Question

Coefficient of friction between two blocks is $\mu =0.6$. The blocks are given velocities in the directions shown in the figure below. Find the common velocity attained by both the blocks. [Take $g=10{\text{m/s}}^2$]

• A. $15\text{m/s}$
• B. $4\text{m/s}$
• C. $2\text{m/s}$
• D. $6\text{m/s}$

Answer 1

The correct option is B $4\text{m/s}$

Given, $m_1=2\text{kg},m_2=1\text{kg}$
Friction force, $f=\mu m_{1}g$
$=0.6\times 2\times 10$
$=12\text{N}$

Taking motion towards left to be positive,
Thus, acceleration of $2\text{kg}$ block, $a_1=\frac{f}{m_1}=\frac{12}{2}=6{\text{m/s}}^2$ (towards left)
Acceleration of $1\text{kg}$ block,
$a_2=\frac{f}{m_2}=\frac{-12}{1}=-12{\text{m/s}}^2$ (towards right)

Let the common velocity of the blocks be $V$.
For $2\text{kg}$ block,
$V=-u_1+a_{1}t$
$V=-3+6t......\left(1\right)$
For $1\text{kg}$ block,
$V=u_2+a_{2}t=18-12t....\left(2\right)$

From equation $\left(\right(1)\times 2+$ $\left(2\right))$, we get
$-6+12t+18-12t=3V$
$3V=12\Rightarrow V=4\text{m/s}$