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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 \(...........J\).

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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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