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 . If they collide head-on completely in elastically, the time taken for the combined mass to reach the ground, in units of is :

Engineering

Physics

Collision

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 $\sqrt{2 gh}$ . 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 :

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

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

$\frac{1}{2}$

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

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time for collision $t_{1} = \frac{h}{\sqrt{2 gh}}$

After t₁

$V_{A} = 0 - {gt}_{1} = - \sqrt{\frac{gh}{2}}$

and $V_{B} = \sqrt{2 gh} - {gt}_{1} = \sqrt{gh} [\sqrt{2} - \frac{1}{\sqrt{2}}]$

at the time of collision

${\vec{P}}_{i} = {\vec{P}}_{f}$

$\Rightarrow m {\vec{V}}_{A} + m {\vec{V}}_{B} = 2 m {\vec{V}}_{f}$

$\Rightarrow - \sqrt{\frac{gh}{2}} + \sqrt{gh} [\sqrt{2} - \frac{1}{\sqrt{2}}] = 2 {\vec{V}}_{f}$

V_f = 0

and height from ground $h - \frac{1}{2} {gt}_{1}^{2} = h - \frac{h}{4} = \frac{3 h}{4}$

so time $= \sqrt{2 \times \frac{(\frac{3 h}{4})}{g}} = \sqrt{\frac{3 h}{2 g}}$