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A particle of mass $m$ is moving in $y z$-plane with a uniform velocity $v$ with its trajectory running parallel to +ve $y$-axis and intersecting $z$-axis at $z=a$ in figure. The change in its angular momentum about the origin as it bounces elastically from a wall at $y=$ constant is
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The correct answer is:
$2 m v a \hat{e}_x$
$2 m v a \hat{e}_x$
The initial velocity is $v_i=v \hat{e}_y$ and after reflection from the wall, the final velocity is $v_f=-v \hat{e}_y$. The trajectory is at constant distance $a$ on $z$ axis and as particle moves along $y$ axis, its $y$ component changes.
So position vector (moving along $y$-axis),
$$
\vec{r}=y \hat{e}_y+a \hat{e}_z
$$
Hence, the change in angular momentum is
$$
\vec{r} \times m\left(v_f-v_j\right)=2 m v a \hat{e}_x
$$
So position vector (moving along $y$-axis),
$$
\vec{r}=y \hat{e}_y+a \hat{e}_z
$$
Hence, the change in angular momentum is
$$
\vec{r} \times m\left(v_f-v_j\right)=2 m v a \hat{e}_x
$$
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