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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

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

A particle of mass mm is dropped from a height hh above the ground. Simultaneously another particle of the same mass is thrown vertically upwards from the ground with a speed of 2gh.Iftheycollidehead−oncompletelyinelastically,thenthetimetakenforthecombinedmasstoreachthegroundis\sqrt{2gh}. If they collide head-on completely inelastically, then the time taken for the combined mass to reach the ground is

Options

  1. A3h4g\sqrt{\dfrac{3h}{4g}}
  2. B3h2g\sqrt{\dfrac{3h}{2g}}
  3. Ch2g\sqrt{\dfrac{h}{2g}}
  4. Dh4g\sqrt{\dfrac{h}{4g}}

Correct answer

B. 3h2g\sqrt{\dfrac{3h}{2g}}

Step-by-step solution

The time for collisiont1=h2gh{t}_{1} = \dfrac{h}{2gh}After t1{t}_{1},vA=0−gt1=−gh/2VB=2gh−gt1=gh[2−12]Atthetimeofcollision{v}_{A} = 0 - g{t}_{1} = - \sqrt{gh / 2} {V}_{B} = \sqrt{2gh} - g{t}_{1} = \sqrt{gh} \left[\sqrt{2} - \dfrac{1}{2}\right]At the time of collisionP→i=P→f⇒mV→A+mV→B=2mV→f⇒−gh2+gh[2−12]=2V→fVf=0\begin{array}{ll}\overset{→}{P} i = \overset{→}{P}f \\ ⇒ m\overset{→}{V}A + m\overset{→}{V}B = 2m\overset{→}{V}f \\ ⇒ - \sqrt{\dfrac{gh}{2}} + \sqrt{gh}\left[\sqrt{2} - \dfrac{1}{2}\right] =2\overset{→}{V}f \\ {V}_{f} = 0\end{array}$andheightfromgroundand height from ground= h - \dfrac{1}{2} g{t}_{1}^{2} = h - \dfrac{h}{4} = \dfrac{3h}{4}sotimeso time= \sqrt{2 \times \left(\dfrac{3h}{4}\right)g} = \sqrt{\dfrac{3h}{2g}} \text{\text{ and } }$
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