Physics - Center of Mass Question with Solution | TestHub

Initially in equilibrium: $m_{cat}=1$ kg. With $f$ the rod's force on the cat: $f-g=a_1$; for block + rod: $4g+f-2T=4a_2$ and $T-2g=5\frac{a_2}{2}$. Solving: $a_1=24a_2$, so average speeds obey $V_1=24V_2$.

Relative speed: $V_1+V_2=2$ m/s, so $V_2=\frac{2}{25}$ m/s, $V_1=\frac{48}{25}$ m/s and $t_0=\frac{1}{V_1+V_2}=\frac{1}{2}$ s.

(B) Distance travelled by block: $\frac{4}{25}\times \frac{1}{2}=\frac{2}{25}$ - (P)

(C) $Y_{com}=\frac{1\times \frac{48}{25}-4\times \frac{2}{25}}{10}\times \frac{1}{2}=\frac{2}{25}$ - (P)

(D) $X=\frac{5\times \frac{4}{25}\times \frac{1}{2}}{9}=\frac{2}{45}$ - (Q)

(A) $\tan \varphi =\frac{m_{cat}{Y}_{cat}}{M_{block+rod}X}=\frac{24/25}{5\times 2/25}=\frac{12}{5}$ - (R)

Hence A - R, B - P, C - P, D - Q.