Question

A 1200 kg car traveling 45.0o south of east at 18.0 m/s collides with a 850 kg car traveling north at 23.0 m/s. The cars stick together. What is the velocity of the wreckage just after the collision?

Answer 1

components of velocity V1x =V cos theta

V1x = 18* cos 45 = 12.72 m/s

V1y = - 12 sin 45 = -12.72 m/s

V2x = 0

V2y = 23 m/s

so use fro conservation of momentum

m1v1x + m2v2x = (m1+m2) Vx

Vx = (1200 * 12.72) /(1200 + 850)

Vx = 7.44 m/s

Vy = -(1200 * 12.72) + (850 *23)/(1200+850)

Vy = 2.1 m/s

net speed V^2 = Vx^2 + Vy^2

V^2 = 7.44^2 + 2.1 ^2

V = 7.72 m/s