First slide

Angular momentum

Question

A particle of mass m is moving in YZ-plane with a uniform velocity v with its trajectory running parallel to + ve Y-axis and intersecting Z-axis at z = a as shown in figure. The change in its angular momentum about the origin as it bounces elastically from a wall at y = constant is [NCERT Exemplar)

Moderate

A

$m v a {\hat{e}}_{x}$
B

$2 m v a {\hat{e}}_{x}$
C

$y m v {\hat{e}}_{x}$
D

$2 y m v {\hat{e}}_{x}$

Solution

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 described as position vector $r = y {\hat{e}}_{y} + a {\hat{e}}_{z}$
Hence, the change in angular momentum
$= r \times m (v_{f} - v_{i})$
$= (y {\hat{e}}_{y} + a {\hat{e}}_{z}) \times (- m v {\hat{e}}_{y} - m v {\hat{e}}_{y})$
$= ({\hat{y}}_{y} + a {\hat{e}}_{z}) \times (- 2 m v {\hat{e}}_{y}) [∵ {\hat{e}}_{y} \times {\hat{e}}_{y} = 0 and {\hat{e}}_{z} \times {\hat{e}}_{y} = - {\hat{e}}_{x}]$
$= 2 m v a {\hat{e}}_{x}$

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