JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Conservation of Momentum

A body of mass 2 kg moving with a speed of

4 m / s

. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is

\frac{x}{10} m / s

. Then the value of

x

is ____.

Answer: 25

Solution

For a head-on elastic collision with the target at rest,

v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u

Here

v_{1}^{'} = \frac{u}{4}

, so

4 (m_{1} - m_{2}) = m_{1} + m_{2}

, which gives

m_{2} = \frac{3 m _{1}}{5} = 1.2

kg. No external horizontal force acts, so the centre of mass keeps the velocity it had before the collision:

v_{c m} = \frac{m _{1} u}{m _{1} + m _{2}} = \frac{2 \times 4}{3.2} = 2.5 m/s

Written as

\frac{x}{10}

m/s this gives

x = 25

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Conservation of Momentum

Solution

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