Question

The masses of blocks $A$ and $B$ are $m$ and $M$ respectively. Between $A$ and $B$, there is a constant frictional force $F$ and $B$ can slide on a smooth horizontal surface. $A$ is set in motion with velocity while $B$ is at rest. What is the distance moved by $A$ relative to $B$ before they move with the same velocity?

• A. $\frac{mM{v}_0^2}{F(m-M)}$
• B. $\frac{mM{v}_0^2}{2F(m-M)}$
• C. $\frac{mM{v}_0^2}{F(m+M)}$
• D. $\frac{mM{v}_0^2}{2F(m+M)}$

Answer 1

Answer: (D)

$\frac{mM{v}_0^2}{2F(m+M)}$

For the block $A$ and $B$ as shown below
Equations of motion
$a_A=\frac{F}{M}$ (in $-x$ direction)
$a_B=\frac{F}{M}$ (in $+x$ direction)
Relative acceleration, of $A$ w.r.t $B$,
$a_{A,B}=a_A-a_B=-\frac{F}{m}-\frac{F}{M}$
$=-F\left(\frac{M+m}{Mn}\right)$ (along $-x$ direction)
Initial relative velocity of $A$ w.r.t. $B$,
$u_{AB}=v_0$
using equation $v^2=u^2+2as$
$0=v_0^2-\frac{2F(m+M)S}{Mm}\Rightarrow S=\frac{Mm{v}_0^2}{2F(m+M)}$
i.e., Distance moved by $A$ relative to $B$
$S_{AB}=\frac{Mm{V}_0^2}{2F(m+M)}$.