Physics - NLM Question with Solution | TestHub

Let $a_M$ be the acceleration of $M$ and $a_{P1}$ be the acceleration of pulley $P_1$. Since the string connecting $M$ and $P_1$ is inextensible, $a_M=a_{P1}$.

The string over pulley $P_1$ has one end fixed to the ceiling and the other end attached to mass $m$. Thus, the acceleration of $m$, $a_m$, is twice the acceleration of $P_1$, so $a_m=2a_{P1}=2a_M$.

Let $T_1$ be the tension in the string connecting $M$ and $P_1$, and $T_2$ be the tension in the string passing over $P_1$. For block $M$: $T_1=M{a}_M$. For pulley $P_1$ (massless): $T_1=2T_2$. For block $m$: $mg-T_2=m{a}_m$. Substitute $T_1=2T_2$ into the first equation: $2T_2=M{a}_M⟹T_2=M{a}_M/2$. Substitute $a_m=2a_M$ and $T_2$ into the third equation: $mg-M{a}_M/2=m(2a_M)$.$mg=(2m+M/2)a_M=(4m+M)/2\cdot a_M$. Given $M=4m$: $mg=(4m+4m)/2\cdot a_M=4m{a}_M⟹a_M=g/4$. P: Acceleration of block $M=a_M=g/4$. Q: Acceleration of block $m=a_m=2a_M=2(g/4)=g/2$. R: Tension $T_1=M{a}_M=4m(g/4)=mg$. S: Tension $T_2=M{a}_M/2=4m(g/4)/2=mg/2$.