NEETPhysicsCenter of Mass, Momentum and Collision
Statement-I: A point particle of mass (m ) moving with speed (v ) collides with stationary point particle of mass (M ). If the maximum energy loss possible is given as (f ( 1 2 m v^2 ) ) then (f= ( m M+m ) ). Statement-II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.
Options
- ABoth Statement-I and Statement-II are correct.
- BBoth Statement-I and Statement-II are incorrect.
- CStatement-I is correct and Statement-II is incorrect.
- DStatement-I is incorrect and Statement-II is correct.
Correct answer
D. Statement-I is incorrect and Statement-II is correct.
Step-by-step solution
Initial kinetic energy (K_i= 1 2 m v^2 ) If after collision the two masses stick (perfectly inelastic), common velocity by momentum conservation is (V= m v m+M ) Final kinetic energy then is (K_f= 1 2 (m+M) V^2= 1 2 (m+M) m^2 v^2 (m+M)^2 = 1 2 m^2 v^2 m+M ) Energy lost: ( K=K_i-K_f= 1 2 m v^2- 1 2 m^2 v^2 m+M = 1 2 m v^2 (1- m m+M )= 1 2 m v^2 M m+M ) So the maximum fractional loss (f ) (such that ( K=f 1 2 m v^2 )) is (f= M m+M ) Thus Statement-I (which asserts (f= m m+M )) is wrong. Statement-II is correct becaus