A mass m moving horizontally (along the x- ax | vedclass

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(A) According to the principle of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collisio

Vedclass · Accepted Answer

$\frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$

Vedclass · Answer

(A) According to the principle of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision.Initial momentum of mass $m$ is $\vec{p}_1 = m v \hat{i}$.Initial momentum of mass $3m$ is $\vec{p}_2 = 3m (2v \hat{j}) = 6m v \hat{j}$.Total initial momentum $\vec{p}_{total} = m v \hat{i} + 6m v \hat{j}$.Let the final velocity of the combined mass $(m + 3m = 4m)$ be $\vec{v}_c$.By conservation of momentum: $4m \vec{v}_c = m v \hat{i} + 6m v \hat{j}$.Dividing by $4m$,we get $\vec{v}_c = \frac{m v}{4m} \hat{i} + \frac{6m v}{4m} \hat{j}$.Therefore,$\vec{v}_c = \frac{1}{4} v \hat{i} + \frac{3}{2} v \hat{j}$.

$A$ mass $m$ moving horizontally (along the $x-$ axis) with velocity $v$ collides and sticks to a mass of $3m$ moving vertically upward (along the $y-$ axis) with velocity $2v$. The final velocity of the combination is

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