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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) According to the law of conservation of linear momentum,the total momentum before the collision must equal the total momentum after the collision.Let $\vec{v}'$ be the final velocity of the combined mass.The initial momentum of mass $m$ is $\vec{p}_1 = m v \hat{i}$.The initial momentum of mass $3m$ is $\vec{p}_2 = (3m)(2v) \hat{j} = 6mv \hat{j}$.The total initial momentum is $\vec{p}_{initial} = m v \hat{i} + 6mv \hat{j}$.The total mass after the collision is $M = m + 3m = 4m$.The final momentum is $\vec{p}_{final} = (4m) \vec{v}'$.Equating the two,we get:$4m \vec{v}' = m v \hat{i} + 6mv \hat{j}$$\vec{v}' = \frac{mv \hat{i} + 6mv \hat{j}}{4m}$$\vec{v}' = \frac{1}{4}v \hat{i} + \frac{6}{4}v \hat{j} = \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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