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Standard XII

Physics

Second Law of Motion

A ball of mas...

Question

A ball of mass $m$ moving with speed $v$ strikes the vertical wall perpendicularly and bounce back with same speed. Collision is perfectly elastic. Calculate the magnitude of the change in momentum of the ball.

m v

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2 m v

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\frac{m v}{2}

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\frac{m v}{4}

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zero

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Solution

The correct option is B $2 m v$
The ball bounces back with the same speed i.e $v$

$∴$ $\to u = - v^i$ and $\to v = v^i$

Thus change in momentum $Δ \to P = [m v - (- m v)]^i = 2 m v^i$

$⟹$ $| Δ \to P | = 2 m v$

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A ball of mass m moving with speed v strikes the vertical wall perpendicularly and bounce back with same speed. Collision is perfectly elastic. Calculate the magnitude of the change in momentum of the ball.

The correct option is B 2mvThe ball bounces back with the same speed i.e v∴ →u=−v^i and →v=v^iThus change in momentum Δ→P=[mv−(−mv)]^i=2mv^i⟹ |Δ→P|=2mv

A ball of mass $m$ moving with speed $v$ strikes the vertical wall perpendicularly and bounce back with same speed. Collision is perfectly elastic. Calculate the magnitude of the change in momentum of the ball.

The correct option is B $2 m v$
The ball bounces back with the same speed i.e $v$

$∴$ $\to u = - v^i$ and $\to v = v^i$

Thus change in momentum $Δ \to P = [m v - (- m v)]^i = 2 m v^i$

$⟹$ $| Δ \to P | = 2 m v$