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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

Two particles of masses M and 2M are moving with speeds of 10ms^ -...

Two particles of masses

M

and

2M are

moving with speeds of

10ms^{- 1}

and

5ms^{- 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 10ms^ - 1 and 5ms^ - 1 , as shown in the figure. They

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

Options

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

Correct answer

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

Step-by-step solution

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

From conservation of linear momentum,
Along

x

-axis:

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}}

= 19.72ms^{- 1}

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