A body of mass $2 kg$ makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body ?

1.2 kg

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1 kg

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1.5 kg

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1.8 kg

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Solution

The correct option is A $1.2 kg$
Given:
$m_{1} = 2 kg, u_{1} = u, u_{2} = 0,$
$v_{1} = u / 4$

By conservation of linear momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$2 u + 0 = 2 \times \frac{u}{4} + m_{2} v_{2}$
$m_{2} v_{2} = \frac{3 u}{2}$

For elastic collision
$v_{seperation} = v_{approach}$
$v_{2} - \frac{u}{4} = u \Rightarrow v_{2} = \frac{5 u}{4}$

Now,
$m_{2} \times \frac{5 u}{4} = \frac{3 u}{2}$
$\Rightarrow m_{2} = 1.2 kg$

Alternate Solution:
For head on elastic collision we have
$v_{1} = \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}}$
Here
$m_{1} = 2 kg, u_{1} = u, u_{2} = 0,$
$v_{1} = u / 4$

$∴ \frac{u}{4} = \frac{(2 - m_{2}) u}{2 + m_{2}} \Rightarrow m_{2} = 1.2 kg$

Hence option $(D)$ is correct.

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A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body ?

A body of mass $2 kg$ makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body ?