COMEDK2022PhysicsCenter of Mass, Momentum and Collision
A bullet of mass m hits a mass M and gets embedded in it. If the block rises to a height h as a result of this collision, the velocity of the bullet before collision is
Options
- Av= 2 g h
- Bv= 2 g h [1+ ( m M ) ]
- Cv= 2 g h [1+ ( M m ) ]
- Dv= 2 g h [1- ( m M ) ]
Correct answer
C. v= 2 g h [1+ ( M m ) ]
Step-by-step solution
Let V be the common velocity of bullet and block just after collision. By conservation of momentum: mv = (m + M)V V = mv m+M After collision, the combined mass rises to height h . By conservation of energy: 1 2 (m+M)V^2 = (m+M)gh V = 2gh Substituting back: 2gh = mv m+M v = m+M m 2gh = 2gh (1 + M m )