Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

112 of 130

Question 112 of 130

A particle of mass mm is dropped from a height hh 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.\sqrt{2gh}. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of hg\sqrt{\dfrac{h}{g}} is:

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Answer: (D) 32\sqrt{\dfrac{3}{2}}

time for collision t1=h2gh{t}_{1}=\dfrac{h}{\sqrt{2gh}}
After t1{t}_{1}
VA=0−gt1=−gh2{V}_{A}=0-g{t}_{1}=-\sqrt{\dfrac{gh}{2}}
and VB=2gh−gt1=gh[2−12]{V}_{B}=\sqrt{2gh}-g{t}_{1}=\sqrt{gh}\left[\sqrt{2}-\dfrac{1}{\sqrt{2}}\right]
at the time of collision
p⃗i=pf⃗{\vec{p}}_{i}=\vec{{p}_{f}}
⇒mV⃗A+mV⃗B=2mV⃗f\Rightarrow m{\vec{V}}_{A}+m{\vec{V}}_{B}=2m{\vec{V}}_{f}
⇒−gh2+gh[2−12]=2V⃗f\Rightarrow -\sqrt{\dfrac{gh}{2}}+\sqrt{gh}\left[\sqrt{2}-\dfrac{1}{\sqrt{2}}\right]=2{\vec{V}}_{f}
Vf=0{V}_{f}=0
and height from ground
=h−12gt12=h−h4=3h4=h-\dfrac{1}{2}g{t}_{1}^{2}=h-\dfrac{h}{4}=\dfrac{3h}{4}
so time ,
t=2×(3h4)g=32hgt=\sqrt{2\times \dfrac{\left(\dfrac{3h}{4}\right)}{g}}=\sqrt{\dfrac{3}{2}}\sqrt{\dfrac{h}{g}}

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