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