The masses of blocks A and B are m and M resp | vedclass

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(D) The correct option is $(d)$.For block $A$,the frictional force $F$ acts in the direction opposite to its motion. Thus,the acceleration of $A$ is $

Vedclass · Accepted Answer

$\frac{mMv_0^2}{2F(m+M)}$

Vedclass · Answer

(D) The correct option is $(d)$.For block $A$,the frictional force $F$ acts in the direction opposite to its motion. Thus,the acceleration of $A$ is $a_A = -\frac{F}{m}$.For block $B$,the frictional force $F$ acts in the direction of motion. Thus,the acceleration of $B$ is $a_B = \frac{F}{M}$.The relative acceleration of $A$ with respect to $B$ is $a_{AB} = a_A - a_B = -\frac{F}{m} - \frac{F}{M} = -F \left( \frac{M+m}{Mm} \right)$.The initial relative velocity of $A$ with respect to $B$ is $u_{AB} = v_0$. When they move with the same velocity,the final relative velocity $v_{AB} = 0$.Using the kinematic equation $v_{AB}^2 = u_{AB}^2 + 2 a_{AB} S_{AB}$,where $S_{AB}$ is the relative distance:$0 = v_0^2 + 2 \left( -F \frac{M+m}{Mm} \right) S_{AB}$$S_{AB} = \frac{v_0^2}{2F \frac{M+m}{Mm}} = \frac{Mmv_0^2}{2F(m+M)}$.

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