A billiard ball moving with a speed of $5 m/s$ collides with an identical ball originally at rest. if the first ball stops after collision, then the second ball will move forward with a speed

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Solution

Step 1: Given
Speed of moving ball initially, $v_{1_{i}} = 5 m/s$
Speed of ball in rest initally, $v_{2_{i}} = 0 m/s$
Speed of moving the ball after the collision, $v_{1_{f}} = 0 m/s$
Speed of resting ball after the collision, $v_{2_{f}}$
Let $m_{1}$ and $m_{2}$ be the mass of the two balls. Since they are identical, $m_{1} = m_{2}$
Step 2: Formulas used
Since there is no external force involved, the law of conservation of momentum can be used. It is given as,
$P_{i} = P_{f}$
where $P_{i}$ is initial momentum and $P_{f}$ is final momentum.
Step 3: Solution
Total initial momentum was,
$\begin{array}{rcl} P_{i} & = & m_{1} v_{1_{i}} + m_{2} v_{2_{i}} \\ = & 5 m_{1} + 0 \\ = & 5 m_{1} \end{array}$
Total final momentum was,
$\begin{array}{rcl} P_{f} & = & m_{1} v_{1_{f}} + m_{2} v_{2_{f}} \\ = & m_{1} \cdot 0 + m_{2} v_{2_{f}} \\ = & m_{2} v_{2_{f}} \end{array}$
We have,
$\begin{matrix} P_{i} = P_{f} \\ \Rightarrow & 5 m_{1} = m_{2} v_{2_{f}} \\ \Rightarrow & v_{2_{f}} = 5 m/s & [∵ m_{1} = m_{2}] \end{matrix}$
Therefore, the velocity of the second ball after the collision is $5 m/s$ .

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