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Two spheres $A$ and $B$ of masses $m_1$ and $m_2$ respectively collide. $A$ is at rest initially and $B$ is moving with velocity $y$ along $x$-axis. After collision $B$ has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction. The mass $A$ moves after collision in the direction
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Verified Answer
The correct answer is:
$\theta=\tan ^{-1}\left(\frac{1}{2}\right)$ to the $x$-axis
$\text {Here, } \mathbf{p}_i=m_2 v \mathbf{i}+m_2 \times 0$
$\mathbf{p}_f=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_2$
Law of conservation of momentum
$\begin{gathered}
\mathbf{p}_i=\mathbf{p}_f \\
m_2 v \mathbf{i}=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_1 \\
\mathbf{v}_1=\frac{m_2}{m_1} v \mathbf{i}+\frac{m_2}{m_1} \frac{v}{2} \mathbf{j}
\end{gathered}$
From this equation we can find
$\begin{aligned}
\tan \theta & =\frac{1}{2} \\
\theta & =\tan ^{-1}\left(\frac{1}{2}\right) \text { to the } x \text {-axis. }
\end{aligned}$
$\mathbf{p}_f=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_2$
Law of conservation of momentum
$\begin{gathered}
\mathbf{p}_i=\mathbf{p}_f \\
m_2 v \mathbf{i}=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_1 \\
\mathbf{v}_1=\frac{m_2}{m_1} v \mathbf{i}+\frac{m_2}{m_1} \frac{v}{2} \mathbf{j}
\end{gathered}$
From this equation we can find
$\begin{aligned}
\tan \theta & =\frac{1}{2} \\
\theta & =\tan ^{-1}\left(\frac{1}{2}\right) \text { to the } x \text {-axis. }
\end{aligned}$
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