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)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
mvm+M
b
mM+m
c
Mmm+M
d
Mm+M
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
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+M