A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface.A crane suddenly drops $858 \mathrm {~kg}$ 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

The conservation of momentum principle applies here. Before the ore is dropped, 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 momentum is the combined momentum of the cart and the ore. The formula for momentum is $p = mv$, where $m$ is mass and $v$ is velocity.Before the ore is dropped:- Momentum of the cart = $1200 \, \text{kg} \times 10.8 \, \text{m/s} = 12960 \, \text{kg} \cdot \text{m/s}$.After the ore is dropped:- The total mass = $1200 \, \text{kg} + 858 \, \text{kg} = 2058 \, \text{kg}$.- Let $v'$ be the velocity of the cart after the ore is dropped.- The total momentum after the ore is dropped must equal the total momentum before the ore is dropped, so $2058 \, \text{kg} \times v' = 12960 \, \text{kg} \cdot \text{m/s}$.Solving for $v'$:$$v' = \frac{12960}{2058} \approx 6.30 \, \text{m/s}$$.Therefore, the cart moves at 6.30 m/s just after being loaded with the ore.

A 1200-Kg Ore Cart Is Rolling at 10 858 kg858 \mathrm {~kg}858 kg

A 1200-Kg Ore Cart Is Rolling at 10 $858 \mathrm {~kg}$