Question

A 500g air-track glider attached to a spring with spring constant 10 N/m is sitting at rest on a frictionless air track. A 250 g glider is pushed toward it from the far end of the track at a speed of 120 cm/s. It collideswith and sticks to the 500g glider. What are the amplitude and period of the subsequent oscillations?

Answer 1

To solve this problem it is necessary to apply the conservation equations of the moment for an inelastic impact or collision. In turn, it is necessary to apply the equations related to the conservation of potential energy and kinetic energy.

Mathematically this definition can be expressed as

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Where,

$v_{1,2} =$ Initial velocity of each object

$m_{1,2} =$Mass of each object

$v_f =$Final velocity

Our values are given as

$m_1 = 0.25kg\\m_2 = 0.5kg\\v_1 = 1.2m/s\\v_2 = 0m/s$

Replacing we can find the value of the final velocity, that is

$0.750v_f = 0.25*1.20+0.5*0$

$v_f = 0.4m/s$

From the definition of the equations of simple harmonic motion the potential energy of compression and equilibrium must be subject to

$KE_{compressed}+PE_{compressed} =KE_{equilibrium} +PE_{equilibrium}$

Since there is no kinetic energy due to the zero speed in compression, nor potential energy at the time of equilibrium at the end, we will have to

$0+\frac{1}{2}KA^2 = \frac{1}{2}(m_1+m_2)v_f^2+0$

Re-arrange to find A

$A = \sqrt{\frac{m_1+m_2}{k}v_f}$

$A = \sqrt{\frac{0.25+0.5}{10}(0.4)}$

$A = 0.11m$

Finally, the period can be calculated through the relationship between the spring constant and the total mass, that is,

$T = 2\pi \sqrt{\frac{m_1+m_2}{k}}$

$T = 2\pi \sqrt{\frac{0.5+0.25}{10}}$

$T = 1.7s$