A mass $m$ moving horizontally (along the $x$-axis) with velocity $v$ collides and sticks to mass of $3 m$…

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

Solution

From the law of conservation of linear momentum

$\begin{aligned}
m v+(3 m)(2 v) & =(4 m) v^{\prime} \\
m v \hat{i}+6 m v \hat{j} & =4 m v^{\prime} \\
v^{\prime} & =\frac{1}{4} v \hat{i}+\frac{6}{4} v \hat{j} \\
y^{\prime} & =\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}
\end{aligned}$

Asked in: NEET 2011 (Mains)

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