Question

A particle of mass 5.00 kg is attached to a spring with a force constant of 130 N/m. It is oscillating on a frictionless, horizontal surface with an amplitude of 3.00 m. A 9.00-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) What is the new amplitude of the vibrating system after the collision?m

(b) By what factor has the period of the system changed?

(c) By how much does the energy of the system change as a result of the collision?J

(d) Account for the change in energy.

Answer 1

(a)

at equilibrium, the object will have maximum speed

v = wA

v = sqrt ( k /m) * A

v = sqrt ( 130 / 5) * 3

v = 15.3 m/s

now, when another object is dropped and stick, this is an inelastic collision,

so,

5 * 15.23 = ( 5 + 9)v

v = 5.46 m/s

so,

new amplitude

1/2kA² = 1/2 (5 + 9 ) v²

A = 1.792 m

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(b)

old period

T = 2$\pi$ sqrt ( m / k)

T = 2$\pi$ sqrt ( 5 / 130)

T = 1.232 sec

New period

T = 2$\pi$ sqrt ( 14 / 130)

T = 2.062 sec

so,

period has changed by a factor of 2.062 / 1.232 = 1.67

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(c) old energy = 1/2 * 5 * 15.3² = 585.22 J

new energy = 1/2 * 14 * 5.46² = 208.7 J

so,

energy has changed by 376.5 J

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(d)

The lost mechanical energy has turned into internal energy in the completely inelastic collision