Question

A 11.0kg mass is traveling to the right with a speed of 2.40m/s on a smooth horizontal surface when it collides with and sticks to a second 11.0kg that is initially at rest but is attached to a light spring with force constant 100N/m. Find the amplitude of the subsequent oscillations?

Answer 1

Conservation of momentum, mv, means that m1 v1 = m2 v2, where m1 is the 11.0 kg mass and m2 is the combined mass, 22kg. We see that the two together start with half the velocity of the incoming mass, so v2 = 1.2 m/s.

Conservation of energy states that the sum of the kinetic energy (KE) and potential energy (PE) will be unchanged.

Initially, we have only kinetic energy, KE=1/2)(m2)(v2)^2 = (0.5)(22)(1.2)^2.
The maximum potential energy is when the velocity is zero, at the amplitude (A) of the motion, and kinetic energy has been converted to potential energy,
PE= (1/2) k A^2. Setting KE=PE gives A, for which I found A=0.56 m, but you should check my calculations. [The distance from one extreme to the other would be 2A, which I would not call the amplitude, but some might.]