Question

A cue ball with mass 170g hits a stationary number 8 ball, which has mass 160g. The cue ball

and the 8 ball move with a speed of 0.2m/s and 0.3m/s respectively after the collision. What
was the speed of the cue ball before the collision?

Answer 1

0.49 m/s

Step-by-step explanation:

The law of conservation of linear momentum states that the sum of momentum in a system before and after collision are same. Momentum is a product of mass and velocity of an object hence in this case

`m_cu_c+m_8u_8=m_cv_u+m_8v_8`

Where m represent mass, u and v represent the initial and final velocities respectively, subscripts c and 8 represent cue ball and number 8 ball respectively.

Since number 8 ball is initially at rest, its initial velocity is zero. Replacing mass of cue ball with 170 g while mass of number 8 ball with 160g, then taking final velocity of cue ball as 0.2 m/s and final velocity of 8 ball as 0.3 m/s then we get

`170u_c+160*0=170*0.2+170*0.3\\u_c\approx 0.49 m/s`