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 1500 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 afterwards. 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 60.0 degrees south of east from the point of impact.How fast were the enmeshed cars moving just after the collision?

Answer 1

Since the cars are moving perpendicularly we can look at solving using conservation of momentum applied one direction at a time.
 
So in the y or north/south direction we have car B traveling at 17 m/s, and car A traveling at 0 m/s (since it's only traveling east/west).
 
Conservation of momentum states for an inelastic collision:

m1v1i+m2v2i=(m1+m2)vf

Once we know the final velocity in the y direction we know it would be a component of the total final velocity in two dimensions:
See the attached image

Now just solve for the hypotenuse or total final velocity using trigonometry.

Answer 2

Answer:

$v = 4.2 \hat i - 7.28 \hat j$

$v = 8.41 m/s$

Explanation:

let say car A is moving with speed v in direction West to East

so its momentum is given as

$P_1 = 2000 v\hat i$

Car B is moving with velocity 17 m/s from North to South

So here we will have

$P_2 = 1500(17)(-\hat j)$

so we have initial total momentum given as

$P = 2000 v\hat i - 25500\hat j$

now the direction of the momentum is given as 60 degree South of East

so it is

$tan60 = \frac{P_y}{P_x}$

$\sqrt3 = \frac{25500}{2000v}$

$v = 7.36 m/s$

now velocity of two cars just after collision is given as

$P_1 + P_2 = (m_1 + m_2) v$

$2000(7.36)\hat i - 25500\hat j = (2000 + 1500) v$

$v = 4.2 \hat i - 7.28 \hat j$

magnitude of velocity is

$v = \sqrt{4.2^2 + 7.28^2}$

$v = 8.41 m/s$