Question

A 1.72 kg ball rolls on a frictionless horizontal surface at 4.19 m/s and collides with a stationary block of mass 1.23 kg which is attached to horizontal spring whose force constant is 1695 N/m. Right after impact, the ball bounces on the opposite direction at 1.59 m/s. To what distance will the spring be compressed?

Answer 1

Given:

$m_1=1.72\:{\rm kg},\; m_2=1.23\:{\rm kg}$

$v_{1x}=4.19\:{\rm m/s},\; v_{2x}=0$

$v_{1x}'=-1.59\:{\rm m/s},\; k=1695\:\rm N/m$

The law of conservation of momentum says

$$m_1v_{1x}+m_2v_{2x}=m_1v_{1x}'+m_2v_{2x}'$$

Hence, the velocity of the block after collision with ball

$$v_{2x}'=\frac{m_1v_{1x}+m_2v_{2x}-m_1v_{1x}'}{m_2}$$

$$v_{2x}'=\frac{1.72*1.23-1.72*(-1.59)}{4.19}=1.16\:\rm m/s$$

The law of conservation of energy says

$$\frac{m_2v_{2x}'^2}{2}=\frac{k(\Delta x)^2}{2}$$

Thus, the compression of the spring

$$\Delta x=\sqrt{\frac{m_2}{k}}v_{2x}=\sqrt{\frac{4.19}{1695}}*1.16=5.76*10^{-2}\:\rm m$$