A 5.5 kg cart moving horizontally on a frictionless surface initially at 4.2 ms-1, has a 2.9 kg load dropped vertically in it, increasing the cart's total mass.The duration of the drop took 1.2 s.Based on the conservation of momentum, calculate the final velocity of the now combined cart and load.

Question

BlackTom AI · Accepted Answer

Conservation of horizontal momentum applies because the load is dropped vertically and there are no external horizontal forces on the cart+load system on a frictionless surface; therefore the system's horizontal momentum is conserved.

1) Relevant concepts
- Momentum p = m v; for an isolated system (no external horizontal impulse) total horizontal momentum before = after.
- The dropped load has zero initial horizontal velocity (dropped vertically), so it contributes no initial horizontal momentum.
- The time duration of the drop (1.2 s) does not change the momentum result because no external horizontal force acts; the collision is inelastic (kinetic energy not conserved), but momentum is.

2) Step-by-step reasoning and calculations
- Initial cart mass m1 = 5.5 kg, initial speed v1 = 4.2 m/s.
- Dropped load mass m2 = 2.9 kg, initial horizontal speed v2 = 0 m/s.
- Total initial horizontal momentum:
  p_initial = m1 * v1 + m2 * v2 = (5.5 kg)(4.2 m/s) + (2.9 kg)(0 m/s)
  p_initial = 23.1 kg·m/s + 0 = 23.1 kg·m/s.
- After the load lands, combined mass M = m1 + m2 = 5.5 kg + 2.9 kg = 8.4 kg.
- By momentum conservation: p_initial = M * v_final.
  So v_final = p_initial / M = 23.1 kg·m/s ÷ 8.4 kg.
- Compute v_final:
  23.1 ÷ 8.4 = 2.75 m/s.

3) Why this is correct
- Momentum accounting exactly yields 2.75 m/s because the added mass had no initial horizontal momentum; the cart must slow so the combined momentum equals the original 23.1 kg·m/s. The time of contact does not alter the final momentum because horizontal external impulse is zero.

4) Why the other choices are incorrect
- a (3.30 m/s): would come from dividing 23.1 by ~7 kg, not the actual combined mass 8.4 kg.
- b (8.88 m/s) and d (7.97 m/s): both exceed the initial speed (4.2 m/s), which would require additional horizontal momentum being added—impossible here since the load had zero horizontal momentum.

Conclusion: The final velocity is 2.75 m/s, which matches the provided correct answer (c. 2.75 m/s).

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