Question

A 675 g air-track glider attached to a spring with spring constant 9.50 N/m is sitting at rest on a frictionless air track. A 600 g glider is pushed toward it from the far end of the track at a speed of 160 cm/s . It collides with and sticks to the 675 g glider. What are the amplitude and period of the subsequent oscillations?

Answer 1

To solve this problem we will proceed to find the period given by the spring, said period must be calculated considering the totality of the masses. At the same time, we will also calculate the final velocity resulting from the impact. With speed and period it will be possible to finally find the amplitude.

The period is defined as,

`T = 2\pi \sqrt{(m_1+m_2)/(k)}`

Here,

m = mass

k = Spring constant

`T = 2\pi \sqrt{((675g+600g)(1kg)/(1000g))/(9.5N/m)}`

`T = 2.301s`

According to conservation of momentum

Initital momentum = Final momentum

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

`(675g)(0)+(600g)((1kg)/(1000g))(160cm/s)((1m)/(100cm)) = (675g+600g)((1kg)/(1000g))v_f`

`v_f = 0.7529m/s`

Now the amplitude of oscillation is

`v = (2\pi A)/(T)`

Here,

v = Velocity

A = Amplitud

T= Period

`A = (vT)/(2\pi)`

`A = ((0.7529)( 2.301))/(2\pi)`

`A = 0.275m`

Therefore the ampliude is 0.275m and the Period is 2.301 of the subsequent oscillations.