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

Two solids \(A\) and \(B\) of mass \(1\, kg\) and \(2\, kg\) respectively are moving with equal linear momentum. The ratio of their kinetic energies \((K.E.)_{ A }:( K.E. )_{ B }\) will be \(\frac{ A }{1},\) so the value of \(A\) will be ..... .

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Answer & Solution

Correct Answer

(B) \(2\)

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Detailed explanation

Kinetic energy \(K =\frac{ P ^{2}}{2 m },\left( P _{ A }= P _{ B }\right)\)

\(K \propto \frac{1}{ m }\)

\(\frac{ K _{ A }}{ K _{ B }}=\frac{ m _{ B }}{ m _{ A }}\)

\(=\frac{2}{1}\)

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