Ask Question
7 April, 12:19

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?

+2
Answers (1)
  1. 7 April, 15:04
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers