Particle A makes a head on elastic collision with another stationary particle B. They fly apart in opposite directions with equal speeds. The mass ratio will be-

\frac{1}{3}

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\frac{1}{2}

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\frac{1}{4}

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\frac{2}{3}

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Solution

The correct option is B $\frac{1}{3}$
For an elastic collision, $e = \frac{R V O S}{R V O A} = - \frac{v_{1} - v_{2}}{u_{1} - u_{2}} = 1$

Let
$u_{1}$ be the velocity of A before collision,
$u_{2}$ be the velocity of B before collision $= 0$
$v_{1}$ be the speed of A after collision
$v_{2}$ be the speed of B after collision
From conservation of momentum,
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$m_{1} u_{1} = m_{1} v_{1} - m_{2} v_{1}$ .................................(1)

$e = - \frac{v_{1} - v_{2}}{u_{1} - u_{2}} = 1$

$v_{2} - (- v_{1}) = u_{1} - u_{2}$
$2 v_{1} = u_{1}$ ..............................................(2)
Putting (2) in (1)
$2 m_{1} v_{1} = m_{2} v_{1} - m_{1} v_{1}$
$\Rightarrow 3 m_{1} v_{1} = m_{2} v_{1}$

$\Rightarrow \frac{m_{1}}{m_{2}} = \frac{1}{3}$

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