MCAT® ExamChemical and Physical Foundations of Biological SystemsMedium
A student is performing a collision experiment with two carts on a frictionless track. Cart A (mass = 1.0 kg) is moving at 2.0 m/s and collides head-on with Cart B (mass = 2.0 kg), which is initially at rest. After the collision, Cart A rebounds with a speed of 0.5 m/s in the opposite direction. What is the final velocity of Cart B?
- A1.0 m/s
- B0.75 m/s
- C1.25 m/s
- D1.5 m/s
Show answer & explanationAnswer & explanation
Correct answer: C. 1.25 m/s
This is a problem involving conservation of momentum. The total momentum before the collision must equal the total momentum after the collision. Initial momentum: (1.0 kg)(2.0 m/s) + (2.0 kg)(0 m/s) = 2.0 kg·m/s. Final momentum: (1.0 kg)(-0.5 m/s) + (2.0 kg)(v_B_final). Equating them: 2.0 kg·m/s = -0.5 kg·m/s + (2.0 kg)(v_B_final). 2.5 kg·m/s = (2.0 kg)(v_B_final). v_B_final = 2.5 / 2.0 = 1.25 m/s. (Note: The negative sign for Cart A's final velocity indicates opposite direction).
Why the other options are wrong
- A. Incorrect calculation, possibly an error in algebraic manipulation or not accounting for the initial momentum of Cart A.
- B. Incorrect calculation, perhaps neglecting the direction change of Cart A or miscalculating the final momentum.
- D. Incorrect calculation, likely an arithmetic error in solving for v_B_final.
Conservation of Momentum
The total momentum of an isolated system remains constant if no external forces act on it.
- Momentum (p) = mass (m) × velocity (v).
- Total momentum before collision = Total momentum after collision.
- p_initial = p_final (m1v1_initial + m2v2_initial = m1v1_final + m2v2_final).
Memory trick: Momentum's a Keeper: It always stays the same before and after the big bang!