Answer to Question #291093 in Physics for King

Question

A snooker ball X of mass 0.3kg moving with a velocity 5m/s hit a stationary ball of mass 0.4kg.Y moves off with a velocity of 2m/s at 30degree to the initial direction of X.Find the velocity v of X and its direction after hitting Y.

Answer 1

The momentum is conserved along x- and y-axes.

Given:

$\theta=30°\\ v_x=5\text{ m/s}\\ u_y=2\text{ m/s}\\ \phi=?\\ u_x=?$

Solution

$$Ox: m_xv_x=m_yu_y\cos\theta+m_xu_x\cos\phi,\\ Oy: 0=m_yu_y\sin\theta+m_xu_x\sin\phi.\\\space\\ m_xu_x\cos\phi=m_xv_x-m_yu_y\cos\theta,\\ m_xu_x\sin\phi=-m_yu_y\sin\theta.\\\space\\ \tan\phi=\dfrac{m_yu_y\sin\theta}{m_yu_y\cos\theta-m_xv_x}=-0.5,\\\space\\ \phi=-26.6°,\\ u_x=2.98\text{ m/s}.$$