Question

Block X and block Y travel toward each other along a horizontal surface with block X traveling in the positive direction. Block X has a mass of 4kg and a speed of 2ms. Block Y has a mass of 1kg and a speed of 1 ms. A completely inelastic collision occurs in which momentum is conserved. What is the approximate speed of block X after the collision

Answer 1

Answer:

  v = 1.4 m / s

Explanation:

To solve this problem we use the law of conservation of momentum, for this we define a system formed by the two blocks in such a way that the force during the collision have been internal

initial instant. Just before the crash

           p₀ = M v₁ - m v₂

final instant. Right after the crash

          p_f = (M + m) v

the moment is preserved

          p₀ = p_f

          M v₁ - m v₂ = (M + m) v

           v = $\frac{1}{M+m}$  (M v₁ - mv₂)

let's calculate

           v = $\frac{1}{4+1}$  (4 2 - 1 1)

           v = 1.4 m / s

in the same direction as the largest block (M)