A particle of mass 4 m explodes into three pi | vedclass

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(C) Initially,the particle of mass $4 m$ is at rest,so its initial momentum is zero.Let the velocity of the heavier mass $(2 m)$ be $\vec{v} = v_x \ha

Vedclass · Accepted Answer

$\sqrt{13} ms^{-1}$

Vedclass · Answer

(C) Initially,the particle of mass $4 m$ is at rest,so its initial momentum is zero.Let the velocity of the heavier mass $(2 m)$ be $\vec{v} = v_x \hat{i} + v_y \hat{j}$.According to the law of conservation of linear momentum,the total final momentum must be zero:$\vec{P}_{initial} = \vec{P}_{final} = 0$$m(4 \hat{i}) + m(6 \hat{j}) + 2m(v_x \hat{i} + v_y \hat{j}) = 0$Dividing by $m$:$4 \hat{i} + 6 \hat{j} + 2v_x \hat{i} + 2v_y \hat{j} = 0$Equating the components:$4 + 2v_x = 0 \Rightarrow v_x = -2 ms^{-1}$$6 + 2v_y = 0 \Rightarrow v_y = -3 ms^{-1}$The magnitude of the velocity of the heavier mass is:$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} ms^{-1}$

$A$ particle of mass $4 m$ explodes into three pieces of masses $m, m$ and $2 m$. The equal masses move along $X$-axis and $Y$-axis with velocities $4 ms^{-1}$ and $6 ms^{-1}$ respectively. The magnitude of the velocity of the heavier mass is

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