Why does the heavier car stop sooner?

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    Car Friction
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Homework Help Overview

The discussion revolves around the stopping distances of two cars, one heavier than the other, traveling at the same speed on a snowy or icy road. Participants explore the relationship between weight, momentum, friction, and stopping distance, questioning how these factors interact under different conditions.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants examine the role of momentum and friction in determining stopping distances, with some questioning whether a heavier car would stop sooner due to increased friction or if its greater momentum would counteract this effect. Others simplify the scenario by considering locked wheels and the implications of friction being independent of weight.

Discussion Status

The discussion is active, with various interpretations being explored regarding the effects of weight and friction on stopping distances. Some participants provide insights into the physics of momentum and kinetic energy, while others express confusion about the relationship between weight and frictional force.

Contextual Notes

Participants are operating under the assumption that both cars are identical in all respects except for weight, and they are considering scenarios without anti-lock braking systems (ABS). There is also a focus on the material composition of the road surface affecting friction.

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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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