Physics - Center of Mass Question with Solution | TestHub

For P: Since there are no external horizontal forces, the COM of the man-plank system remains stationary. If the man walks a distance $L$ relative to the plank, the plank moves a distance $x_p$. The man's displacement relative to ground is $L+x_p$. Equating initial and final COM positions: $m(0)+M(L/2)=m(L+x_p)+M(L/2+x_p)$, which simplifies to $0=mL+(m+M)x_p$, so $x_p=-\frac{mL}{m+M}$. (P)-(4). For Q: With no external horizontal forces, the velocity of the COM of the man-plank system remains zero if it was initially at rest. (Q)-(1). For R: The acceleration of the COM of a system is given by $a_{COM}=F_{ext}/M_{total}$. Here, $F_{ext}=F$ and $M_{total}=m+M$. So $a_{COM}=\frac{F}{m+M}$. (R)-(2). For S: The velocity of the rocket (which is the COM of the rocket-fuel system) is given by the rocket equation $v(t)=v_e\ln (\frac{M_0}{M(t)})$, where $M(t)=M_0-kt$. Thus, $v(t)=v_e\ln (\frac{M_0}{M_0-kt})$. (S)-(3).