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

Answer 1

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

Draw a free body diagram of the given problem
Find the distance travelled by both the particle.
Formula Used: $s=ut+\frac{1}{2}a{t}^2$
Time of collision
$t_0=\frac{h}{\sqrt{2gh}}=\sqrt{\frac{h}{2g}}\dots \left(i\right)$
$s_1=\frac{1}{2}g{t}_0^2=\frac{1}{2}g\cdot \frac{h}{2g}=\frac{h}{4}$
$s_2=\frac{3h}{4}$
Speed of $\left(A\right)$ just before collision $v_1\downarrow$
$=g{t}_0$
Substituting the value from equation $\left(i\right)$
$=\sqrt{\frac{gh}{2}}$
And speed cof $\left(B\right)$ just before collision $v_2\downarrow$
$=\sqrt{2gh}-\sqrt{\frac{gh}{2}}$
Find the velocity of center of mass
After collision velocity of center of mass
$v_{cm}=\frac{m\left(\sqrt{2gh-\sqrt{\frac{gh}{2}}}\right)-m\sqrt{\frac{gh}{2}}}{2m}=0$
So from their, time of fall ' $t$ '
$\frac{3h}{4}=\frac{1}{2}g{t}^2$
$t=\sqrt{\frac{3}{2}\frac{h}{g}}$