A steel ball of radius $2 c m$ is at rest on a frictionless surface. Another ball of radius $4 c m$ moving at a velocity of $81 c m / s e c$ collides elastically with first ball. After collision the smaller ball moves with speed of

81 c m / s e c

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63 c m / s e c

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144 c m / s e c

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Solution

The correct option is C $144 c m / s e c$
Ratio in radius of steel balls = $1 / 2$
Mass= Density x Volume
$V \propto r^{3} ⟹ M \propto r^{3}$

So, ratio in their masses $= \frac{1}{8}$

Let $m_{1} = 8 m$ and $m_{2} = m$

Applying momentum consercation:

$v_{2} = \frac{2 m_{1} u_{1}}{m_{1} + m_{2}} = \frac{2 \times 8 m \times 81}{8 m + m} = 144 c m / s$

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A steel ball of radius 2cm is at rest on a frictionless surface. Another ball of radius 4cm moving at a velocity of 81cm/sec collides elastically with first ball. After collision the smaller ball moves with speed of

The correct option is C 144cm/sec Ratio in radius of steel balls =1/2 Mass= Density x VolumeV∝r3⟹M∝r3So, ratio in their masses =18Let m1=8m and m2=m Applying momentum consercation:v2=2m1u1m1+m2=2×8m×818m+m=144cm/s

A steel ball of radius $2 c m$ is at rest on a frictionless surface. Another ball of radius $4 c m$ moving at a velocity of $81 c m / s e c$ collides elastically with first ball. After collision the smaller ball moves with speed of