Two particles of equal mass& m are moving with velocities v ^ i and v ( ^ i + ^ j 2 )& respectively.& If the particles collide completely inelastically, then the energy lost in the process is

Correct answer: D. mv^ 2 8 Conserving momentum mv ^ i + m((v)/(2) ^ i + (v)/(2) ^ j ) = 2m ( v _ 1 ^ i + v _ 2 ^ j ) On solving v _ 1 = (3v)/(4) and v _ 2 = (v)/(4) Change in K.E. [(1)/(2) mv ^ 2 + (1)/(2)m((v)/(2) 2 )2] - [(1)/(2) (2m ) ( 9v^ 2 16 + (v2)/(16))] = 3 mv ^ 2 4 - 5 mv ^ 2 8 = (mv2)/(8)

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

Two particles of equal mass& m are moving with velocities v ^ i and...

Two particles of equal mass&

m

are moving with velocities

v\overset{^}{i}

and

v \left(\dfrac{\overset{^}{i} + \overset{^}{j}}{2}\right)& respectively.& If the particles collide completely inelastically, then the energy lost in the process is

Options

A $\dfrac{3mv^{2}}{4}$
B $\dfrac{2mv^{2}}{3}$
C $\dfrac{mv^{2}}{3}$
D $\dfrac{mv^{2}}{8}$

Correct answer

D. $\dfrac{mv^{2}}{8}$

Step-by-step solution

Conserving momentum

$mv\overset{^}{i} + m\left(\dfrac{v}{2}\overset{^}{i} + \dfrac{v}{2}\overset{^}{j}\right) = 2m \left({v}_{1}\overset{^}{i} + {v}_{2}\overset{^}{j}\right)On solving $ {v}_{1} = $\dfrac{3v}{4} and$ {v}_{2} = $\dfrac{v}{4}Change in K.E. \left[\dfrac{1}{2}{mv}^{2} + \dfrac{1}{2}m\left(\dfrac{v}{2}\sqrt{2}\right)2\right] - \left[\dfrac{1}{2} (2m ) \left(\dfrac{9v^{2}}{16} + \dfrac{v2}{16}\right)\right]=\dfrac{3{mv}^{2}}{4} - \dfrac{5{mv}^{2}}{8} = \dfrac{mv2}{8}$ $$

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