Car A, with mass of 1735kg, was traveling north at 45.5km/h and car B, with a mass of 2540kg, was traveling west at 37.7km/h when they collided at an intersection. if the cars stuck together after the collision, what was their combined momentum? Was the collision elastic or inelastic?

It is NOT an elastic collision, they stuck together. Combined momentum= sum of initial momentums. Do this as vector addition.

Since the directions are north and west you directly use the Pythagoras theorem to find the resultant speed?? and the tan of the two two get the angle??

To find the combined momentum of the two cars after the collision, we need to calculate the momentum of each car individually and then add them together.

The momentum of an object is given by the formula p = m * v, where p is momentum, m is mass, and v is velocity.

Car A:
Mass (m1) = 1735 kg
Velocity (v1) = 45.5 km/h

First, we need to convert the velocity of Car A from km/h to m/s:
v1 = 45.5 km/h * (1000 m/1 km) * (1 h/3600 s) = 12.64 m/s

Now, we can calculate the momentum of Car A:
p1 = m1 * v1 = 1735 kg * 12.64 m/s = 21,906.4 kg·m/s

Car B:
Mass (m2) = 2540 kg
Velocity (v2) = 37.7 km/h

We need to convert the velocity of Car B from km/h to m/s:
v2 = 37.7 km/h * (1000 m/1 km) * (1 h/3600 s) = 10.47 m/s

Now, we can calculate the momentum of Car B:
p2 = m2 * v2 = 2540 kg * 10.47 m/s = 26,568.8 kg·m/s

The combined momentum of Car A and Car B after the collision is given by:
p_combined = p1 + p2 = 21,906.4 kg·m/s + 26,568.8 kg·m/s = 48,475.2 kg·m/s

The collision is inelastic because the cars stick together after the collision. In an elastic collision, the objects would separate after the collision.

Therefore, the combined momentum of the cars after the collision is 48,475.2 kg·m/s, and the collision is inelastic.

Given:

Mass of Car a (Ma)= 1735kg, Vai= 45.5 km/hour= 12.64m/s

Mass of Car b (Mb) = 2540kg, Vbi= 37.7km/hour= 10.47m/s

pi = pf

MaVai+ MbVbi = Vf (m1+m2)

(175)(12.64) + (2540)(10.47) = pf

pf = 28805.8 J

Elastic collision because Pi=Pf