Question
A 66.0 kg man is ice-skating due north with a velocity of 5.70 m/s when he collides with a 36.0 kg child. The man and child stay together hand have a velocity of 2.20 m/s at an angle of 33.0° north of east immediately after the collision. What was the velocity of the child just before the collision?
Direction north of east?
Direction north of east?
Answers
drwls
Apply the law of conservation of momentum and subtract the man's momentum from the combined momentum after collision.
vx = child's intial moentum east
vy = child's initial mometum north
(2.2)(102)*sin33 - (66)*(5.7) = 36*vy
(2.2)(102)*cos33 - 0 = 36*vx
The two e
vx = child's intial moentum east
vy = child's initial mometum north
(2.2)(102)*sin33 - (66)*(5.7) = 36*vy
(2.2)(102)*cos33 - 0 = 36*vx
The two e
Anonymous
i have not idea