Question

A 54.0-kg skater is traveling due east at a speed of 2.60 m/s. A 74.5-kg skater is moving due south at a speed of 6.65 m/s. They collide and hold on to each other after the collision, managing to move off at an angle θ south of east, with a speed of vf. Find the following.

(a) the angle θ =________ °

(b) the speed v_f, assuming that friction can be ignored =_____________ m/s

Answer 1

m1 = 54 kg     v1 = 2.6 i m/s

m2 = 74.5 kg     v2 = -6.65 j m/s

after collison

final velocity vf = vx i - vy j

from momentm conservation

momentum before collison momentum after collision

m1*v1 + m2*v2 = (m1+m2)*vf

(54*2.6i) - (74.5*6.65 j) = (54+74.5)*(vx i - vy j)

140.4i - 495.425 j = 128.5*(vx i - vyj)

vx = 140.4/128.5 = 1.09 m/s

vy = 495.425/128.5 = 3.85 m/s

angle theta = tan^-1(vy/vx)

theta = 74.2 degrees

(b) magnitude = sqrt(vx^2 + vy^2 ) = sqrt(1.09^2+3.85^2) = 4 m/s