A mass M moving with velocity V along X-axis | vedclass

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(B) According to the law of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collisio

Vedclass · Accepted Answer

$\frac{V}{3}\hat{i} + 2V\hat{j}$

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(B) According to the law of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision.Initial momentum of mass $M$: $\vec{p}_1 = M(V\hat{i}) = MV\hat{i}$.Initial momentum of mass $2M$: $\vec{p}_2 = 2M(3V\hat{j}) = 6MV\hat{j}$.Total initial momentum: $\vec{p}_{initial} = MV\hat{i} + 6MV\hat{j}$.After the collision, the masses stick together to form a single body of mass $(M + 2M) = 3M$.Let the final velocity be $\vec{V}_{final}$.Final momentum: $\vec{p}_{final} = (3M)\vec{V}_{final}$.Equating initial and final momentum: $MV\hat{i} + 6MV\hat{j} = 3M\vec{V}_{final}$.Dividing by $3M$: $\vec{V}_{final} = \frac{MV\hat{i} + 6MV\hat{j}}{3M} = \frac{V}{3}\hat{i} + 2V\hat{j}$.

$A$ mass $M$ moving with velocity $V$ along $X$-axis collides and sticks to another mass $2M$ which is moving along $Y$-axis with velocity $3V$. The velocity of the combination, after the collision is

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