Questions
A bullet of mas m moving horizontally with velocity v hits a block of wood of mass M, resting on a smooth horizontal plane. The fraction of energy of the bullet dissipated in the collision it self is (assume collision to be inelastic)
detailed solution
Correct option is D
Applying the law of conservation of momentum, we havemV=(m+M)V V=mvm+M Loss of K.E. =12mv2−12(m+M)V2 =12mv2−12(m+M)mvM+m2 =12mv21−mm+M=12mv2Mm+M Fraction of K.E. dissipated = Loss of K.E. Initial K.E. =Mm+MTalk to our academic expert!
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A light body A and a heavy body B have equal linear momentum. Then the KE of the body A is
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