Combination Problem: Work and Conservation of Momentum

[PeachBanana]

Homework Statement

A 930 kg sports car collides into the rear end of a 2300 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.9 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.80, calculates the speed of the sports car at impact.

Homework Equations

Force friction = μ * Normal Force
W = Force of friction * distance

The Attempt at a Solution

The force of friction = (0.80)(9.8 m/s^2)(3230 kg)
W = 25323.2 m*kg/s^2 * 2.8 m
W = 7.0904 x 10 ^4 J

Now I am stuck. W can't equal ΔK because this is an inelastic collision.

 

[gneill]

Homework Statement

A 930 kg sports car collides into the rear end of a 2300 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.9 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.80, calculates the speed of the sports car at impact.

Homework Equations

Force friction = μ * Normal Force
W = Force of friction * distance

The Attempt at a Solution

The force of friction = (0.80)(9.8 m/s^2)(3230 kg)
W = 25323.2 m*kg/s^2 * 2.8 m
W = 7.0904 x 10 ^4 J

Now I am stuck. W can't equal ΔK because this is an inelastic collision.

Check your work calculation -- the problem statement had 2.9 m for the sliding distance.

What energy was dissipated by the friction? Where did it come from?

 

[PeachBanana]

So the energy came from the impact of the two cars. Using 2.9 m instead of 2.8 m, W = 7.3437 x 10^4 J.

W = combined masses of cars * combined velocity
W / combined masses = combined velocity

7.3437 x 10^4 J / 3230 kg = about 23 m/s

 

[gneill]

So the energy came from the impact of the two cars. Using 2.9 m instead of 2.8 m, W = 7.3437 x 10^4 J.

W = combined masses of cars * combined velocity
W / combined masses = combined velocity

7.3437 x 10^4 J / 3230 kg = about 23 m/s

No, M*V is not work (energy), it's momentum. Different units.

 

[PeachBanana]

So what that should have said was

7.3437 x 10 ^ 4 kg * m /s = 3230 kg * x

7.3437 x 10^4 kg*m/s / 3230 kg = 23 m/s

 

[gneill]

So what that should have said was

7.3437 x 10 ^ 4 kg * m /s = 3230 kg * x

7.3437 x 10^4 kg*m/s / 3230 kg = 23 m/s

The equivalent units of Joules is kg*m²/s². I'll ask again, slightly differently, which form of energy was dissipated by the friction?

 

[PeachBanana]

Was it kinetic energy? I thought I had the answer correct because my class uses online homework and it said it was right but I'm probably still not understanding the concept correctly.

 

[gneill]

Was it kinetic energy? I thought I had the answer correct because my class uses online homework and it said it was right but I'm probably still not understanding the concept correctly.

I hate to say it, but your "correct" result happened to be a fluke of the numbers in this problem

Anyways, yes, it was kinetic energy that was dissipated. That means that the two locked-together cars started with some initial velocity immediately after the impact. Can you find that velocity (using the appropriate equation for KE)?

 

[PeachBanana]

lol that's funny how that worked. So I had to call someone to help me figure this out.

So kinetic energy is not conserved during the actual collision but after the collision it is.

W = ΔK

The final kinetic energy of the two cars is zero because they do not move.

7.3437 x 10 ^ 4 J = 0.5 (3230 kg) * v^2 ( I didn't put in the negative because the question asked for magnitude only )

v = 6.7 m/s (velocity of system after collision)

Now apply conservation of momentum.

mass sports car * velocity sports car + mass suv * velocity suv = combined mass * velocity

(930 kg)(v) = (3230 kg)(6.7 m/s)

v = 23 m/s

 

[gneill]

Yup. Muuuuch better