A 0.60 kg object moving at 4.0 m/s to the right collides head-on with a 0.40 kg object moving at 6.0 m/s to the left. After the collision, the 0.60 kg object moves to the left at 2.0 m/s. What is the velocity of the 0.40 kg object immediately after the collision? (Take right as positive.)
Show the answer and explanation
Correct answer
B. 3.0 m/s to the right
Principle or equation
In an isolated system, total momentum is conserved: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f.
Why this answer is correct
Let right be positive. Initial momentum: (0.60)(4.0) + (0.40)(-6.0) = 2.4 - 2.4 = 0. After collision, v₁f = -2.0 m/s. Then 0 = (0.60)(-2.0) + (0.40)v₂f => -1.2 + 0.40 v₂f = 0 => v₂f = 3.0 m/s. Positive means to the right.
Example
If a 1 kg ball at 2 m/s hits a 2 kg ball at rest and the 1 kg ball rebounds at 1 m/s, then 1(2)+0 = 1(-1)+2v => v = 1.5 m/s.
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