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 ..... .
Pick an option to check your answer.
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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