Question

Two stones are dropped down simultaneously from different heights with initial speed $u_1=0$ and $u_2=30\text{m/s}$ respectively. If the initial distance between them is $10\text{m}$ and mass of the first stone is twice that of the second stone, find the velocity of their $COM$ after $2\text{s}$. [Take $g=10{\text{m/s}}^2$]

• A. $20\text{m/s}$
• B. $30\text{m/s}$
• C. $50\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is B $30\text{m/s}$

Given, $u_1=0,u_2=30\text{m/s}$
Let the mass of the second stone be $m$. Then, mass of the first stone will be $2m$.
Acceleration of $COM$ is $g{\text{m/s}}^2$
Initial velocity of $COM$,
$u_{com}=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}$
$=\frac{2m\times 0+m\times 30}{m+2m}=10\text{m/s}$
Velocity of $COM$ after $2\text{s}$ will be
$v_{com}=u_{com}+a_{com}\times t$
$v_{com}=10+10\times 2=30\text{m/s}$