Question

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

• 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 1

The correct option is A $\frac{1}{4}v\hat{i}+\frac{3}{2}v\hat{j}$

From momentum conservation
$m\nu \hat{i}+3m\left(2\nu \right)\hat{j}=\left(4m\right)\overrightarrow{\nu }$
$\overrightarrow{\nu }=\frac{\nu }{4}\hat{i}+\frac{6}{4}\nu \hat{j}$
$=\frac{\nu }{4}\hat{i}+\frac{3}{2}\nu \hat{j}$