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Q.

A particle of mass $m$ moving with velocity $v$ makes an elastic one–dimensional collision with a stationary particle of mass $m$ They come in contact for a very small time $t_{0} .$ Their force of interaction increases from zero to $F_{0}$ linearly in time $0.5 t_{0},$ and decreases linearly to zero in further time $0.5 t_{0}$ as shown in figure. The magnitude of $F_{0}$ is

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a

$\frac{2 m u}{t_{0}}$

b

$\frac{m u}{2 t_{0}}$

c

$\frac{m u}{t_{0}}$

d

None of these

answer is B.

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Detailed Solution

In elastic collision between equal masses velocities are interchanged. Therefore, change in momentum in any one particle is $m u .$
Now, $| Δ p | = | i m p u l s e |$
$= a r e a u n d e r F - t g r a p h ∴ m u = \frac{1}{2} \times t_{0} \times F_{0} ∴ F_{0} = \frac{2 m u}{t_{0}}$

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