Momentum, collision problem

[aquirk]

Homework Statement

A car with a mass of 1680 kg is traveling directly northeast (45° between north and east) at a speed of 14 m/s (31 mph), and collides with a smaller car with a mass of 1300 kg that is traveling directly south at a speed of 13 m/s (29 mph). The two cars stick together during the collision. With what speed and direction does the tangled mess of metal move right after the collision?
Magnitude __________ m/s
Direction _________ degrees South of east

Homework Equations

m1v1i+m2v2i = v1(m1+m2)

The Attempt at a Solution

i tried plugging my values into this equation to figure out the answer but it doesn't seem to give me the right answer. I am thinking maybe I am doing it wrong becuase there is an angle involved but I am not sure.

 

[AEM]

Homework Statement

A car with a mass of 1680 kg is traveling directly northeast (45° between north and east) at a speed of 14 m/s (31 mph), and collides with a smaller car with a mass of 1300 kg that is traveling directly south at a speed of 13 m/s (29 mph). The two cars stick together during the collision. With what speed and direction does the tangled mess of metal move right after the collision?
Magnitude __________ m/s
Direction _________ degrees South of east

Homework Equations

m1v1i+m2v2i = v1(m1+m2)

The Attempt at a Solution

i tried plugging my values into this equation to figure out the answer but it doesn't seem to give me the right answer. I am thinking maybe I am doing it wrong becuase there is an angle involved but I am not sure.

You're on the right track using conservation of momentum, but you seem to have forgotten that momentum is a vector quantity and therefore you will have an equation for the x-components and an equation fro the y-components. So, yes you need to make use of the angle.

 

[nlightner]

I can provide a response to this problem by using the principles of momentum and conservation of momentum. In this situation, the total momentum of the system before the collision is equal to the total momentum after the collision. This can be represented by the equation m1v1i + m2v2i = (m1+m2)v, where m1 and m2 are the masses of the two cars, v1i and v2i are their initial velocities, and v is the velocity of the tangled mess of metal after the collision.

Using this equation, we can solve for the final velocity (v) by plugging in the given values:
(1680 kg)(14 m/s) + (1300 kg)(-13 m/s) = (1680 kg + 1300 kg)v
v = (23520 kg m/s + (-16900 kg m/s)) / (2980 kg)
v = 2.5 m/s

The magnitude of the final velocity is 2.5 m/s. To determine the direction, we can use the concept of vector addition. The initial velocities of the two cars can be represented as vectors, with the northeast direction being 45° and the south direction being 180°. The final velocity vector can be found by adding these two vectors together. This can be done using trigonometric functions, specifically the law of cosines.

Cosine law: c^2 = a^2 + b^2 - 2abcosC
Where c is the final velocity, a is the initial velocity of the first car, b is the initial velocity of the second car, and C is the angle between the two initial velocities.

Plugging in the values:
c^2 = (14 m/s)^2 + (13 m/s)^2 - 2(14 m/s)(13 m/s)cos(180°-45°)
c = 12.4 m/s

Therefore, the final velocity vector has a magnitude of 12.4 m/s and a direction of (180°-45°) = 135° south of east.

In conclusion, the tangled mess of metal will have a final velocity of 2.5 m/s in the direction of 135° south of east after the collision.