A 10 kg bomb explodes into three fragments having masses in the ratio of $1 : 1 : 3$ . Two fragments having masses in the ratio $1 : 1$ move at right angles to each other with a velocity of $15 m / s$ . Find the K.E. of heavy fragment.

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Solution

By momentum conservation in $x -$ direction :
$\Rightarrow 6 v_{x} - 2 \times 15 = 0$
$\Rightarrow v_{x} = 5 m / s$
In $y -$ direction :
$\Rightarrow 6 v_{y} - 2 \times 15 = 0$
$\Rightarrow v_{y} = 5 m / s$
$\Rightarrow v = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}} = 5 \sqrt{2} m / s$
K.E. of heavy fragment $= \frac{1}{2} \times 6 \times v^{2}$
$= 150$
Hence, the answer is $150$

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A 10 kg bomb explodes into three fragments having masses in the ratio of 1:1:3. Two fragments having masses in the ratio 1:1 move at right angles to each other with a velocity of 15 m/s. Find the K.E. of heavy fragment.

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