Elementary physics question

[Kuko 4000]

I'm once again exposing my almost total lack of physics knowledge, but that's why I'm here

Q1: A person in Finland is standing on top of a 100 m building. He has two bowling balls with him. They are indentical in size, but the other one weighs 10 kg and the other one weighs 100 kg. The person drops the balls at the same exact time from the same exact height. Which of the balls reaches ground first, or will they reach it at the same time?

Q2: A person in Finland has two identical cars. Except, the other one weighs 4000 kg and the other one weighs 8000 kg. Both of the cars are driving at the speed of 100 km/h. They will brake at the same exact time on the same exact strength and on the same exact surface, ie. everything else is identical except the weight of the cars. Which of the cars stops first?



Thank you.

 

[Guybrush Threepwood]

I'm once again exposing my almost total lack of physics knowledge, but that's why I'm here

Q1: A person in Finland is standing on top of a 100 m building. He has two bowling balls with him. They are indentical in size, but the other one weighs 10 kg and the other one weighs 100 kg. The person drops the balls at the same exact time from the same exact height. Which of the balls reaches ground first, or will they reach it at the same time?

Q2: A person in Finland has two identical cars. Except, the other one weighs 4000 kg and the other one weighs 8000 kg. Both of the cars are driving at the speed of 100 km/h. They will brake at the same exact time on the same exact strength and on the same exact surface, ie. everything else is identical except the weight of the cars. Which of the cars stops first?



Thank you.

Q1: The 100kg ball will hit the ground first. Both balls are acted on by two forces, an accelerating force due to gravity, which depends on the balls mass, and a retarding force due to air resistance, which depends on the surface area and velocity through the air, so the retarding force will be equal for the two balls, but the accelerating force greater for the 100kg ball.

Q2: The lighter car will stop first, if the retarding force is equal then the heavier car has twice the kinetic energy, so will take twice the distance to stop.

 

[Wolfman]

The first question is an issue of acceleration; acceleration will be uniform for both bowling balls, they will fall at the same speed.

The second question is an issue of inertia; the heavier vehicle will have greater inertia, and therefore take more time to stop.

If it helps...the heavier bowling ball will also have greater inertia. This won't be demonstrated by an increase in acceleration, but rather by the fact that it will bury itself deeper in the ground when it lands.

 

[Dave Rogers]

Q1: A person in Finland is standing on top of a 100 m building. He has two bowling balls with him. They are indentical in size, but the other one weighs 10 kg and the other one weighs 100 kg. The person drops the balls at the same exact time from the same exact height. Which of the balls reaches ground first, or will they reach it at the same time?

Guybrush Threepwood has correctly pointed out that the air resistance is the same for both balls, but will decelerate the more massive ball less, therefore that ball will strike the ground first. However, over a 100m drop I suspect that the difference in impact times would be very small, and careful experiment design would be needed to measure it at all.

Q2: A person in Finland has two identical cars. Except, the other one weighs 4000 kg and the other one weighs 8000 kg. Both of the cars are driving at the speed of 100 km/h. They will brake at the same exact time on the same exact strength and on the same exact surface, ie. everything else is identical except the weight of the cars. Which of the cars stops first?

This one's more interesting. It depends on whether the brakes are applied hard enough to initiate a skid. If not, then GT is correct in that the force on the two cars is always the same, so the lighter car stops first, in half the distance of the heavier car. However, if the brakes lock the wheels of both cars, then their deceleration is caused by the dynamic friction force between the tyre and the road surface, which is directly proportional to the weight carried by the tyre; so in a skid, the force on the lighter car is only half the force on the heavier, and both will therefore stop in the same distance.

It's also possible for the heavier car to stop first. If the force applied by the brakes is greater than the limiting friction force between the tyres of the lighhter car and the road, but not greater than that between the tyres of the heavier car and the road (which is twice as large), then the lighter car will skid, but the heavier car won't. The lighter car will be decelerated by at most the dynamic friction force, whereas the heavier will be decelerated by up to the limiting friction force, which is greater than the dynamic friction force. Therefore, if the brakes are applied by the same amount on both cars, such that the braking force isn't quite high enough to cause the heavier car to skid, then the heavier car stops first.

Dave

 

[jasonpatterson]

For the first question, it would depend on whether they wanted you to consider ideal conditions or not, which I guess you would know from the context of the source of the question. If it is usually one that requires you to consider things like air resistance, then the more massive ball lands first. If not, they land at the same time.

Good analysis on the second question Dave. For the heavy car to stop first, the static coefficient of friction would have to be greater than twice the kinetic for these tires on the given road. If not, the required frictional acceleration to get the sliding light car to stop would be greater than that of the non-sliding heavy car. Strange tires, but I wouldn't be surprised if such a thing existed (perhaps on an icy road...)

 

[edd]

I'm once again exposing my almost total lack of physics knowledge, but that's why I'm here

Q1: A person in Finland is standing on top of a 100 m building. He has two bowling balls with him. They are indentical in size, but the other one weighs 10 kg and the other one weighs 100 kg. The person drops the balls at the same exact time from the same exact height. Which of the balls reaches ground first, or will they reach it at the same time?

Q2: A person in Finland has two identical cars. Except, the other one weighs 4000 kg and the other one weighs 8000 kg. Both of the cars are driving at the speed of 100 km/h. They will brake at the same exact time on the same exact strength and on the same exact surface, ie. everything else is identical except the weight of the cars. Which of the cars stops first?



Thank you.

All this mention of Finland has me wondering about the validity of the Copernican Principle on scales of order 10⁶m.

 

[dlorde]

Guybrush Threepwood has correctly pointed out that the air resistance is the same for both balls, but will decelerate the more massive ball less, therefore that ball will strike the ground first. However, over a 100m drop I suspect that the difference in impact times would be very small, and careful experiment design would be needed to measure it at all.

I read his reply as incorrectly stating that the retarding force due to air resistance will be equal, but the accelerating force due to gravity would be greater for the more massive ball...

I guess I would call your reading of it 'generous', but perhaps he didn't express himself clearly enough for me

 

[Dave Rogers]

Good analysis on the second question Dave. For the heavy car to stop first, the static coefficient of friction would have to be greater than twice the kinetic for these tires on the given road.

No, it just has to be greater than the kinetic, which is usually the case. Maximum deceleration of the heavier car is the static coefficient of friction times g, because the mass cancels out; deceleration of the lighter car is the kinetic coefficient times g, for the same reason. Think in terms of angle of friction and it's obvious.

Dave

 

[Dave Rogers]

I read his reply as incorrectly stating that the retarding force due to air resistance will be equal, but the accelerating force due to gravity would be greater for the more massive ball...

I see what you mean, and it's not exactly clear; however, it's true that the accelerating force will indeed be greater for the heavier ball, it just won't produce a greater acceleration.

Dave

 

[Guybrush Threepwood]

I read his reply as incorrectly stating that the retarding force due to air resistance will be equal, but the accelerating force due to gravity would be greater for the more massive ball...

I guess I would call your reading of it 'generous', but perhaps he didn't express himself clearly enough for me

No, I think your reading of what I wrote was correct, but it's the same as Dave's reading. The gravitational force on the large ball is ~981N and on the small one ~98.1N. The acceleration due to gravity is the same for both balls, 9.81 ms^-2
In a vacuum the balls would accelerate equally, but when there is a drag force dependent only on velocity (since the shape and surface area are the same for both balls), it reduces the total downward force on the lighter ball much more rapidly..

 

[CORed]

Guybrush Threepwood has correctly pointed out that the air resistance is the same for both balls, but will decelerate the more massive ball less, therefore that ball will strike the ground first. However, over a 100m drop I suspect that the difference in impact times would be very small, and careful experiment design would be needed to measure it at all.



This one's more interesting. It depends on whether the brakes are applied hard enough to initiate a skid. If not, then GT is correct in that the force on the two cars is always the same, so the lighter car stops first, in half the distance of the heavier car. However, if the brakes lock the wheels of both cars, then their deceleration is caused by the dynamic friction force between the tyre and the road surface, which is directly proportional to the weight carried by the tyre; so in a skid, the force on the lighter car is only half the force on the heavier, and both will therefore stop in the same distance.

It's also possible for the heavier car to stop first. If the force applied by the brakes is greater than the limiting friction force between the tyres of the lighhter car and the road, but not greater than that between the tyres of the heavier car and the road (which is twice as large), then the lighter car will skid, but the heavier car won't. The lighter car will be decelerated by at most the dynamic friction force, whereas the heavier will be decelerated by up to the limiting friction force, which is greater than the dynamic friction force. Therefore, if the brakes are applied by the same amount on both cars, such that the braking force isn't quite high enough to cause the heavier car to skid, then the heavier car stops first.

Dave

I disagree. If the stopping distance is friction limited, the stopping distances will be equal. The heavier car will have twice the frictional force, but it will be decelerating twice the mass.

 

[RossFW]

I disagree. If the stopping distance is friction limited, the stopping distances will be equal. The heavier car will have twice the frictional force, but it will be decelerating twice the mass.

Assuming the two cars have identical brakes both acting at maximum efficiency, this is not correct.

Brakes dissapate energy over time. More energy (which is so with a more massive car) takes more time to dissapate. Light cars stoip quicker than heavy cars- a fact born out in everyday life.

 

[Dilb]

Assuming the two cars have identical brakes both acting at maximum efficiency, this is not correct.

Brakes dissapate energy over time. More energy (which is so with a more massive car) takes more time to dissapate. Light cars stoip quicker than heavy cars- a fact born out in everyday life.

The energy dissipation should not be the limiting factor for a decent set of brakes. Brakes on a consumer vehicle are engineered so they can easily apply as much force as we like, enough to lock the wheels and keep them locked at highway speed. I can lock the wheels in a Chevy minivan (~3000 kg) at 100 km/hr, so brakes should not be the limiting factor unless they are broken. The issue is the road-tire friction force. Tires do not have a constant coefficient of friction, it decreases as the load increases by a small amount. Breaking distances for heavier vehicles is maybe 20% more as a result.

Of course, I don't know what sort of breaks a 4000 kg car would have, let alone an 8000 kg one.

Regarding the 10 kg bowling ball, it should be pretty close to terminal velocity after 100m, so seeing air resistance shouldn't be a challenge. 100m is roughly 30 stories, an impressive height to be working with. Of course, a 100 kg bowling ball is pretty impressive too.

 

[russell_morris]

I see what you mean, and it's not exactly clear; however, it's true that the accelerating force will indeed be greater for the heavier ball, it just won't produce a greater acceleration.

Dave

Forgive me if I'm too far away from my undergrad physics classes (10 years now...), but the acceleration of the ball is only one part of the problem, isn't it? The ball also exerts an (exceptionally small, but nonzero) force on the earth, which causes an (exceptionally small, but nonzero) acceleration of the earth towards the ball.

Doesn't this mean that - at least in the hypothetical case where the earth and bowling balls are spheres of uniform density - the heavier ball will make contact with the earth a very, very, very small amount of time before the lighter ball makes contact?

In other words, the two bowling balls each fall with the same acceleration towards the earth, but the earth "falls" a tad more quickly towards the heavier bowling ball than it does towards the lighter one.

 

[jasonpatterson]

No, it just has to be greater than the kinetic, which is usually the case. Maximum deceleration of the heavier car is the static coefficient of friction times g, because the mass cancels out; deceleration of the lighter car is the kinetic coefficient times g, for the same reason. Think in terms of angle of friction and it's obvious.

Dave

Ack, my bad. I was considering the frictional force on the brake calipers of the heavier car as opposed to the tires on the road of the lighter car at one point and didn't mentally switch to tires and tires for this. Thanks.

Forgive me if I'm too far away from my undergrad physics classes (10 years now...), but the acceleration of the ball is only one part of the problem, isn't it? The ball also exerts an (exceptionally small, but nonzero) force on the earth, which causes an (exceptionally small, but nonzero) acceleration of the earth towards the ball.

Doesn't this mean that - at least in the hypothetical case where the earth and bowling balls are spheres of uniform density - the heavier ball will make contact with the earth a very, very, very small amount of time before the lighter ball makes contact?

In other words, the two bowling balls each fall with the same acceleration towards the earth, but the earth "falls" a tad more quickly towards the heavier bowling ball than it does towards the lighter one.

Wow, that would be right, they should if you drop them separately. As you mention, the time difference would be very very very very small and completely unmeasurable by any reasonable means. Earth, with a mass of 6x10²⁴kg, would wind up accelerating toward the ball at a whopping 1.6x10^-22 m/s² in the case of the heavier ball and a paltry 1.6x10^-23 m/s² for the lighter. Over the time of the fall we would expect Earth to move 1.6x10^-21m in the heavy ball case and 1.6x10^-22m in the light ball case. That would mean that the light ball had to travel a tiny bit farther and thus would land later. In the problem though, the balls were dropped at the same time, so the total acceleration of Earth upward toward them would be the combination of the two values.

 

[MattusMaximus]

Assuming the two cars have identical brakes both acting at maximum efficiency, this is not correct.

Brakes dissapate energy over time. More energy (which is so with a more massive car) takes more time to dissapate. Light cars stoip quicker than heavy cars- a fact born out in everyday life.

RossFW is correct. Here's why...

The negative work done by the brakes in stopping the car is equal to the loss of kinetic energy of the car. Assume all the KE is lost, so...

loss of KE = Work

mv²/2 = Fd

where m is the mass of the car, v is its speed, F is the braking force, and d is the stopping distance. Solving for d yields...

d = mv² / 2F

Thus, if the braking forces are the same for each car and the initial speeds are also the same, then the stopping distance is directly proportional to the car's mass. This quickie calculation also assumes the brakes don't lock, there is no skidding, constant braking forces, etc.

 

[RossFW]

Thanks Mattus,

Not only that, braking systems almost always atop a vehicle more quickley, and therefore MUST be dissapating energy more quickley, when they DON'T lock up.

Perhaps a better way of saying it is that the MINIMUM braking distance will be shorter for the lighter car.

 

[Ziggurat]

Thus, if the braking forces are the same for each car, then the stopping distance is directly proportional to the car's mass. This quickie calculation also assumes the brakes don't lock, there is no skidding, constant braking forces, etc.

This is a bad assumption. As previously mentioned, braking force for a car is generally limited by the maximum friction force between the road and the tires (kinetic if they skid, static if they don't), which depends upon the normal force and hence the weight of the cars, and not by the limitations of the brakes themselves. If the weights are not the same, the braking forces will generally not be either.

 

[Ziggurat]

Perhaps a better way of saying it is that the MINIMUM braking distance will be shorter for the lighter car.

No. The minimum distance is determined by the maximum static friction, which is proportional to the mass of the car. So while momentum and KE scale with mass, so does maximum friction force, so they will have identical lower bounds for stopping.

 

[Dr. Trintignant]

The minimum distance is determined by the maximum static friction, which is proportional to the mass of the car.

Also a bad assumption. Were this true, bicycle tires would be perfectly sufficient for race cars, since the pressure times the contact patch size is equal no matter what the width of the tire (keeping the mass of the vehicle constant).

In practice, race cars use as large a contact patch as practical (wide tires at low pressure), because static friction is not proportional to pressure for the rubber/road interface, at least at relatively low pressures. In many ways, it's almost a mechanical interface, like a gear on a toothed track.

- Dr. Trintignant