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# A billiard ball moving with a speed of $5\text{m/s}$ 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: GivenSpeed of moving ball initially, ${v}_{{1}_{i}}=5\text{m/s}$Speed of ball in rest initally, ${v}_{{2}_{i}}=0\text{m/s}$Speed of moving the ball after the collision, ${v}_{{1}_{f}}=0\text{m/s}$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 usedSince 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: SolutionTotal 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,$\begin{array}{rcl}{P}_{f}& =& {m}_{1}{v}_{{1}_{f}}+{m}_{2}{v}_{{2}_{f}}\\ & =& {m}_{1}·0+{m}_{2}{v}_{{2}_{f}}\\ & =& {m}_{2}{v}_{{2}_{f}}\end{array}$We have,$\begin{array}{ccc}& {P}_{i}={P}_{f}& \\ ⇒& 5{m}_{1}={m}_{2}{v}_{{2}_{f}}& \\ ⇒& {v}_{{2}_{f}}=5\text{m/s}& \left[\because {m}_{1}={m}_{2}\right]\end{array}$Therefore, the velocity of the second ball after the collision is $5\text{m/s}$.

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