Question

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

• A. $\frac{3}{2}v\hat{j}+\frac{1}{4}v\hat{j}$
• B. $\frac{1}{4}v\hat{j}+\frac{3}{2}v\hat{j}$
• C. $\frac{1}{3}v\hat{j}+\frac{2}{3}v\hat{j}$
• D. $\frac{2}{3}v\hat{j}+\frac{1}{3}v\hat{j}$

Answer 1

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

The given problum is a perfect exemple of inelastic collision.
By using Newton's secound law (Net momentum conservation)

so, $\overrightarrow{v_x}=\frac{v}{4}\hat{i}\overrightarrow{v_y}=\frac{6v}{4}\hat{j}=\frac{3v}{2}\hat{j}$
hence,
$\overrightarrow{v}=\frac{v}{4}\hat{i}+\frac{3v}{2}\hat{j}$

so, option B is correct.