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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
A
$\frac{2}{3}v\hat{i} + \frac{1}{3}v\hat{j}$
B
$\frac{3}{2}v\hat{i} + \frac{1}{4}v\hat{j}$
C
$\frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$
D
$\frac{1}{3}v\hat{i} + \frac{2}{3}v\hat{j}$
Detailed Solution
The masses stick together, so the collision is perfectly inelastic; linear momentum is conserved.
Initial momentum = $m(v\hat{i}) + 3m(2v\hat{j}) = mv\hat{i} + 6mv\hat{j}$
Final momentum = $(m + 3m)\vec{V} = 4m\vec{V}$
$4m\vec{V} = mv\hat{i} + 6mv\hat{j}$
$\vec{V} = \frac{mv\hat{i} + 6mv\hat{j}}{4m}$
$\vec{V} = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$
Initial momentum = $m(v\hat{i}) + 3m(2v\hat{j}) = mv\hat{i} + 6mv\hat{j}$
Final momentum = $(m + 3m)\vec{V} = 4m\vec{V}$
$4m\vec{V} = mv\hat{i} + 6mv\hat{j}$
$\vec{V} = \frac{mv\hat{i} + 6mv\hat{j}}{4m}$
$\vec{V} = \frac{1}{4}v\hat{i} + \frac{3}{2}v\hat{j}$