A bomb of mass $5 k g$ explodes at rest into three pieces of mass in the ratio $1 : 1 : 3$ . The piece with equal masses fly in mutually normal direction with speed of $21 m s^{- 1}$ . Velocity of the third piece is :

9.12 m s^{- 1}

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7.98 m s^{- 1}

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9.87 m s^{- 1}

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8.97 m s^{- 1}

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Solution

The correct option is D $9.87 m s^{- 1}$
Resultant momentum of two piece of equal masses is

$p = \sqrt{(21 \times 1)^{2} + (21 \times 1)^{2}}$

$= 21 \sqrt{2} k g m s^{- 1}$

From conservation of momentum $3 V = 21 \sqrt{2} ($ where $V$ is velocity of $3 k g$ piece $)$

or $V = 7 \sqrt{2} = 9.87 m s^{- 1}$

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A bomb of mass 5 kg explodes at rest into three pieces of mass in the ratio 1:1:3. The piece with equal masses fly in mutually normal direction with speed of 21 ms−1. Velocity of the third piece is :

The correct option is D 9.87 ms−1Resultant momentum of two piece of equal masses is p=√(21×1)2+(21×1)2=21√2 kgms−1From conservation of momentum 3V=21√2(where V is velocity of 3 kg piece)or V=7√2=9.87 ms−1

A bomb of mass $5 k g$ explodes at rest into three pieces of mass in the ratio $1 : 1 : 3$ . The piece with equal masses fly in mutually normal direction with speed of $21 m s^{- 1}$ . Velocity of the third piece is :