Question

A mass $m$ moving horizontally (along the x-axis) with velocity $v$ collides and sticks to the 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

Answer: (A)

As the two masses stick together after collision, hence it is inelastic collision. Therefore, only momentum is conserved.

$\therefore mv\hat{i}+3m\left(2 v\right)\hat{j}=\left(4 m\right) \, \overset{ \rightarrow }{v}$
$\overset{ \rightarrow }{v}=\frac{v}{4}\hat{i}+\frac{6}{4}v\hat{j}=\frac{v}{4}\hat{i}+\frac{3}{2}v\hat{j}$