Question

Two cars collide at an intersection. Car A, with a mass of 2000 kg , is going from west to east, while car B , of mass 1400 kg , is going from north to south at 17.0 m/s . As a result of this collision, the two cars become enmeshed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65.0 ∘ south of east from the point of impact.

1. How fast were the enmeshed cars moving just after the collision?

2. How fast was car A going just before the collision?

Answer 1

ma = 2000 Kg
mb = 1400 Kg

va = ?
vb = 17 m/s

Inital Momentum in X direction = ma * Va
Inital Momentum in Y direction = mb * Vb


Let the Final Velocity = Vf @ 65^o

Final Velocity in X direction = Vf cos(65)
Final Velocity in Y direction = Vf sin(65)

Final Momentum in X direction = (ma + mb) * Vf cos(65)
Final Momentum in Y direction = (ma + mb) * Vf sin(65)

Using Momentum Conservation

In X direction =
ma * Va =  (ma + mb) * Vf cos(65)
2000 * Va = (2000 + 1400) * Vf * cos(65) --------1

In Y direction =
mb * Vb = (2000 + 1400) * Vf * sin(65)
1400 * 17 = 3400 * Vf * sin(65)
Vf = 7.72 m/s

Substituing Value in eq 1
2000 * Va = (2000 + 1400) * 7.72 * cos(65)
Va = 5.55 m/s

1) Final Velocity of Cars Just after the Collision, Vf = 7.72 m/s
2) Speed of Car A just before Collision, Va = 5.55 m/s