Conservation of Momentum - Collisions

[doug1]

Homework Statement

A truck (4 000 kg) runs a red light and enters an intersection traveling at 81 km/h [E]. The truck collides with a car (2
000 kg) that was traveling at 54 km/h [N20oW]. Immediately after the collision the truck was traveling at 60 km/h [N
30o E]. Determine the velocity vector of the car immediately after the collision.

Homework Equations

I used the law of conservation of momentum. I used the concept that momentum is conserved in each component direction.

The Attempt at a Solution

The answer that I got was a velocity of approx 99 km/h [E32S]. Can anyone confirm this?

 

[mfb]

This would violate energy conservation, unless the truck or the car have some active system to kick away other vehicles.
I don't understand the notation for the direction, but I would expect that momentum is not conserved either.

 

[doug1]

The notation [N30E] means 30 degrees east of north for example.

Momentum is conserved in this question, but energy is not

 

[mfb]

I agree that kinetic energy does not have to be conserved, but kinetic energy cannot increase in the process.

Working in units of 1000kg*km/h:
to east:
truck initial 4*81=364
car initial -2*54*sin(20°)
truck final 4*60*sin(30°)
car final ~207 or 104 km/h
Hmm... looks wrong.

to north:
truck initial 0
car initial 2*54*cos(20°)
truck final 4*60*cos(30°)
car final -106 or -53km/h

Total energy initially: 2*81^2+1*54^2=16000 (in 1000kg*(km/h)^2)
Total energy finally: 2*60^2+1*(104^2+53^2) = 20800
Clearly this cannot happen in a car accident. The same problem appears with your result of 99km/h total speed as well.

 

[Nickie]

Your solution using the law of conservation of momentum is correct. The final velocity of the car can be calculated using the equation:

m1v1 + m2v2 = (m1 + m2)vf

where m1 and v1 are the mass and initial velocity of the truck, m2 and v2 are the mass and initial velocity of the car, and vf is the final velocity of the combined system after the collision.

Plugging in the values given in the problem, we get:

(4000 kg)(81 km/h) + (2000 kg)(54 km/h) = (4000 kg + 2000 kg)vf

Solving for vf, we get:

vf = (4000 kg)(81 km/h + (2000 kg)(54 km/h)) / (4000 kg + 2000 kg)

vf = (324000 kg*km/h + 108000 kg*km/h) / 6000 kg

vf = (432000 kg*km/h) / 6000 kg

vf = 72 km/h

Since the final velocity is in the direction of the combined momentum, we can use trigonometry to find the direction. The angle of the final velocity can be found using:

tanθ = (m1v1y + m2v2y) / (m1v1x + m2v2x)

where v1x and v1y are the x and y components of the initial velocity of the truck, and v2x and v2y are the x and y components of the initial velocity of the car.

Plugging in the values, we get:

tanθ = ((4000 kg)(81 km/h)(cos 0°) + (2000 kg)(54 km/h)(cos 20°)) / ((4000 kg)(81 km/h)(sin 0°) + (2000 kg)(54 km/h)(sin 20°))

tanθ = (324000 kg*km/h + 94800 kg*km/h) / (0 kg*km/h + 37800 kg*km/h)

tanθ = (418800 kg*km/h) / (37800 kg*km/h)

tanθ = 11.1

Solving for θ, we get:

θ = tan^-1(11.1) = 83.2°

Therefore, the final velocity of the car