Question

A 1 kg block with initial velocity 4 m/s collides elastically with a stationary 2 kg block. After the collision, the 1 kg block has a velocity of -1.33 m/s. Determine the speed of the 2 kg block after the collision.

Answer 1

Answer:

  v₂ = 2,665 m / s

Explanation:

This is an exercise of conservation of the momentum, we define a system formed by the two cars, so that the forces during the collision are internal and the moment is conserved.

Initial instant. Before the crash

           p₀ = m v_o

Final moment. Right after the crash

           p_f = m v₁ + M v₂

where the first car uses subscript 1 and has a mass m = 1 kg and the second car uses the subscript 2 with a mass M = 2kg

how the moment is preserved

        p₀ = p_f

         m v_o = m v₁ + M v₂

They indicate that the collision is elastic, therefore the kinetic energy is conserved, the velocity of the first car is v1 = -1.33 me / S

     

         v₂ = $\frac{ m ( v_o - v_1)}{M}$

we calculate

        v₂ = $\frac{ 1 ( 4 + 1.33)}{2}$

        v₂ = 2,665 m / s