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 √(2gh). If they collide head-on completely

Step-by-step solution

1. 1

   Initial conditions

   Particle 1: $m$, dropped from $h$. Particle 2: $m$, thrown up with $u_2 = \sqrt{2gh}$.
2. 2

   Kinematics & Momentum

   Use $s = ut + \frac{1}{2}at^2$ and conservation of momentum for inelastic collision.
3. 3

   Collision time & height

   Let $t_c$ be collision time. $y_1 = h - \frac{1}{2}gt_c^2$, $y_2 = u_2t_c - \frac{1}{2}gt_c^2$. At collision, $y_1 = y_2$.
4. 4

   Calculate collision time

   $h - \frac{1}{2}gt_c^2 = u_2t_c - \frac{1}{2}gt_c^2 \implies h = u_2t_c$

   $$h = \sqrt{2gh} t_c \implies t_c = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$$
5. 5

   Velocities before collision

   $v_1 = -gt_c = -g\sqrt{\frac{h}{2g}} = -\sqrt{\frac{gh}{2}}$

   $$v_2 = u_2 - gt_c = \sqrt{2gh} - g\sqrt{\frac{h}{2g}} = \sqrt{2gh} - \sqrt{\frac{gh}{2}} = \sqrt{\frac{gh}{2}}$$
6. 6

   Combined velocity after collision

   Conservation of momentum: $mv_1 + mv_2 = (m+m)V$. $m(-\sqrt{\frac{gh}{2}}) + m(\sqrt{\frac{gh}{2}}) = 2mV$.

   $$0 = 2mV \implies V = 0$$
7. 7

   Collision height

   $y_c = h - \frac{1}{2}gt_c^2 = h - \frac{1}{2}g\left(\frac{h}{2g}\right) = h - \frac{h}{4} = \frac{3h}{4}$
8. 8

   Time to reach ground

   Combined mass starts from $y_c = \frac{3h}{4}$ with $V=0$. Time $T'$ to fall: $y_c = 0 \cdot T' + \frac{1}{2}gT'^2$.

   $$\frac{3h}{4} = \frac{1}{2}gT'^2 \implies T'^2 = \frac{3h}{2g} \implies T' = \sqrt{\frac{3h}{2g}}$$
9. 9

   Final time in units

   $T' = \sqrt{\frac{3}{2}} \sqrt{\frac{h}{g}}$. In units of $\sqrt{\frac{h}{g}}$, the time is $\sqrt{\frac{3}{2}}$.