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Conservation of Momentum

Let P1 and P2...

Question

Let $P_{1}$ and $P_{2}$ be the momenta of two identical particles before elastic collision and $P_{1}^{'}$ and $P_{2}^{'}$ be their momenta after collision. If $P_{2} = 0$ , which of the following is true?

K E_{1} + K E_{2} = K E_{1}^{'} + K E_{2}^{'}

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P_{1} = P_{1}^{'} + P_{2}^{'}

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P_{1}^{2} = (P_{1}^{'})^{2} + (P_{2}^{'})^{2}

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All of these

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Solution

The correct option is D All of these
As it is elastic collision, kinetic energy is conserved.

$i . e . K E_{1} + K E_{2} = K E_{1}^{'} + K E_{2}^{'} . . . (1)$

Momentum will also be conserved, as there is no external force acting.

$i . e . P_{1} + P_{2} = P_{1}^{'} + P_{2}^{'}$

As, $P_{2} = 0$

$∴ P_{1} = P_{1}^{'} + P_{2}^{'}$

Since, kinetic energy is given by, $K E = \frac{P^{2}}{2 m}$ and $P_{2} = 0 \Rightarrow u_{2} = 0 \Rightarrow K E_{2} = 0$

From $(1)$ ,

$\frac{P_{1}^{2}}{2 m} = \frac{{P_{1}^{'}}^{2}}{2 m} + \frac{{P_{2}^{'}}^{2}}{2 m}$

$\Rightarrow P_{1}^{2} = (P_{1}^{'})^{2} + (P_{2}^{'})^{2}$

Hence, option $(D)$ is the correct answer.

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Conservation of Momentum

Standard IX Physics

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Let P1 and P2 be the momenta of two identical particles before elastic collision and P′1 and P′2 be their momenta after collision. If P2=0, which of the following is true?

Let $P_{1}$ and $P_{2}$ be the momenta of two identical particles before elastic collision and $P_{1}^{'}$ and $P_{2}^{'}$ be their momenta after collision. If $P_{2} = 0$ , which of the following is true?