A spherical ball A of mass $4 k g$ , moving along a straight line strikes another spherical ball B of mass $1 k g$ at rest. After the collision, A and B move with velocities $v_{1} m s^{- 1}$ and $v_{2} m s^{- 1}$ respectively making angles of $30^{\circ}$ and $60^{\circ}$ with respect to the original direction of motion of ball A. The ratio $\frac{v_{1}}{v_{2}}$ will be:

\frac{\sqrt{3}}{4}

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\frac{4}{\sqrt{3}}

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\frac{1}{\sqrt{3}}

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\sqrt{3}

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Solution

The correct option is A $\frac{\sqrt{3}}{4}$
Since there is no momentum in the line perpendicular to the direction of motion of ball A before collision, after collision also there would be no momentum along that line. So by conservation of momentum along that line we have:
$0 = 4 v_{1} sin 30 ° - v_{2} sin 60 ° \Rightarrow \frac{v_{1}}{v_{2}} = \frac{\sqrt{3}}{4}$
So, option A is the correct answer.

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Similar questions

Q. A spherical ball

A

of mass

4 k g

, moving along a straight line strikes another spherical ball

B

of mass

1 k g

at rest. After the collision,

A

and

B

move with velocities

v_{1} m s^{- 1}

and

v_{2} m s^{- 1}

respectively making angles of

30^{\circ}

and

60^{\circ}

with respect to the original direction of motion of

A

. The ratio

\frac{v_{1}}{v_{2}}

will be:

A spherical ball A of mass $4 k g$ , moving in a straight line strikes (elastically) another spherical ball B of mass $1 k g$ at rest. After the collision, balls A and B move with velocities $V_{1}$ m/s and $V_{2}$ m/s respectively, making angles of $30^{\circ}$ and $60^{\circ}$ with respect to the original direction of motion of A. The ratio $\frac{V_{1}}{V_{2}}$ will be: