Question

The figure given below shows a block of mass $m=10\text{kg}$ placed on another block of mass $M=20\text{kg}$. The block of mass $m\text{kg}$ is given a velocity of $u=10\text{m/s}$ and the block of mass $M$ is at rest on a smooth horizontal surface. The coefficient of friction between the two blocks is $\mu =0.3$. Then what will be the velocity of $COM$ of the system when the system reaches common speed?

• A. $\frac{5}{3}\text{m/s}$ towards right
• B. $\frac{10}{3}\text{m/s}$ towards right
• C. $5\text{m/s}$ towards right
• D. $3\text{m/s}$ towards right

Answer 1

The correct option is B $\frac{10}{3}\text{m/s}$ towards right

Given, friction coefficient between the two blocks $=0.3$
Initial velocity of upper block $u=10\text{m/s}$
Mass of upper block $m=10\text{kg}$
Mass of lower block $M=20\text{kg}$

As we can see, there is no external force acting on the two block system. (friction between the two blocks is an internal force because we have taken both the blocks as a system). Hence, linear momentum will be conserved for the system.
i.e $p_i=p_f$
or $mu+M\left(0\right)=(m+M)v$
where, $v$ = common velocity of both the blocks
$\Rightarrow v=\left(\frac{m}{m+M}\right)u=\left(\frac{10}{10+20}\right)\times 10=\frac{10}{3}\text{m/s}$

So, the system of 2 blocks will move with a common velocity of $v=\frac{10}{3}\text{m/s}$ towards right. Hence, the $COM$ of the system will also move with the same velocity.

Alternate solution:
No external force is acting on the system Hence velocity of $COM$ will not be changed.
Initial velocity of $COM$ = final velocity of $COM$
i.e $v_{com}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}=\frac{10\times 10+20\times \left(0\right)}{10+20}$
$=\frac{10}{3}\text{m/s}$