A particle of mass m is dropped from a height h above the ground.


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:

  1. (1) $\sqrt{\frac{1}{2}}$

  2. (2) $\sqrt{\frac{3}{4}}$

  3. (3) $\frac{1}{2}$

  4. (4) $\sqrt{\frac{3}{2}}$

Correct Option: , 4



Let 1 ve ue ume taken uy me parucie uroppea нom height $\mathrm{h}$ to collide with particle thrown upward.


$v^{2}-u^{2}=2 g h$

$\Rightarrow v^{2}-0^{2}=2 g h$

$\Rightarrow v=\sqrt{2 g h}$

Downward distance travelled

$S_{1}=\frac{1}{2} g t^{2}=\frac{1}{2} g \cdot \frac{h}{2 g}=\frac{h}{4}$

Distance of collision point from ground

$s_{2}=h-\frac{h}{4}=\frac{3 h}{4}$

Speed of (A) just before collision

$v_{1}=g t=\sqrt{\frac{g h}{2}}$

And speed of(B) just before collision

$v_{2}=\sqrt{2 g h}-\sqrt{\frac{g h}{2}}$

Using principle of conservation of linear momentum

$m v_{1}+m v_{2}=2 m v_{f}$

$\frac{\Rightarrow v_{f}=m\left(\sqrt{2 g h}-\sqrt{\frac{g h}{2}}\right)-m \sqrt{\frac{g h}{2}}=0}{2 m}$

After collision, time taken $\left(t_{1}\right)$ for combined mass to reach the ground is

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

$\Rightarrow t_{1}=\sqrt{\frac{3 h}{2 g}}$

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