A block of mass m moves on a horizontal rough | vedclass

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(D) $1$. The angular momentum $L$ of a particle about a point is given by $L = r 	imes p$. For the center of mass of the block moving with velocity $

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All of the above.

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(D) $1$. The angular momentum $L$ of a particle about a point is given by $L = r 	imes p$. For the center of mass of the block moving with velocity $v$ at a height $h$ above point $A$ on the surface,the initial angular momentum is $L = mvh$. Thus,option $A$ is correct.$2$. Since the surface is rough,a kinetic friction force $f_k = \mu_k N$ acts on the block in the direction opposite to its motion. This force causes a deceleration,so the velocity $v$ of the block decreases as time passes. Thus,option $B$ is correct.$3$. The torque $	au$ about point $A$ is given by $	au = r 	imes F$. The friction force acts at the surface,and the normal force acts through the center of mass. The torque due to friction about point $A$ is $	au = f_k 	imes h 
eq 0$. Since there is a non-zero external torque acting on the system about point $A$,the angular momentum is not conserved. Thus,option $C$ is correct.$4$. Since $A$,$B$,and $C$ are all correct,the correct choice is $D$.

$A$ block of mass $m$ moves on a horizontal rough surface with initial velocity $v$. The height of the centre of mass of the block is $h$ from the surface. Consider a point $A$ on the surface.

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