Time to stop body in motion depend on its mass?

  • Thread starter kallypso
  • Start date
hi, so here is the situation, and also my question:

you have two of identical cars, moving with same velocity. only difference between them is that you loaded the first car with a lot of package, so it's mass is greater than the mass of a second car.
so, what question is: when you hit the break, and want to stop the car, will it take you more time, and will you go larger distance until your velocity is zero, when you are driving the car loaded with stuff, or the mass doesn't matter at all?

so, the same question but slightly different situation: you have a truck,loaded with stuff, and a car. truck has much greater mass than a car. they both have same velocity.. so when they decide to stop, will it take a truck to travel larger distance, until it's velocity is zero?

physically, i think the mass doesn't have influence (am i right?), but in real situation, what does it depend on? does different mass means different time /distance traveled to stop?or does it depend only on what kind of vehicle you are driving ( and their ability to accelerate/ decelerate ?) or it depends on both mass and accelerate/deccelerate ability?

i hope i didn't make it too complicated :)
 
Well... the mass of these vehicles does play its part. This difference in distance you speak of, or the time it takes these vehicles to stop in comparison is based on a number of factors. It depends on the following; the mass of the vehicle, the traction (friction slows objects) of the vehicles tires assuming it has tires, the linear velocity, and the ability of the vehicles brakes. A more massive vehicle will have more momentum then lesser vehicles with the same linear velocity and therefore it will take more energy to stop its motion.

So yes the mass of the vehicle does indeed matter.
 
2,014
85
It's all Newtons second law.

F = ma.

With the same brakes (ie the same retarding force) an object with more mass will have a lower deceleration.
 

olivermsun

Science Advisor
1,244
117
As Mr. Chiappone says, a couple things are in play here.

Say your braking is limited by the maximum rate that your brakes can dissipate energy as heat. Then, more mass (at the same velocity) will mean more kinetic energy, since KE = 1/2 mv^2, and hence longer to dissipate the KE.

As a mitigating factor, you might expect that a truck that weighs (e.g.,) 4 times as much as a car would have bigger brakes -- hence, the truck would dissipate more KE per time than the car, and it wouldn't be quite as bad as the additional mass suggests.

On the other hand, suppose your car's braking is limited by the adhesion between the tires and the road (which is more likely for a reasonably-performing car). If you model the adhesion as a rolling friction coefficient * normal force, then you might expect a car carrying more packages to have just as much extra friction as it has extra mass to slow down. In practice it isn't quite as much as ideal friction, but you get the idea.
 

Borek

Mentor
27,999
2,509
Many years ago I owned http://en.wikipedia.org/wiki/Fiat_126. I was mostly riding alone, sometimes with my wife - we are both slim, so my reflexes were suited for (almost) empty car. One day I was riding with three friends of mine - each well over 200 pounds. Car was about 1300 pounds, that meant passengers added almost 50% to the mass. At some moment I had to stop at the red light - and we were very close to a crash. I wasn't aware that braking distance was substantially longer, if not for the fact that lane to the left was empty we would hit the back of the truck in front of us.

So yes, time to stop body motion depends on the mass.
 
tnx for answers:)
so, to conclude, if i 'm looking the situation on the road, it depends on the mass, on the tires&road (friction), on brakes etc.

i was also thinking of what would happen in ideal. case, in lab for example:
two balls of different mass moving with same constant velocity, same friction coefficient. now there aren't any of those other conditions on their motion as we had with vehicles on the road. if i exert same force on both of them (in opposite direction to their velocity, so i get deceleration), i apply Newton law F=ma, as xxChrisxx said, and expect ball with bigger m to have lower a. ok that's all clear. what was buggin me, is when i look at equations:
v(t)=v(0)+at,
x(t)=x(0)+v(0)t+1/2 at^2

there is no mass included, but i see what a dummy i was, sure it depends on mass, because of F=ma => a=F/m.

when i look at first situation, the one on the road, two different vehicles,same velocities but different mass and other, the only situation when their stopping time and distance traveled would be the same if they both would have same acceleration rate.
because this acc.rate depends on mass, tires,friction,brakes, engine etc., the vehicles of bigger mass,with tires that have bigger friction coeff. etc, need to have stronger engine, that could produce the same acc rate.
so i suppose that truck has somehow stronger engine than a car, and tires are different ( i don't know much about it :) what i want to know are trucks produced that way that when the driver wants to stop, it does indeed take more time and more distance traveled to stop then a car? or some truck have engine strong enough to stop the same?
 
and Borek, cute car it was :smile:
 
i have some troubles with the same issue, consider we have a truck and it is loaded with two packages m1 and m2 where m1>m2, both packages of the same material, so they have the same coefficient of both static and dynamic friction, suppose that μd is the coefficient of dynamic friction. now the truck suddenly stopped, applying newton second law for m1
m1a1 = - μd n1, where n1 is the normal force
but n1 = m1g
then a1 = -μdg
the same thing goes for m2
then a2 = -μdg
then a1 = a2 = a

but both packages have the same initial speed vi
and the same final speed vf = 0

v2f = v2i+2ax
then x=v2i/2a

both of them will move the same distance (x) until come to stop !!!
i know that there is something wrong on my analysis
please tell me what is wrong
 
649
2
Mass does play its part, depending on the tire and brake capacity. But I find that the significance of car mass in the context of driving is often misunderstood.

For example, consider two cars of masses m1 < m2. Let's say both have enough braking capacity to lock the wheels (i.e. there is no shortage of braking capacity). Say that they both have ideal ABS brakes.

Also assume that the friction between the tires and the road follows the usual simple relationship, i.e. is proportional to the normal force from the ground, with the same friction coefficient, in both cases. (Btw, this is where the idealisation differs most from reality..)

Then the two cars would have the exact same stopping distance, since the frictional force in both cases would be proportional to the car mass with the same proportionality coefficient in both cases. This is no paradox. It is actually what happens in this case.

However, what usually happens is that the more massive car will cause the tires to deform more while braking and they will perform less optimally. Therefore, the effective frictional coefficient will be lower for the more massive car, and its stopping distance will be a bit longer.

Btw, also remember that the mass distribution in the car also matters... If most of the breaking force occurs with the front wheels, and you have loaded up the back of the car, a more detailed analysis will show that the braking distance will be significantly longer, because the additional mass will not cause an increase in the maximum possible friction between the front wheels and the ground. So it depends on the distribution of the mass in the car, and the distribution of braking force among the wheels.
 
Mass does play its part, depending on the tire and brake capacity. But I find that the significance of car mass in the context of driving is often misunderstood.

For example, consider two cars of masses m1 < m2. Let's say both have enough braking capacity to lock the wheels (i.e. there is no shortage of braking capacity). Say that they both have ideal ABS brakes.

Also assume that the friction between the tires and the road follows the usual simple relationship, i.e. is proportional to the normal force from the ground, with the same friction coefficient, in both cases. (Btw, this is where the idealisation differs most from reality..)

Then the two cars would have the exact same stopping distance, since the frictional force in both cases would be proportional to the car mass with the same proportionality coefficient in both cases. This is no paradox. It is actually what happens in this case.

However, what usually happens is that the more massive car will cause the tires to deform more while braking and they will perform less optimally. Therefore, the effective frictional coefficient will be lower for the more massive car, and its stopping distance will be a bit longer.
i understood from your statement that coefficient of friction will change and it will depend on the surface area for the tires (deformation will enlarge the surface area), while i wonk that it doesn't depend on the surface area

and in my example no brakes, the two packages of the same material, so one of them has a larger surface area in contact with the bed of the truck, and a gain i know that coefficient of friction don't depend on the surface area
???
 
649
2
i understood from your statement that coefficient of friction will change and it will depend on the surface area for the tires (deformation will enlarge the surface area), while i wonk that it doesn't depend on the surface area
The friction coefficient will be different in real life for the two cases. But the effect I was talking about was not really directly related to contact area. I'm saying that the friction coefficient is lower for the heavier car because the tires are deformed in other ways when the car is braking aggressively. I didn't mean the simple increase in contact area when the car is loaded up. The contact area doesn't even increase linearly with the mass of the car either. If it did, perhaps the difference would be smaller. I meant that the tire stretches while braking, and also that the rubber heats up more at the contact interface, for the heavier car.

E.g. the heating of the rubber at the contact interface while breaking can be compensated for by using larger tires. Also, rubber itself has a limited available friction force against the ground before it is ripped apart. This happens more for the heavier car, unless this is compensated for by using larger tires or a better rubber composition.

and in my example no brakes, the two packages of the same material, so one of them has a larger surface area in contact with the bed of the truck, and a gain i know that coefficient of friction don't depend on the surface area
???
Your analysis was correct, within the assumptions that you made, i.e. that the friction coefficient was the same for both masses. The two masses will travel the same distance in your idealised example. That's part of the reason it makes sense to talk about "coefficient of friction" in the first place.
 
476
0
Many years ago I owned http://en.wikipedia.org/wiki/Fiat_126. I was mostly riding alone, sometimes with my wife - we are both slim, so my reflexes were suited for (almost) empty car. One day I was riding with three friends of mine - each well over 200 pounds. Car was about 1300 pounds, that meant passengers added almost 50% to the mass. At some moment I had to stop at the red light - and we were very close to a crash. I wasn't aware that braking distance was substantially longer, if not for the fact that lane to the left was empty we would hit the back of the truck in front of us.

So yes, time to stop body motion depends on the mass.
Very curious way of experimenting :rolleyes:
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top