A 70 kg boy sits in a 30 kg canoe at rest on the water. He holds two cannonballs (huh. okay.) each of mass 10 kg. He picks them up and throws them both together over the stern of his canoe (velocity = 5.0 m/s). A 50 kg girl sits in a 50 kg canoe, also at rest on the water. She also holds two 10 kg cannonballs. However she throws them over the stern of her canoe one at a time, each ball leaving her hands with a velocity of 5.0 m/s. Assuming negligible friction between the water and the canoe (a poor assumption), calculate the final velocity of each canoe.
Yeah, long. Can anyone just walk me through the process of solving this? Thanks.
2007-04-25 16:04:51 · 2 answers · asked by Ariel 2 in Science & Mathematics ➔ Physics
Momentum is conserved.
How much momentum has left the canoe in each case? (m * v) [cannonballs]
Therefore, in order to conserve momentum, what must be true of the momentum of the canoe after the cannonballs have been thrown? (- m*v) [canoe]
Neglecting friction with the water means that there is no other force acting on the canoe in the horizontal plane and therefore no other impulse in the horizontal directions. The lapse in time between throwing the cannonballs for the girl would not affect the momentum transfer to the canoe in any case, but the lack of water friction means that there is no counteracting impulse even though her events take longer than the events of the boy.
It's that simple...
2007-04-25 16:17:17 · answer #1 · answered by AnswerMan 4 · 0⤊ 0⤋
Conservation of momentum.
2 * 10 kg * 5 m/s = 100 kgm/s
100kg * velocity = 100 kgm/s
Since boy and boat mass 100 kg and momentum is conserved the boat will end up moving at 1 m/s.
Since girl and boat mass 100 kg and momentum is conserved the boat will end up moving at 1 m/s.
It does not matter if she throws them one at a time or both together.
2007-04-25 23:11:42 · answer #2 · answered by J C 5 · 0⤊ 0⤋