A block of mass m slides down on a wedge of m | vedclass

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(D) $1$. For the wedge of mass $M$,the horizontal force is provided by the horizontal component of the normal reaction $N_1$. The horizontal component

Vedclass · Accepted Answer

$m\vec a_2 = -M\vec a_1$

Vedclass · Answer

(D) $1$. For the wedge of mass $M$,the horizontal force is provided by the horizontal component of the normal reaction $N_1$. The horizontal component of $N_1$ is $N_1 \sin 	heta$. Thus,$N_1 \sin 	heta = M|\vec a_1|$. This confirms option $C$ is correct.$2$. For the block of mass $m$,the equation of motion perpendicular to the inclined surface is $mg \cos 	heta - N_1 = m a_{1y}$,where $a_{1y}$ is the vertical component of the block's acceleration relative to the wedge. Resolving the acceleration,we find $N_1 = m(g \cos 	heta - |\vec a_1| \sin 	heta)$. This confirms option $B$ is correct.$3$. For the entire system,the vertical acceleration of the center of mass is downward. Thus,the normal force $N_2$ from the ground must be less than the total weight $(M+m)g$. This confirms option $A$ is correct.$4$. The center of mass of the system moves only due to external forces. The horizontal external force on the system is zero. Therefore,the horizontal acceleration of the center of mass is zero: $M\vec a_1 + m\vec a_{2x} = 0$. Option $D$ states $m\vec a_2 = -M\vec a_1$,which implies the block's total acceleration is related to the wedge's acceleration in a way that ignores the relative motion component. This is incorrect.

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