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.

Step-by-step 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 * 2 m/s + 0.15 kg * 0 m/s = 0.2 kg * v'_(1) + 0.15 kg * 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) = (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.