Two cars collide at an intersection. car a , with a mass of 2000kg , is going from west to east, while car b , of mass 1400kg , is going from north to south at 14.0m/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 65.0 degrees south of east from the point of impact.

Complete 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 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65∘ south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car A going just before the collision?

a) 6.36 m/s b) 4.57 m/s

Explanation:

a) Assuming no external forces acting during the collision, total momentum must be conserved.

As momentum is a vector, we can decompose it along two directions perpendicular each other.

Just for convenience, we choose as our x-axis to the W-E direction, and as our y-axis, the direction N-S.

If we know that total momentum must be conserved, same must be true for both components, px and py.

Applying the information provided (both cars become enmeshed after the collision, moving at an angle of 65º south of east from the point of impact), we have:

px = ma * va = (ma+mb) * vab * cos 65º  (1)

py = mb * vb = (ma + mb) * vab * sin 65º (2)

Replacing by the values of ma, mb, and sin 65º, we can solve for vab, as follows:

vab = 1,400 kg* 14.0 m/s / (3,400 Kg * sin 65º) = 6.36 m/s

b) Replacing vab from above in (1), and solving for va, we have:

va = 3,400 kg* 6.36 m/s* cos 65º / 2,000 Kg = 4.57 m/s

increase masses, decrease separation