Question

The billiard ball moves at V1 speed and hits another billiard ball at rest. After hitting the first ball continues to move in the same direction and towards as fast as 1/3 of the initial speed.

Calculate the velocity of the first ball before hitting, if the velocity of the second ball after hitting is 1Om / s. (stroke to be taken inelastic)

Answer 1

According to the momentum conservation law, the sum of the momentums of the balls before the collision $p$ is equal to the corresponding sum after the collision $p'$.

Since the second ball did not move before the collision, obtain:

$$p = mv_1$$

where $m$ is the mass of the first (and the second as well) ball, and $v_1$ is the required speed.

The momentum after the collision:

$$p' = m\dfrac{v_1}{3} + mv_2$$

where $v_1/3$ is the speed of the first ball after the collision, and $v_2 = 10m/s$ is the speed of the second ball. Equating $p$ and $p'$, obtain:

$$mv_1 = m\dfrac{v_1}{3} + mv_2\\ v_1 = \dfrac{v_1}{3} + v_2\\ \dfrac{2v_1}{3} = v_2\\ v_1 = \dfrac{3v_2}{2}\\ v_1 = \dfrac{3\cdot 10m/s}{2} = 15m/s$$

Answer. 15 m/s.