A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s. A second block, sliding at a faster 4.0 m/sm/s, collides with the first from behind and sticks to it. The final velocity of the combined blocks is 1.8 m/sm/s?

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block=?

v₁ = velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃ = final velocity of combined block=1.8m/s

applying the formula above

(2.2 * 1.2) + (m₂ * 4) = (2.2 + m₂) * 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

m₂ = 0.6kg