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 14.0 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 ∘ south of east from the point of impact. How fast were the enmeshed cars moving just after the collision? How fast was car A going just before the collision?

Answer 1

This describes a perfectly inelastic collision

car A was moving only in x and Car B was moving only in y, which is convenient

Car B had y-momentum of (1400 kg) (14 m/s) = 19600 kg-m/s in the south direction, which must therefore also be the y-momentum of the Cars AB enmeshed mass

if the cars' angle of motion together was 60^o south of east, the cars' momentum must also be in this direction. That momentum has a y-component of 19600 kg-m/s, so, using trigonometry, the x-component must be

(19600 kg-m/s) / tan(65)

then, knowing the x- and y-components of the momentum, the total momentum can be found using the Pythagorean Theorem; it would be the square root of the sum of the squares of the components.

Divide that momentum by the combined mass of the cars, 3400 kg, to answer part A.

To answer part B, just recognize that the x-momentum of Cars AB is the same as the original momentum of Car A. You already know that momentum, so simply divide it by the mass of Car A, 2000 kg, to get the car's original speed