Question

Block B of mass $2\text{kg}$ is placed on smooth horizontal plane. Block A of mass $1\text{kg}$ is placed on block B. The coefficient of friction between A and B is 0.40. The block A is imparted a velocity $16\text{m/s}$ at t = 0. Find the time at which momentum of the two blocks are equal (in seconds). $(g=10{\text{m/s}}^2)$.

Answer 1

$P_1=m_1{V}_1=\left(1\right)(16-\mu gt)$ till velocities are equal.
$P_2=m_2{V}_2=\left(2\right)\left(\frac{\mu m_{1}g}{m_2}\right)t=\mu \left(1\right)gt$ till velocities are equal.
For $P_1=P_2,16-\mu gt=\mu gt$
$t=\frac{16}{\left(2\right)\left(0.4\right)\left(10\right)}=2\text{sec}$