A shell of mass is at rest initially. It explodes into three fragments having mass in the ratio . If the fragments having equal mass fly off along mutually perpendicular directions with speed , the speed of the third (lighter) fragment is [202

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Solution

Q.
A shell of mass $m$ is at rest initially. It explodes into three fragments having mass in the ratio $2 : 2 : 1$ . If the fragments having equal mass fly off along mutually perpendicular directions with speed $v$ , the speed of the third (lighter) fragment is [2022]

1 $v$

2 $\sqrt{2} v$

3 $2 \sqrt{2} v$

4 $3 \sqrt{2} v$

Ans.
(3)

Given, mass of the shell = $m$

Ratio of masses of the fragments is 2 : 2 : 1

Therefore, masses of the three fragments are

$m_{1} = \frac{m}{2}, m_{2} = \frac{m}{2}$ and $m_{3} = \frac{m}{4}$

Now fragments with equal masses i.e., $m_{1}$ and $m_{2}$ fly off perpendicularly with speeds $v_{1} = v_{2} = v .$ Let the velocity of third fragment be $v' .$

Applying law of conservation of momentum,

$m_{1} v_{1} \hat{i} + m_{2} v_{2} \hat{j} + m_{3} \vec{v} = 0$

$\frac{m v}{2} \hat{i} + \frac{m v}{2} \hat{j} + \frac{m}{4} \vec{v'} = 0 \Rightarrow \vec{v'} = 2 v (- \hat{i} - \hat{j})$

$| \vec{v} |$ $= 2 v \sqrt{{(- 1)}^{2} + {(- 1)}^{^{2}}} = 2 \sqrt{2} v$

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