JEE Main 201910 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 10 April
Shift: Morning Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2019 · 10 April · Morning Shift · Physics · Center of Mass, Momentum and Collision

Two particles of masses M and 2M are moving with speeds of 10m s ^...

Two particles of masses

M

and

2M are

moving with speeds of

10m{s}^{- 1}

and

5m{s}^{- 1}

, as shown in the figure. They collide at the origin and after that they move along the indicated directions with speeds

{v}_{1}

and

{v}_{2}

, respectively. The values of

{v}_{1}

and

{v}_{2}

are, nearly

Two particles of masses M and 2M are moving with speeds of 10m s ^ - 1 and 5m s ^ - 1 , as shown in the figure. They

Options

A $6.5m{s}^{- 1}$ and $3.2m{s}^{- 1}$
B $3.2m{s}^{- 1}$ and $12.6m{s}^{- 1}$
C $13.02m{s}^{- 1}$ and $19 . 7m{s}^{- 1}$
D $3.2m{s}^{- 1}$ and $6.3m{s}^{- 1}$

Correct answer

C. $13.02m{s}^{- 1}$ and $19 . 7m{s}^{- 1}$

Step-by-step solution

Two particles of masses M and 2M are moving with speeds of 10m s ^ - 1 and 5m s ^ - 1 , as shown in the figure. They
From conservation of linear momentum,
Along $x$ -axis:

${\text{10Mcos30}}^{0} + 5\left(\text{2M}cos{45}^{0}\right) = \text{M}{v}_{1}cos{30}^{0} + 2\text{M}{v}_{2}cos{45}^{0}$
$10M\dfrac{\sqrt{3}}{2} + 10M\dfrac{1}{\sqrt{2}} = M\dfrac{\sqrt{3}}{2}{v}_{1} + 2M{v}_{2}\dfrac{1}{\sqrt{2}}$
$\dfrac{\sqrt{3}}{2}{v}_{1} + \sqrt{2}{v}_{2} = 5 ( \sqrt{3} + \sqrt{2} )$ … (1)
Along $y$ -axis:
$- 10M\dfrac{1}{2} + 10M\dfrac{1}{\sqrt{2}} = M{v}_{1}\dfrac{1}{2} - 2M{v}_{2}\dfrac{1}{\sqrt{2}}$
$\dfrac{{v}_{1}}{2} - \sqrt{2}{v}_{2} = 5 ( \sqrt{2} - 1 )$ … (2)
From equations (1) and (2)
${v}_{1}\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\right) = 5\left(\sqrt{3} + \sqrt{2}\right) + 5 ( \sqrt{2} - 1 )$
${v}_{1} = 13.02m / s$
${v}_{2} = \dfrac{5\sqrt{3} + 5\sqrt{2} - 13.02 \times \dfrac{\sqrt{3}}{2}}{\sqrt{2}}$
$= 19.72m{s}^{- 1}$

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