Kinetics Problem: Head-On Collision Between Two Cars

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SUMMARY

The discussion focuses on solving a kinetics problem involving a head-on collision between two cars of equal mass. The initial kinetic energy (Ki) is calculated as 512.5J, with 75% of the energy lost, resulting in a final kinetic energy (Kf) of 128.125J. The velocities of the two cars are derived from the equations of momentum conservation and energy balance, leading to the conclusion that the first car's velocity is -8.5 m/s and the second car's velocity is 13.5 m/s. The participants emphasize the importance of using symbolic representation for the equations to solve for the unknown velocities accurately.

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  • Understanding of kinetic energy calculations (Ki and Kf)
  • Familiarity with momentum conservation principles
  • Ability to solve systems of equations with two unknowns
  • Knowledge of quadratic equations and their applications in physics
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  • Learn about the center of momentum frame in collision analysis
  • Study the derivation and application of the quadratic formula in physics problems
  • Explore energy loss mechanisms in collisions, particularly in automotive contexts
  • Investigate advanced topics in collision dynamics, such as elastic and inelastic collisions
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Graador
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Homework Statement
If two cars of equal mass hit each other head on, car A travelling at 25 m/s towards the right and car B travelling at 20 m/s to the left, what will be the speed of each car post-collision if 75% of the energy is lost in bending the metal of the cars.
Relevant Equations
m1v1 + m2v2 = m1V1 + m2V2
e = Kf/Ki
Since the mass of both vehicles is the same, it's possible to calculate Ki which happens to be 512,5J and from there, multiply it by 0,25 since 75% of the energy is gone and I end up with 128,125J.

Now my problem is that for the velocity, I have: 25,0 + -20,0 = V1 + V2 which is 5,00 = V1 + V2

When I take Kf, I have: 128,125 = V1(sq)/2 + V2(sq)/2 so 256,25 = V1(sq) + V2(sq)
If I try and get the square root, I end up with 16,0 = V1 + V2 which doesn't fit with the first equation which said 5,00 = V1 + V2

After some trial and error, I found that the velocity of the first car is -8,5 m/s and the second 13,5 m/s which when added up, equal 5,00 and when their squares are added up and divided by two, equal 127,25 which is close enough. I just don't understand how to find both velocities from the actual equations themselves because all I find is a quadratic formula that gives me wrong answers.
 
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You will have to solve a system of two equations and two unknowns. This is best done when you first write down equations using symbols, then substitute the numbers after solving the system symbolically. Your energy balance equation seems incorrect.
 
kuruman said:
You will have to solve a system of two equations and two unknowns. This is best done when you first write down equations using symbols, then substitute the numbers. Your energy balance equation seems incorrect.
So if my energy equation is correct and my first equation is also correct (hard to calculate that one wrong), what am I supposed to do? Is there an equation that I have not considered? I already tried swapping things from on formula to another but they don't equal the same thing for V1 + V2 so that didn't work.
 
Write down two equations using symbols for the the different quantities that you have. Use different symbols for different quantities. The first equation is momentum conservation, the second equation is the energy balance equation which says that the initial kinetic energy of the cars is equal to the final kinetic energy of the cars plus the energy loss in bending metals. The two unknowns are the final velocities of the cars.
 
Graador said:
755 of the energy is lost in bending
...
calculate Kf which happens to be 512,5J
...
256,25 = V1(sq) + V2(sq)
If I try and get the square root, I end up with 16,0 = V1 + V2
I take it that should be 75%.
...
You mean Ki, and since you do not know the mass it is 512,5m ms-2, where m is the mass of each car.
...
##\sqrt{x^2+y^2}\neq x+y## (unless one or both is zero).
 
Graador said:
So if my energy equation is correct and my first equation is also correct (hard to calculate that one wrong), what am I supposed to do? Is there an equation that I have not considered? I already tried swapping things from on formula to another but they don't equal the same thing for V1 + V2 so that didn't work.
This may be simpler in the centre of momentum frame, where you have a symmetry of the motion.
 

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