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

• Correct answer
  A

  $\frac{1}{4} v \hat{\mathbf{i}}+\frac{3}{2} y \hat{\mathbf{j}}$
• B

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

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

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

Solution

From the law of conservation of linear momentum $\begin{aligned} m v+(3 m)(2 v) & =(4 m) v^{\prime} \\ m v \hat{i}+6 m v \hat{j} & =4 m v^{\prime} \\ v^{\prime} & =\frac{1}{4} v \hat{i}+\frac{6}{4} v \hat{j} \\ y^{\prime} & =\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j} \end{aligned}$

Hence, the correct answer is option: 'A'.