A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface.A crane suddenly drops 858 kg. Of ore vertically into the cart.How fast does the cart move just after being loaded with the ore?
A)6.30 m/s
B)3.80 m/s
C)4.20 m/s
D)5.70 m/s

Question

Quizplus · Accepted Answer

This problem can be solved using the principle of conservation of momentum. The total momentum before the ore is dropped is equal to the total momentum after the ore is dropped. Before the ore is dropped, the momentum is $p_{initial} = m_{cart} \times v_{cart} = 1200 \, \text{kg} \times 10.8 \, \text{m/s}$. After the ore is dropped, the total mass is $m_{total} = m_{cart} + m_{ore} = 1200 \, \text{kg} + 858 \, \text{kg}$. The final velocity $v_{final}$ can be found by setting the initial momentum equal to the final momentum: $1200 \times 10.8 = (1200 + 858) \times v_{final}$. Solving for $v_{final}$ gives $v_{final} = 6.30 \, \text{m/s}$.

A 1200-Kg Ore Cart Is Rolling at 10