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 inelastically, the time taken for the combined mass to reach the ground, in units of $\sqrt{\frac{h}{g}}$ is

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

Answer 1

The correct option is D $\sqrt{\frac{3}{2}}$

Let $S_1$ be distance travelled by particle being dropped from height $h$ before collision
and $S_2$ be distance travelled by vertically projected particle before collision
Here,
$S_1=\frac{1}{2}g{t}^2$
$S_2=ut-\frac{1}{2}g{t}^2$
Given that $u=\sqrt{2gh}$
We know that,
$S_1+S_2=h$
$\sqrt{2gh}t=h$
$t=\sqrt{\frac{h}{2g}}$




Velocity of dropped particle just before collision is $v_1=gt=\sqrt{\frac{hg}{2}}$
Velocity of projected particle just before collision is $v_2=u-gt=\sqrt{2gh}-\sqrt{\frac{hg}{2}}$

For inelastic collission, using principle of conversation of linear momentum
$m{v}_1+m{v}_2=2m{v}_f$

$\Rightarrow v_f=\frac{m(\sqrt{2gh}-\sqrt{\frac{gh}{2}})-m\sqrt{\frac{gh}{2}}}{2m}=0$

ie after collision combined mass as zero velocity.
Distance travelled by this combined mass after collision before reaching ground is $S_2=h-S_1=h-\frac{h}{4}=\frac{3h}{4}$
After collison, time taken $\left(t_1\right)$ for combined mass to reach the ground is

$\Rightarrow \frac{3h}{4}=\frac{1}{2}g{t}_1^2$

$\Rightarrow t_1=\sqrt{\frac{3h}{2g}}$