Why do two cars rolling downhill reach the finish line together without friction?

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keemosabi
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Homework Statement


Two cars start 10 meters up a hill. Car A has a mass of 9 grams and Car B has a mass of 2 grams. Additionally, both cars must travel the same distance to the finish line.
a) If friction is ignored which car will win the race?
b) If friction is not ignored, which car will win the race.


Homework Equations


u + k = u + k


The Attempt at a Solution


a) Both cars will finish at the same time because at any point on the track, both cars will have equal velocities.
b) Car B will finish first because it is lighter, so it will have less of a friction force that will slow it down less. As a result, Car B will move faster and thus reach the finish line first.

Is this right?
 
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Discussion

What do the equations say?

With friction or without what is the dependence on mass?
 
LowlyPion said:
What do the equations say?

With friction or without what is the dependence on mass?

With mgh = 1/2mv^2 the mass would cancel out, so is there so dependence on mass? This means the cars finish at the same time, right?

Would friction play any role, or no?
 
keemosabi said:
With mgh = 1/2mv^2 the mass would cancel out, so is there so dependence on mass? This means the cars finish at the same time, right?

Would friction play any role, or no?

Write out the equation for the case with friction.
 
LowlyPion said:
Write out the equation for the case with friction.
Is the first part correct?

Fnet = ma
uN = ma
n(mg)=ma
ng=a

So the mass is canceled out when friction is and is not ignored?
 
keemosabi said:
Is the first part correct?

Fnet = ma
uN = ma
n(mg)=ma
ng=a

So the mass is canceled out when friction is and is not ignored?

The mass canceled out in the first one, so yes it would have to be correct wouldn't it?

What happens in the case where you have

Potential at the top = kinetic at the bottom + work lost to friction ?
 
LowlyPion said:
The mass canceled out in the first one, so yes it would have to be correct wouldn't it?

What happens in the case where you have

Potential at the top = kinetic at the bottom + work lost to friction ?
Friction will do more work on the heavier car, so more energy will be lost. As a result, there will be less kinetic energy to contribute to the velocity of the heavier car than the lighter car.
 
keemosabi said:
Friction will do more work on the heavier car, so more energy will be lost. As a result, there will be less kinetic energy to contribute to the velocity of the heavier car than the lighter car.

What does the equation say?

m*g*h = 1/2*m*v2 - μ*m*g*cosθ*d
 
LowlyPion said:
What does the equation say?

m*g*h = 1/2*m*v2 - μ*m*g*cosθ*d
Can you cancel the mass out of each term?
 
keemosabi said:
Can you cancel the mass out of each term?

That would be what it looks like to me.

And d must be the same according to the problem.