A 1200 kg ore cart is rolling at 10.8 m/s across a flat surface.A crane dumps 691 kg of ore (vertically)into the cart.How fast does the cart move after being loaded with ore? Assume that frictional forces may be neglected.

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Quizplus · Accepted Answer

Momentum is conserved, so the total momentum before the ore is dumped is equal to the total momentum after the ore is dumped.$$(1200	ext{ kg})(10.8	ext{ m/s})+(691	ext{ kg})(0	ext{ m/s})=(1200	ext{ kg}+691	ext{ kg})(v)$$$$v=\frac{(1200	ext{ kg})(10.8	ext{ m/s})+(691	ext{ kg})(0	ext{ m/s})}{1200	ext{ kg}+691	ext{ kg}}=6.8	ext{ m/s}$$

A 1200 Kg Ore Cart Is Rolling at 10