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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

  1. Av^2 9 g
  2. Bv^2 2 g
  3. Cv^2 18 g
  4. 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 =

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