Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be
PhysicsCenter of Mass Momentum and CollisionJEE Main
Options:
  • A $\frac{1}{2} m v^2 \times \frac{m}{(m+M)}$
  • B $\frac{1}{2} m v^2 \times \frac{M}{(m+M)}$
  • C $\frac{1}{2} m v^2 \times \frac{(M+m)}{M}$
  • D $\frac{1}{2} M v^2 \times \frac{m}{(m+M)}$
Solution:
2067 Upvotes Verified Answer
The correct answer is: $\frac{1}{2} m v^2 \times \frac{m}{(m+M)}$
By conservation of momentum, $\quad m v+M \times 0=(m+M) v$ Velocity of composite block $V=\left(\frac{m}{m+M}\right) v$ K.E. of composite block $=\frac{1}{2}(M+m) V^2$ $=\frac{1}{2}(M+m)\left(\frac{m}{M+m}\right)^2 v^2=\frac{1}{2} m v^2\left(\frac{m}{m+M}\right)$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.