Conservation of momentum and energy problem

In summary: Therefore, solution 2 where the "following" car has the faster speed is not physically possible. In summary, in an elastic collision between two identical bumper cars, with the initial speed of the leading car being 5.60 m/s and the following car being 6.00 m/s, the final speeds of the cars can either be v=5.6 m/s and w=6.0 m/s or v=6.0 m/s and w=5.6 m/s. However, the second solution is not physically possible as the following car cannot have a faster speed than the leading car.
  • #1
photon184739
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In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling in the same direction. The initial speed of the leading bumper car is 5.60 m/s and that of the trailing car is 6.00 m/s. Assuming that the mass of the drivers is much, much less than that of the bumper cars, what are their final speeds?

Name "v" the unknown final speed of the 'leading' bumper care and "w" to the unknown final speed of the 'following' bumper car.

400kg*5.60 m/s + 400kg*6.00 m/s = 400kg*v + 400kg*w (conservation of momentum)
1/2*(400kg)*(5.60 m/s)^2 +1/2*(400kg)*(6.00 m/s)^2 = 1/2*400kg*v^2 + 1/2*400kg*w^2 (conservation of energy)

5.60 m/s + 6.00 m/s = v + w
5.60 m/s + 6.00 m/s = v^2 + w^2

The above system of 2 equations has the following 2 solutions:

https://www.wolframalpha.com/input/?i=5.6+6.0=v+++w,+5.6^2+++6.0^2+=+v^2+++w^2

solution 1: v=5.6, w=6.0
solution 2: v=6.0, w=5.6

Why are there 2 solutions? What additional constraint is needed to know that the leading bumper car's final speed v = 6.0 and the following bumper car's speed w = 5.6?
 
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  • #2
All the maths does is find values of ##v## and ##w## so that the total momentum and energy are equal to their initial values. Do you notice anything about one solution for ##v## and ##w## compared to the initial values of velocity?
 
  • #3
photon184739 said:
Summary:: conservation of momentum and energy problem

Why are there 2 solutions?
Both energy and momentum are conserved either if they collide elastically or if they miss and don’t collide at all.
 
  • #4
In addition to any other responses you get, if you need Wolfram Alpha to solve two equations in two unknowns, you need to up your algebra game if you want to succeed in physics. (There's also at least one more mistake)
 
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  • #5
Vanadium 50 said:
(There's also at least one more mistake)
Indeed, although it appears to be a typo since the equations as written have complex roots. @photon184739, maths is a lot easier to understand on forums if you use LaTeX, and hence you are less likely to make typos. Check out the guide linked below the reply box.
 
  • #6
photon184739 said:
Summary:: conservation of momentum and energy problem

Why are there 2 solutions? What additional constraint is needed to know that the leading bumper car's final speed v = 6.0 and the following bumper car's speed w = 5.6?
The additional constraint is imposed by common sense. As long as there has been a collision, the following car can only be "following" if it is moving slower than the "leading" car.
 

1. What is the conservation of momentum and energy problem?

The conservation of momentum and energy problem is a fundamental principle in physics that states that the total momentum and energy of a closed system remains constant over time. This means that in any interaction or process, the total momentum and energy before the interaction is equal to the total momentum and energy after the interaction.

2. How is the conservation of momentum and energy problem applied in real-world situations?

The conservation of momentum and energy problem is applied in a wide range of real-world situations, from simple collisions between billiard balls to more complex systems such as rocket launches and planetary orbits. It is also used in industries such as transportation and engineering to design efficient and safe systems.

3. What are the equations used to calculate momentum and energy conservation?

The equations used to calculate momentum and energy conservation depend on the specific situation and the type of interaction or process involved. For example, in a collision between two objects, the equations would involve the masses and velocities of the objects. In more complex systems, additional factors such as forces and potential energy may also need to be considered.

4. How does the conservation of momentum and energy relate to Newton's laws of motion?

The conservation of momentum and energy is closely related to Newton's laws of motion, specifically the first and third laws. The first law states that an object will remain at rest or in motion at a constant velocity unless acted upon by an external force. The third law states that for every action, there is an equal and opposite reaction. These laws are fundamental to understanding and applying the conservation of momentum and energy.

5. Are there any exceptions to the conservation of momentum and energy?

The conservation of momentum and energy is a fundamental principle in physics and is considered to be a universal law. However, there are certain situations where it may not apply, such as in the presence of external forces or in cases of nuclear reactions where mass can be converted into energy. In these cases, the conservation of momentum and energy can still be applied, but additional factors must be taken into account.

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