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?

Question

Quizplus · Accepted Answer

Momentum is conserved:$$(1200	ext{ kg})(10.8	ext{ m/s})+(858	ext{ kg})(0	ext{ m/s})=(1200	ext{ kg}+858	ext{ kg})v$$$$v=\frac{(1200	ext{ kg})(10.8	ext{ m/s})}{2058	ext{ kg}}=\boxed{6.30	ext{ m/s}}$$

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

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