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?

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?

Anonymous · Accepted Answer

i have not idea

drwls · Answer

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