MCAT® ExamChemical and Physical Foundations of Biological SystemsHard
A student is performing a collision experiment on a frictionless surface. A 2.0 kg cart moving at 3.0 m/s collides with a stationary 4.0 kg cart. After the collision, the 2.0 kg cart rebounds with a speed of 1.0 m/s in the opposite direction. What is the speed of the 4.0 kg cart after the collision?
- A1.5 m/s
- B2.0 m/s
- C1.0 m/s
- D2.5 m/s
Show answer & explanationAnswer & explanation
Correct answer: B. 2.0 m/s
This is a problem of conservation of momentum. Let m1 = 2.0 kg, v1i = 3.0 m/s, v1f = -1.0 m/s (opposite direction). Let m2 = 4.0 kg, v2i = 0 m/s. We need to find v2f. According to the conservation of momentum: m1v1i + m2v2i = m1v1f + m2v2f (2.0 kg)(3.0 m/s) + (4.0 kg)(0 m/s) = (2.0 kg)(-1.0 m/s) + (4.0 kg)v2f 6.0 kg·m/s + 0 = -2.0 kg·m/s + 4.0 kg·v2f 6.0 = -2.0 + 4.0v2f 8.0 = 4.0v2f v2f = 8.0 / 4.0 = 2.0 m/s. The speed of the 4.0 kg cart after the collision is 2.0 m/s.
Why the other options are wrong
- A. Incorrect calculation. This might arise from an arithmetic error or misapplication of the conservation law.
- C. Incorrect calculation. This would result if the initial momentum was conserved but the signs were misapplied.
- D. Incorrect calculation. This would result from an error in algebraic manipulation.
Conservation of Momentum
In an isolated system (no external forces), the total momentum of the system remains constant before and after a collision or interaction. Mathematically, Σ(mv)initial = Σ(mv)final.
- Applies to all types of collisions (elastic, inelastic).
- Momentum is a vector quantity, so direction is crucial (assign positive/negative signs).
- Requires no net external force acting on the system.
Memory trick: Momentum's My Constant, Before and After Contact.