A 0.80 kg object moving to the right at 3.0 m/s collides head-on with a 1.2 kg object moving to the left at 2.0 m/s. After the collision, the 0.80 kg object rebounds to the left with a speed of 1.0 m/s. What is the velocity (magnitude and direction) of the 1.2 kg object immediately after the collision?
Show the answer and explanation
Correct answer
A. 0.67 m/s to the right
Principle or equation
In an isolated system, total momentum before collision equals total momentum after collision. Momentum is a vector quantity: p = mv, with direction sign convention.
Why this answer is correct
Take right as positive. Initial momentum = (0.80)(3.0) + (1.2)(-2.0) = 2.4 - 2.4 = 0. After collision, momentum of 0.80 kg object = (0.80)(-1.0) = -0.80 kg·m/s. Let v be the velocity of the 1.2 kg object. Total final momentum = -0.80 + 1.2v. Set equal to 0: -0.80 + 1.2v = 0 => v = 0.80/1.2 = 0.67 m/s. Positive means to the right.
Example
A 2 kg ball moving at 3 m/s right hits a 1 kg ball moving at 2 m/s left. After collision, the 2 kg ball moves left at 1 m/s. Find the 1 kg ball's velocity. Initial momentum = 6 - 2 = 4. Final: -2 + 1*v = 4 => v = 6 m/s right.
This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.