NEETPhysicsCenter of Mass, Momentum and Collision
A block of mass m sliding with speed v on a smooth horizontal surface collides head-on and elastically with a stationary block of mass 2m . Immediately after the collision, the first block enters a rough patch with a coefficient of kinetic friction . The stopping distance of the first block after it enters the rough patch is (acceleration due to gravity is g ):
Options
- Av^2 9 g
- Bv^2 2 g
- Cv^2 18 g
- D2v^2 9 g
Correct answer
C. v^2 18 g
Step-by-step solution
For a one-dimensional elastic collision, the velocity of the first block after the collision is given by: v₁ = m₁ - m₂ m₁ + m₂ u₁ Substituting the given values ( m₁ = m , m₂ = 2m , u₁ = v ): v₁ = m - 2m m + 2m v = - v 3 The negative sign indicates that the first block rebounds with a speed of v 3 . When this block enters the rough patch, the kinetic friction provides a retardation a = g . Using the third equation of motion ( v_f^2 = v_i^2 - 2as ) with final velocity v_f = 0 : 0 = ( v 3 )^2 - 2( g)s s = v^2 9 2 g =