Physics - NLM Question with Solution | TestHub

Let $a_1$ be the acceleration of $m_1$ (to the right), $a_P$ the acceleration of $P_2$ (downwards), and $a_2$ the acceleration of $m_2$ (downwards). Due to the string connecting $m_1$ and $P_2$ over $P_1$, we have $a_1=a_P$. Due to the string passing over $P_2$ (one end fixed, other to $m_2$), the length constraint implies $a_2=2a_P$. Thus, $a_2=2a_1$. Let $a_1=a$, so $a_P=a$ and $a_2=2a$. FBD for $m_1$: $T_1=m_{1}a=ma$. (1) FBD for $P_2$: $m_{P}g+2T_2-T_1=m_P{a}_P=ma$. (2) FBD for $m_2$: $m_{2}g-T_2=m_2{a}_2=m(2a)$. (3) From (3), $T_2=mg-2ma$. Substitute this into (2): $mg+2(mg-2ma)-ma=ma⟹3mg-4ma-ma=ma⟹3mg=6ma⟹a=g/2$. Given $g=10m/s^2$, $a=10/2=5m/s^2$. For P: Acceleration of block $m_1$ is $a_1=a=5m/s^2$. So (P)-(4). For Q: Acceleration of block $m_2$ is $a_2=2a=2(5)=10m/s^2$. So (Q)-(6). For R: Tension in the string connecting $m_1$ and $P_2$ is $T_1=ma=m(g/2)=mg/2$. So (R)-(5). For S: Tension in the string passing over $P_2$ is $T_2=mg-2ma=mg-2m(g/2)=mg-mg=0$. So (S)-(3).