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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2021
  • Hard
  • Numerical

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 ..... .

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Answer & Solution

Correct Answer

(C) \(25\)

Step-by-step Solution

Detailed explanation

\(p_{i}=p_{f}\)

\(2 \times 4=2 \times 1+m_{2} \times v_{2}\)

\(m_{2} v_{2}=6\)

by coefficient of restitution

\(1=\frac{v_{2}-1}{4} \Rightarrow v_{2}=5 m / s\)

\(m _{2} \times 5=6\)

\(m _{2}=1.2 kg\)

\(v _{ cm }=\frac{ m _{1} v _{1}+ m _{2} v _{2}}{ m _{1}+ m _{2}}\)

\(v _{ cm }=\frac{2 \times 1+1.2 \times 5}{2+1.2}=\frac{8}{3.2}=\frac{25}{10}\)

\(x =25\)

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