Combination Problem: Work and Conservation of Momentum

In summary: That's the velocity of the cars just after the collision. Now how does that relate to the velocity before?And, you know, now that I look at it, I think you may have been right before. The online homework system may have been wrong. :uhh:We can check using the NOW CORRECTED sliding distance of 2.9 m.We have the final velocity. How can we get the initial velocity?In summary, the police officer calculates the speed of the sports car at impact by finding the energy dissipated by friction during the collision and using conservation of momentum to find the final velocity of the locked-together cars. The initial velocity before the collision can then be found using the
  • #1
PeachBanana
191
0

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.
 
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  • #2
PeachBanana said:

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?
 
  • #3
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
 
  • #4
PeachBanana said:
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.
 
  • #5
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
 
  • #6
PeachBanana said:
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?
 
  • #7
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.
 
  • #8
PeachBanana said:
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 :smile:

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)?
 
  • #9
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
 
  • #10
Yup. Muuuuch better :wink:
 

Related to Combination Problem: Work and Conservation of Momentum

1. What is a combination problem involving work and conservation of momentum?

A combination problem involving work and conservation of momentum is a physics problem that requires the use of both work and momentum equations to solve. These types of problems typically involve objects undergoing a change in velocity due to the application of a force, and also involve the conservation of momentum principle, which states that the total momentum of a system remains constant unless acted upon by an external force.

2. How do I approach solving a combination problem involving work and conservation of momentum?

First, identify the objects involved in the problem and their initial and final velocities. Then, use the work equation (W = Fd) to find the work done on the objects. Next, use the conservation of momentum principle (mv = mv) to set up equations for the initial and final momenta of the objects. Finally, use algebra to solve for any unknown variables.

3. What are some common pitfalls when solving combination problems involving work and conservation of momentum?

One common pitfall is forgetting to consider the direction of the forces and velocities involved. It is important to use vector notation and account for both the magnitude and direction of the forces and velocities to accurately solve the problem. Another pitfall is not properly setting up the conservation of momentum equations, which can lead to incorrect solutions.

4. Can conservation of energy be applied to combination problems involving work and conservation of momentum?

Yes, conservation of energy can also be applied to these types of problems. In addition to using the work equation and conservation of momentum principle, you can also use the energy conservation principle (KE + PE = constant) to solve for unknown variables. This is particularly useful when the problem involves changes in the objects' heights or gravitational potential energy.

5. How can I check my work and make sure my solution is correct?

To check your work and ensure that your solution is correct, you can first double check your calculations and equations for any errors. You can also try solving the problem using a different approach to see if you get the same result. Additionally, you can compare your solution to the expected answer or use a physics simulator to see if your solution matches the predicted outcome.

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