A particle of mass m is moving along the x-ax | vedclass

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(C) According to the law of conservation of linear momentum,the total momentum before the split equals the total momentum after the split.Initial mome

Vedclass · Accepted Answer

$\frac{3}{2}v_0 \hat{i} - v_0 \hat{j}$

Vedclass · Answer

(C) According to the law of conservation of linear momentum,the total momentum before the split equals the total momentum after the split.Initial momentum: $\vec{P}_i = m v_0 \hat{i}$.Final momentum: $\vec{P}_f = (m/3)(2v_0 \hat{j}) + (2m/3)\vec{v}$,where $\vec{v}$ is the velocity of the remaining part.Equating $\vec{P}_i = \vec{P}_f$:$m v_0 \hat{i} = \frac{2m v_0}{3} \hat{j} + \frac{2m}{3} \vec{v}$.Dividing by $m$ and rearranging:$v_0 \hat{i} - \frac{2}{3} v_0 \hat{j} = \frac{2}{3} \vec{v}$.Multiplying by $3/2$:$\vec{v} = \frac{3}{2} v_0 \hat{i} - v_0 \hat{j}$.

$A$ particle of mass $m$ is moving along the $x$-axis with a speed $v_0$. Suddenly,a part of its mass,$m/3$,breaks off and moves parallel to the $y$-axis with a speed $2v_0$. Find the velocity of the remaining part in terms of unit vectors.

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