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(B) The angular momentum $L$ of a particle about a point $P$ is given by $L = r 	imes p$,where $r$ is the position vector and $p$ is the linear momen

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$2mvd \cos 	heta$

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(B) The angular momentum $L$ of a particle about a point $P$ is given by $L = r 	imes p$,where $r$ is the position vector and $p$ is the linear momentum.For a particle moving with velocity $v$ at an angle $	heta$ with the horizontal,the perpendicular distance from point $P$ to the line of motion is $d \cos 	heta$.Initial angular momentum $L_i = m v (d \cos 	heta)$ (in the clockwise direction).Since the collision is perfectly elastic,the ball rebounds with the same speed $v$ at the same angle $	heta$ with the horizontal.Final angular momentum $L_f = -m v (d \cos 	heta)$ (in the counter-clockwise direction).The change in angular momentum $\Delta L = L_f - L_i = -m v d \cos 	heta - (m v d \cos 	heta) = -2m v d \cos 	heta$.The magnitude of the change in angular momentum is $2mvd \cos 	heta$.

$A$ ball of mass $m$ moving with velocity $v$ collides with a wall elastically as shown in the figure. After the impact,the change in angular momentum about point $P$ is:

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