A 56.0-kg skater is traveling due east at a speed of 2.80 m/s. A 77.0-kg skater is moving due south at a speed of 4.40 m/s. They collide and hold on to each other after the collision, managing to move off at an angle south of east, with a speed of vf. Find (a) the angle and (b) the speed vf, assuming that friction can be ignored.

Question

A 56.0-kg skater is traveling due east at a speed of 2.80 m/s. A 77.0-kg skater is moving due south at a speed of 4.40 m/s. They collide and hold on to each other after the collision, managing to move off at an angle  south of east, with a speed of vf. Find (a) the angle  and (b) the speed vf, assuming that friction can be ignored.

bobpursley · Answer

The momentum is conserved. Therefore the final momentum has E and S as before.

Momentum E=56*2.8
momentum S=4.4*77

Use these to get the angle

arctanTheta=MomentumS/momentumE
where theta is the angle S of E.