Ncert Exercises & Solutions

A particle of mass $m$ is moving in $yz\text-$ plane with a uniform velocity $v$ with its trajectory running parallel to the $+\text{ve}$ $y\text{-}$ axis and intersecting $z\text{-}$ axis at $z=a$ in the figure. The change in its angular momentum about the origin as it bounces elastically from a wall at $y$ = constant is:

1.	$mva~\hat e_{x}$	2.	$2mva~\hat e_{x}$
3.	$ymva~\hat e_{x}$	4.	$2ymva~\hat e_{x}$

(b) Hint: After rebounding, only the direction of angular momentum changes.

Step 1: Find the initial and final angular momentum.

The initial velocity is

v_{i} = v {\hat{e}}_{y}

$v_{i} = v {\hat{e}}_{y}$ , and after reflection from the wall, the final velocity is

v_{t} = - v {\hat{e}}_{y}

$v_{t} = - v {\hat{e}}_{y}$ . The trajectory is described as position vector

r = y {\hat{e}}_{y} + a {\hat{e}}_{z}

$r = y {\hat{e}}_{y} + a {\hat{e}}_{z}$ .

Step 2: Find the change in angular momentum.The magnitude of the angular momentum
vector is
Hence, in angular momentum is

r \times m (v_{f} - v_{j}) = 2 {mvae}_{x}

$r \times m (v_{f} - v_{j}) = 2 {mvae}_{x}$ .

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