A body at rest breaks into three fragments. T | vedclass

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(A) Let the mass of each of the two fragments be $m$ and their speed be $u = 30 \; m/s$. The mass of the third fragment is $3m$. Let its velocity be $

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$10\sqrt{2} \; m/s, 135^\circ$

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(A) Let the mass of each of the two fragments be $m$ and their speed be $u = 30 \; m/s$. The mass of the third fragment is $3m$. Let its velocity be $v$ at an angle $	heta$ with the direction opposite to the resultant momentum of the first two fragments.By the law of conservation of linear momentum,the initial momentum is zero,so the final momentum must also be zero.The resultant momentum of the two fragments moving perpendicularly is $P_r = \sqrt{(mu)^2 + (mu)^2} = mu\sqrt{2}$.This resultant momentum is directed at $45^\circ$ to each of the two fragments.The third fragment must move in the opposite direction to balance this,so its momentum $3mv$ must be equal in magnitude to $mu\sqrt{2}$.$3mv = mu\sqrt{2} \Rightarrow v = \frac{u\sqrt{2}}{3} = \frac{30\sqrt{2}}{3} = 10\sqrt{2} \; m/s$.The direction is $180^\circ - 45^\circ = 135^\circ$ relative to the direction of each of the first two fragments.

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