JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2023
- Medium
- Numerical
A body of mass \(5\,kg\) is moving with a momentum of \(10\,kg\,ms ^{-1}\). Now a force of \(2\,N\) acts on the body in the direction of its motion for \(5\,s\). The increase in the Kinetic energy of the body is \(...........J\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(30\)
Step-by-step Solution
Detailed explanation
Given
\(M =5\,kg\)
\(P _{ i }=10\,kg\,m / s \text { (initial momentum) }\)
\(\text { Impulse }= F \Delta t =\Delta P = P _{ f }- P _{ i }\)
\(2 \times 5= P _{ f }-10\)
\(P _{ f }=20\,kg\,m / s \text { (final momentum) }\)
\(\text { Increase in } KE = KE _{ f }- KE _{ i }\)
\(=\frac{ P _{ f }^2}{2 m }-\frac{ P _{ i }^2}{2\,m }\)
\(=\frac{400}{2 \times 5}-\frac{100}{2 \times 5}=40-10=30\,J\)
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