as it passes through its equilibrium point. The two objects stick together.
(a) What is the new amplitude of the vibrating system after the collision?
-----m

(b) By what factor has the period of the system changed?
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(c) By how much does the energy of the system change as a result of the collision?
-------J

Explanation / Answer

A = xm = 5m

so kx2 /2 = mv2 /2  

150 x 52 /2 = 3 x v2 /2

v = 35.36 m/s    at equilibrium point
new velocity of 3 kg + 7kg mass is v'.

so 3 X 35.36 = (3 + 7)v'

v' = 10.61 m/s

a) 150 X x'2 / 2 = (10) x 10.612 /2

x' = 2.74 m

A' = 2.74 m

b) T = 2m/k

so T" / T = 210/k   / 23/k

    = (10/3) = 1.83

c) E = [ 10 X 10.612 /2] - [3 X 35.362 /2 ] = - 1312.634 J

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