First slide

Rectilinear Motion

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 $\sqrt{2 g h}$ . 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 :

Difficult

A

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

$\frac{1}{2}$
C

$\sqrt{\frac{1}{2}}$
D

$\sqrt{\frac{3}{4}}$

Solution

Time for collision $t_{1} = \frac{r e l a t i v e d i s \tan c e}{r e l a t i v e v e l o c i t y} = \frac{h}{\sqrt{2 g h}} [r e l a t i v e a c c e l e r a t i o n i s z e r o]$
After $t_{1}$ , $v e l o c i t y o f A = V_{A} = 0 - g t_{1} = - \sqrt{\frac{g h}{2}}$
and $v e l o c i t y o f B = V_{B} = \sqrt{2 g h} - g t_{1} = \sqrt{g h} [\sqrt{2} - \frac{1}{\sqrt{2}}] = \sqrt{\frac{g h}{2}}$
At the time of collision , apply momentum conservation
   ${\vec{P}}_{i} = {\vec{P}}_{f} \Rightarrow m {\vec{V}}_{A} + m {\vec{V}}_{B} = 2 m {\vec{V}}_{f} \Rightarrow - \sqrt{\frac{g h}{2}} + \sqrt{\frac{g h}{2}} = 2 {\vec{V}}_{f} \Rightarrow V_{f} = 0$

and
Height from ground at time of collision , $h' = h - \frac{1}{2} g t_{1}^{2} = h - \frac{h}{4} = \frac{3 h}{4}$
So time taken for the combined mass to reach the ground =    $\sqrt{2 \frac{h'}{g}} = \sqrt{2 \times \frac{(\frac{3 h}{4})}{g}} = \sqrt{\frac{3 h}{2 g}} = \sqrt{\frac{3}{2}} \sqrt{\frac{h}{g}}$

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