A person stands in stationary canoe and throws a 5.00-kg stone with a velocity of 8.00 m/s at an angle of 30 degrees above the horizontal. The person and canoe have a combined mass of 105 kg. Ignoring air resistance and effects of the water, find the horizontal recoil velocity (magnitude and direction) of the canoe.
2007-01-11 09:05:39 · 2 answers · asked by Someone 1 in Science & Mathematics ➔ Physics
(eqn 1): (mv)1 = (mv)2 → v2 = (mv)1/m2.
The horizontal component of momentum of the stone is 5*8*√3/2. Divide this by m2, the mass of the canoe & person, and you have
v2 = .33 m/s
2007-01-11 09:17:39 · answer #1 · answered by Steve 7 · 1⤊ 0⤋
You can't ignore the water because it is preventing the canoe from going backwards and DOWN at a 30 degree angle. However: The horizontal momentum will be opposite the direction the 5kg mass was thrown, it will be equal to 5kg x 8m/sec x cos30 degrees.
40kg-m/sec = 105 kg x X m/sec.
Trig tables, anyone?
2007-01-11 17:20:13 · answer #2 · answered by Richard S 6 · 0⤊ 0⤋