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) 4.20 m/s
B) 3.80 m/s
C) 5.70 m/s
D) 6.30 m/s

Question

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)  4.20 m/s
B)  3.80 m/s
C)  5.70 m/s
D)  6.30 m/s

Quizplus · Accepted Answer

The speed of the cart after being loaded can be found using the principle of conservation of momentum. Initially, the momentum of the system (cart + ore) is the momentum of the cart since the ore is not yet in the cart. After the ore is dropped into the cart, the total mass of the system increases, but the total momentum of the system remains the same (assuming no external forces act on the system). The initial momentum is $p_{initial} = m_{cart} \times v_{cart} = 1200 \, \text{kg} \times 10.8 \, \text{m/s}$. After the ore is added, the total mass is $m_{total} = m_{cart} + m_{ore} = 1200 \, \text{kg} + 858 \, \text{kg}$. Using the conservation of momentum, $p_{initial} = p_{final}$ or $m_{cart} \times v_{cart} = (m_{cart} + m_{ore}) \times v_{final}$. Solving for $v_{final}$, we find $v_{final} = \frac{1200 \times 10.8}{1200 + 858} = 6.3 \, \text{m/s}$.

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