Question

A particle of mass$m$ is dropped from a height $h$ above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of $\sqrt{2gh}$​. If they collide head on completely in elastically, the time taken for the combined mass to reach the ground, in units of $\sqrt{\frac{h}{g}}$​​ is:

• A. $\sqrt{\frac{3}{2}}$
• B. $\sqrt{\frac{1}{2}}$
• C. $\frac{1}{2}$
• D. $\sqrt{\frac{3}{4}}$

Answer 1

The correct option is A

$\sqrt{\frac{3}{2}}$

Step1: Given data and assumptions.

Mass of particle $A$, $m_A=m$

Mass of particle $B$, $m_B=m$

Height from the ground of particle $A$$=h$

Speed of the particle $B=\sqrt{2gh}$

Time taken by combined mass to reach the ground $t=\sqrt{\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$g$}\right.\right)}$

Step2: Height from ground at time of collision.

We know that,

Time for collision $t_1=\frac{relativedis\tan ce}{relativevelocity}=\frac{h}{\sqrt{2gh}}$

Now,

After time $t_1$, velocity of particle $A$.

$v_A=u_A+g{t}_1$

Where $v_A$ is the final velocity, $u_A$is the initial velocity, $g$ is the acceleration due to gravity

$v_A=0-g{t}_1$ $\left[\because u_A=0\right]$

$\Rightarrow$$v_A=-\sqrt{\frac{gh}{2}}$

And velocity of particle $B$

$v_B=\sqrt{2gh}-g{t}_1$

$\Rightarrow$$v_B=\sqrt{\frac{gh}{2}}$

Again,

According to the law of conservation of linear momentum.

$m_A{v}_A+m_B{v}_B=\left(m_A+m_B\right)v_f$

$\Rightarrow$$-\sqrt{\frac{gh}{2}}+\sqrt{\frac{gh}{2}}=2m{v}_f$

$\Rightarrow$$v_f=0$

And

height from ground at time of collision,

$h\text{'}=h-\frac{1}{2}g{t}_1^2$

$\Rightarrow$$h\text{'}=\frac{3h}{4}$ $\left[\because t_1=\frac{h}{\sqrt{2gh}}\right]$

Step3: Finding the value of time $t$

Now the time taken for the combined mass to reach the ground

$t=\sqrt{2\frac{h\text{'}}{g}}$

$\Rightarrow$$t=\sqrt{2\frac{\left({\displaystyle \raisebox{1ex}{$3h$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}{g}}$ $\left[\because h\text{'}=\frac{3h}{4}\right]$

$\Rightarrow$$t=\sqrt{\frac{3}{2}}\sqrt{\frac{h}{g}}$

Hence, option A is correct. The time taken for the combined mass to reach the ground, in units of $\sqrt{\frac{h}{g}}$​​ is $\sqrt{\frac{3}{2}}$