Why does the heavier car stop sooner?

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SUMMARY

The discussion centers on the stopping distances of two cars, one heavier than the other, on snowy or icy roads. It establishes that while momentum (p = mv) suggests the lighter car should stop sooner, the frictional force, which is independent of weight but dependent on material composition, plays a crucial role. The kinetic energy (KE = (m * v * v)/2) of each car and the relationship between mass and friction are critical in determining stopping distances. Ultimately, both cars will stop at the same time if they are identical in all other respects, as frictional force scales with weight.

PREREQUISITES
  • Understanding of basic physics concepts such as momentum and kinetic energy.
  • Knowledge of friction and its dependence on material composition.
  • Familiarity with equations of motion, particularly F = ma and F = k * mg.
  • Basic grasp of how mass affects energy consumption in vehicles.
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  • Research the impact of different surface materials on friction coefficients.
  • Explore the physics of braking systems, particularly the role of ABS in stopping distances.
  • Investigate the relationship between vehicle weight and fuel efficiency in various driving conditions.
  • Study the effects of angular momentum on vehicle dynamics during braking.
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Physics students, automotive engineers, and anyone interested in understanding vehicle dynamics and safety on slippery surfaces.

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Homework Statement



Given two cars, one heavier than the other but identical in all other respects, traveling at the same speed on a snowy or icy road which one will stop sooner?

Momentum would say that the lighter car should stop sooner but how does the heavier car's increased weight in combination with friction figure into the stopping distance? Shouldn't the heavier car be able to dig into snow more increasing the opposing frictional force? On ice even though neither car can "dig in", wouldn't the heavier car experience a greater frictional force causing it to stop sooner or is the increased momentum greater than this increase friction?

Explanations are fine as are equations. Thanks for the help.
 
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One dependency is how much the angular momentum at the wheels is shifted into linear velocity. . .


p.s. this is a true driver dependency.
 
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How about I simplify the problem.

Let's just say neither car has ABS and to stop both drivers choose to mash on the brakes and lock up the wheels.

So essentially both cars are just sliding blocks.
 
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Frictional force is independent of weight, but dependent on material composition.

Momemtum is mass * velocity. p = mv. Friction will retard speed.

Friction uses energy directly. Kinetic energy(KE) is (m * v * v)/2.
So let us say KE = mK for this example. Friction will leave r*m*k units of energy to continue the work.

With one mass m1 and the other m2, the KE ratio, yields m1/m2 as the deciding factor of who travels the furthest. Example, if both cars are identical, then m1 = m2, and they both stop at the same time. This makes sense because you know heavier cars use more gas to attain the same speeds as lighter cars and since friction is independent of weight, the solution is complete.

The important point is at the beginning of this post.
 
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basePARTICLE said:
Frictional force is independent of weight, but dependent on material composition.

Momemtum is mass * velocity. p = mv. Friction will retard speed.

Friction uses energy directly. Kinetic energy(KE) is (m * v * v)/2.
So let us say KE = mK for this example. Friction will leave r*m*k units of energy to continue the work.

With one mass m1 and the other m2, the KE ratio, yields m1/m2 as the deciding factor of who travels the furthest. Example, if both cars are identical, then m1 = m2, and they both stop at the same time. This makes sense because you know heavier cars use more gas to attain the same speeds as lighter cars and since friction is independent of weight, the solution is complete.

The important point is at the beginning of this post.

This doesn't make sense. Sliding friction depends linearly on weight. F = k.mg
where k is the coefficient of friction. If you combine that with F=m.a to compute the deceleration of the car, you'll see that THAT doesn't depend on mass.
 

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