A $1 k g$ stationary bomb is exploded in three parts having mass ratio $1 : 1 : 3$ . Parts having same mass move in perpendicular directions with velocity $30 m / s$ , then the velocity of bigger part will be:

10 \sqrt{2} m / s

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\frac{10}{\sqrt{2}} m / s

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15 \sqrt{2} m / s

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\frac{15}{\sqrt{2}} m / s

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Solution

The correct option is A $10 \sqrt{2} m / s$
$1 x = 1 k g$
Ration of $m_{1} : m_{2} : m_{3} = 1 : 1 : 3$
$v_{1} = v_{2} = 30 m / s$
$m_{1} = m_{2} = m = \frac{1}{5} k g = 0.2 k g$
$m_{3} = 3 m = 3 \times 0.2 = 0.6 k g$
According to the law of conservation of momentum
$m_{1} v_{1} - m_{3} v_{3} = 0 m_{3} v_{2} - m_{3} v_{3} = 0$
$\Rightarrow v_{3} 1 = \frac{v_{1}}{3} V_{3} 2 = \frac{v_{2}}{3}$
$v_{3} = \sqrt{(v_{2} 1)^{2} + (v_{3} 2)^{2}}$
$= \sqrt{10^{2} + 20^{2}}$
$= 10 \sqrt{2} m / s$ .

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