Question

Two billiard balls having masses of 0.2 kg and 0.15kg approach each other. The first ball having a velocity of 2m/s hit the second ball which is at rest. What will be the velocity of the first ball if the second ball travels at 1.5 m/s after collision?

Answer 1

Answer: The velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.

Explanation:

Given: $m_{1}$ = 0.2 kg,      $m_{2}$ = 0.15 kg

$v_{1}$ = 2 m/s,    $v_{2}$ = 0 m/s,      $v'_{1}$ = ?,          $v'_{2}$ = 1.5 m/s

Formula used is as follows.

$m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}$

where,

v = velocity before collision

v' = velocity after collision

Substitute the values into above formula as follows.

$m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}\\0.2 kg \times 2 m/s + 0.15 kg \times 0 m/s = 0.2 kg \times v'_{1} + 0.15 kg \times 1.5 m/s\\0.4 kg m/s + 0 = 0.2v'_{1} + 0.225 kg m/s\\0.2v'_{1} kg = 0.175 kg m/s\\v'_{1} = \frac{0.175 kg m/s}{0.2 kg}\\= 0.875 m/s$

Thus, we can conclude that the velocity of the first ball is 0.875 m/s if the second ball travels at 1.5 m/s after collision.