A stationary bomb explodes into three pieces. One piece of 2 Kg mass moves with a velocity of $8 m s^{- 1}$ at right angles to the other pieces of mass 1 kg moving with a velocity of $12 m s^{- 1}$ . If the mass of the third piece is 0.5 kg, then its velocity is

10 m s^{- 1}

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20 m s^{- 1}

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30 m s^{- 1}

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40 m s^{- 1}

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Solution

The correct option is D $40 m s^{- 1}$
Stationary so initial $m o m e n t u m$ will be $z e r o .$
momentum of $2 k g$ is $P_{1} = 2 \times 8 = 16 k g m / s$
that if $1 k g$ is $P_{2} = 1 \times 12 = 12 k g m / s$
both being at $r i g h t$ angle so their resultant will be $P = \sqrt{P_{1}^{2} + P_{2}^{2}} = \sqrt{256 + 144} = 20 k g m / s$
the third should have $s a m e$ momentum but in $o p p o s i t e$ direction to make the $n e t$ momentum $z e r o .$
Now its velocity should be $V = \frac{P}{m} = \frac{20}{0.5} = 40 m / s$

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