JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 ..... .
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
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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