Question

A block of mass $m$ is pushed with a velocity $v_0$ along the surface of a trolley car of mass $M$. If the horizontal ground is smooth and the coefficient of kinetic friction between the block of plank is $\mu$. Find the minimum distance $m$ slides relative to plank.

• A. $x=\frac{v_0^2}{(1+\frac{m}{M})\mu g}$
• B. $x=\frac{v_0^2}{2(1+\frac{m}{M})\mu g}$
• C. $x=\frac{v_0^2}{2(1+\frac{m}{M})g}$
• D. $x=\frac{2v_0^2}{(1+\frac{m}{M})\mu g}$

Answer 1

The correct option is B $x=\frac{v_0^2}{2(1+\frac{m}{M})\mu g}$

Work energy theorem in centre of mass reference frame:

$W_{\text{ext}}+W_{\text{int}}=(\Delta K{)}_{\text{cm}}=(K_f-K_i{)}_{\text{cm}}$

Initial kinetic energy of system w.r.t $CM$
$(K_i{)}_{\text{cm}}=\frac{1}{2}\frac{mM}{(m+M)}{v}_0^2$
As $m$ comes to rest relative to $M,$ $|(V_{\text{rel}}{)}_f|=0$
Final kinetic energy of system w.r.t $CM$
$(K_f{)}_{\text{cm}}=0$
hence $0+(-\mu mgx)=0-\frac{1}{2}\frac{mM}{(m+M)}{v}_0^2$
$\Rightarrow x=\frac{v_0^2}{2(1+\frac{m}{M})\mu g}$