Question

A plank of mass $10kg$ rests on a smooth horizontal surface. Two blocks $A$ and $B$ of masses $m_A=2kg$ and $m_B=1kg$ lies at a distance of $3m$ on the plank as shown in the figure. The friction coefficient between the blocks and plank are ${\mu }_A=0.3$ and ${\mu }_B=0.1$. Now a force $F=15N$ is applied to the plank in horizontal direction. Find the time after which block $A$ collides with $B$.

Answer 1

Maximum friction force between the plank $\left(P\right)$ and $A$ is $F_A={\mu }_A{m}_{A}g=6N$
Maximum friction force between the plank $\left(P\right)$ and $B$ is $F_B={\mu }_B{m}_{B}g=1N$
Let system moves together with acceleration a then $a=\frac{F}{m_A+m_B+M}=\frac{15}{13}m/s$
Now; friction force on $m_B,f_B=m_{B}a=\frac{15}{13}N$
Friction force on $m_A,f_A=m_{A}a=\frac{30}{13}N$
As $f_B>F_B\Rightarrow$ Hence $B$ will not move with plank.
$f_A<F_A\Rightarrow$ Hence $A$ will move with plank.
$a_B=\frac{F_B}{m_B}=1{m/s}^2$
$Ap=\frac{F-F_B}{m_A+M}=\frac{7}{6}{m/s}^2$
Now $F.B.D.$ of the plank and $A$ & $B$ are Acceleration of $B$ w.r.t. the plank
$a_{BP}=1-\frac{7}{6}=-\frac{1}{6}{m/s}^2$
Now $L_{BP}=\frac{1}{2}{a_{BP}}^{\frac{2}{L}}\Rightarrow -3=\frac{1}{2}(x-6)$
Time of collision $t=6\sec$
Alternate
In ground frame, at time of collision.
$\frac{1}{2}{A}_P{t}^2=L+\frac{1}{2}{a}_B{t}^2\Rightarrow \frac{1}{2}(A_P-a_B)t^2=L\Rightarrow \frac{1}{2}(\frac{7}{6}-1)t^2=3\Rightarrow t=6\sec$