Question

Two particles are of the same mass $m$. One particle is dropped from a height of $10\text{m}$ and the other particle is projected up with a velocity of $10\text{m/s}$. If both stick to each other at some height, find the maximum height reached by their $COM$. [Take $g=10{\text{m/s}}^2$]

• A. $1.25\text{m}$
• B. $2.5\text{m}$
• C. $10\text{m}$
• D. $5\text{m}$

Answer 1

The correct option is A $1.25\text{m}$

Assume upward motion as positive.
Given, initial velocity of second particle $v_2=10\text{m/s}$ upward
Initial velocity of first particle $v_1=0\text{m/s}$ downward
Let the mass of both particles be $m$.

Acceleration of second particle $a_2=-g$ downwards
Acceleration of first particle $a_1=-g$ downwards

Then, velocity of $COM$ $V_{COM}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$
$V_{com}=\frac{m\left(0\right)+m\left(10\right)}{m+m}=5\text{m/s}$
$V_{com}=5\text{m/s}$ (upwards)

Acceleration of $COM$ $=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$
$=\frac{m(-g)+m(-g)}{m+m}=\frac{-2mg}{2m}$
$=-g$

We can use $V_2^2-V_1^2=2aS${Here$V_1=V_{com}$}
Here, speed of $COM$ at maximum height $V_2=0$
$\Rightarrow 0^2-5^2=2\times -10\times S$
$\Rightarrow S=1.25\text{m}$

Hence the maximum height attained by the $COM$ of the two particle system is $1.25\text{m}$.