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Q.

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

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a

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

b

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

c

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

d

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

answer is A.

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Detailed Solution

From the law of conservation of linear momentum

$mv + (3 m) (2 v) = (4 m) v^{'}$

$mv \hat{i} + 6 mv \hat{j} = 4 {mv}^{'}$

$v^{'} = \frac{1}{4} v \hat{i} + \frac{6}{4} v \hat{j}$

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

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