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A bob at rest explodes in mid-air into two equal fragments. If one of the fragments is moving vertically upward with $V_{0}$ , then

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The total impulse which causes explosion of the bomb is twice the momentum of each

The total impulse which causes explosion of the bomb is zero

The second fragment will move vertically down with a velocity of

V_{0}

The internal force which causes the bomb is

{mv}_{0} {/t}_{0}

, where m is the mass of the bomb and

t_{0}

is the time of explosion

answer is A, C, D.

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Detailed Solution

By linear momentum conservation

$m_{1} = m, m_{2} = m$

$o = {mv}_{0} + mv$

$v = \frac{- {mv}_{0}}{m} = - v_{0}$

Total impulse = 0

$F = \frac{mv}{t} = \frac{m \times v_{0}}{t_{0}}$

$∴$ Internal force $= \frac{{mv}_{o}}{t_{0}}$

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