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

The approximate speed of block X after the collision is 1.4 m/s.

Given information:

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 4 kg and a speed of 2 m/s.

Block Y has a mass of 1 kg and a speed of -1 m/s.

The collision between them is perfectly inelastic. So, the final velocity of both the objects will be the same.

The momentum of the system will be conserved. Let v be the final velocity of both the blocks.

The final velocity v can be calculated as,

Therefore, the approximate speed of block X after the collision is 1.4 m/s.

For more details about collision, refer to the link:

link

  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 =  (M v₁ - mv₂)

let's calculate

           v =  (4 2 - 1 1)

           v = 1.4 m / s

in the same direction as the largest block (M)

magnitude and direction