Question

A 48.5-kg skater is traveling due east at a speed of 2.55 m/s. A 71.0-kg skater is moving due south at a speed of 7.30 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

Using momentum balance in X-direction:

Pix = Pfx

m1*u1x + m2*u2x = (m1 + m2)*Vx

Vx = (m1*u1x + m2*u2x)/(m1 + m2)

u1x = 2.55 m/sec

u2x = 0

Vx = (48.5*2.55 + 71*0)/(48.5 + 71) = 1.035 m/sec

Similarly in y-direction

Piy = Pfy

m1*u1y + m2*u2y = (m1 + m2)*Vy

Vy = (m1*u1y + m2*u2y)/(m1 + m2)

u1y = 0 m/sec

u2y = -7.30 m/sec

Vy = (48.5*0 - 71*7.30)/(48.5 + 71) = -4.337 m/sec

So,

Angle = arctan (-4.337/1.035) = -76.57 deg OR 76.57 deg below x-axis

|V| = sqrt (1.035^2 + 4.337^2) = 4.458 m/sec