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Conservation of Momentum

Two particles...

Question

Two particles of mass $M_{A}$ and $M_{B}$ and there velocities are $V_{A}$ and $V_{B}$ respectively collides. After collision they inter changes their velocities the ratio of $\frac{M_{A}}{M_{B}}$ is :-

\frac{V_{A}}{V_{B}}

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\frac{V_{b}}{V_{A}}

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\frac{V_{A} + V_{B}}{V_{B} - V_{A}}

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Solution

The correct option is D $1$
Before collision,

Momentum. $P_{1} = M_{A} V_{A} + M_{B} V_{B}$

After collision,

Momentum, $P_{2} = M_{A} V_{B} + M_{B} V_{A}$

By conservation of momentum,

$P_{1} = P_{2}$

$M_{A} V_{A} + M_{B} V_{B} = M_{B} V_{A} + M_{A} V_{B}$

Dividing both side by $M_{B}$

$\frac{M_{A}}{M_{B}} V_{A} + V_{B} = V_{A} + \frac{M_{A}}{M_{B}} V_{B}$

$\frac{M_{A}}{M_{B}} (V_{A} - V_{B}) = (V_{A} - V_{B})$

Then,

$\frac{M_{A}}{M_{B}} = 1$

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Two particles of mass MA and MB and there velocities are VA and VB respectively collides. After collision they inter changes their velocities the ratio of MAMB is :-

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Two particles of mass $M_{A}$ and $M_{B}$ and there velocities are $V_{A}$ and $V_{B}$ respectively collides. After collision they inter changes their velocities the ratio of $\frac{M_{A}}{M_{B}}$ is :-