# A ball with a mass of  5 kg is rolling at 18 m/s and elastically collides with a resting ball with a mass of 3 kg. What are the post-collision velocities of the balls?

May 9, 2018

Given

Mass of 1st ball ${m}_{1} = 5$ kg

Mass of 2nd ball ${m}_{2} = 3$ kg

Initial velocity of 1st ball ${u}_{1} = 18 m \text{/} s$

Initial velocity of 2nd ball ${u}_{2} = 0 m \text{/} s$

Let post collision velocity of 1st ball be ${v}_{1}$ and that of 2nd ball be ${v}_{2}$

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

$5 \cdot 18 + 3 \cdot 0 = 5 {v}_{1} + 3 {v}_{2}$

$\implies 5 {v}_{1} + 3 {v}_{2} = 90. \ldots . \left[1\right]$

Again the collision being elastic one the relative velocity of one ball with respect to the other is reversed by the collision and we get

${v}_{2} - {v}_{1} = {u}_{1} - {u}_{2}$

$\implies {v}_{2} - {v}_{1} = 18 - 0$

$\implies {v}_{2} - {v}_{1} = 18. \ldots \ldots \left[2\right]$

By  and  we get

$5 {v}_{1} + 3 \left(18 + {v}_{1}\right) = 90$

$\implies 8 {v}_{1} = 90 - 54 = 36$

$\implies {v}_{1} = \frac{36}{8} = 4.5$ m/s

${v}_{2} = 18 + {v}_{1} = 18 + 4.5 = 22.5$m/s