# Stopping distance

1. Nov 3, 2006

### samdiah

Why does mass have no effect on stopping distance?

2. Nov 3, 2006

### HallsofIvy

Staff Emeritus
Who said it didn't? What assumptions are you making here? If a mass, m, traveling at speed v, is opposed by a constant force F, then F= ma so a, the acceleration (which is negative since F is negative) is F/m and the mass will stop when v- (F/m)t= 0 or when t= mv/F. In that situation the stopping time is proportional to the mass. Is your question about a special situation?

3. Nov 3, 2006

### QuantumCrash

Maybe you are simply looking at the kinematics equation. In that case, you are already given acceleration and hence mass may seem to have no effect on the stopping distance. When in fact it does, as HallsofIvy have demonstrated.

4. Nov 3, 2006

### marc.morcos

hey my first post, try to think of it like this...
A car and a truck DONT have the same stopping distance when initally going the same speed, it maybe possible to have the same stopping distance if you have big brakes on the truck. Ffriction=(mu)*(Normal), so the larger the mass (normal becomes larger) the more frictional force you need (bigger brakes)...
Hence mass Does indeed affect stopping distance.
Hope that helps.

5. Nov 3, 2006

### samdiah

I am looking at a situation where one mass is double the other mass, and the kenetic energy and force were constant or same for both.

Thankyou everyone

6. Nov 4, 2006

### teclo

assuming they have the same coefficient of friction, i don't see why the lighter one would stop any faster than the heavier one.

F = (mu)N = ma
N = (mg)

7. Nov 4, 2006

### QuantumCrash

Well, if you look at N = mg, and if you noticed m in there, the mass actually influences the N, which influences the F, which changes the aceleration...

From F=ma =>larger mass smaller deceleration.
From F=(mu)N => larger mass, larger Normal force, smaller deceleration.

the m's don't cancel out, if that is what you are implying because that would mean the forces that you are comparing are different.

8. Nov 5, 2006

### teclo

the m's do cancel out, though. the equation reads (mu)*m*g = m*a, solving for the acceleration you get a = mu*g. this came up in physics 1 a few years ago and i was super confused.

of course if there is a braking force applied (not just coasting to a stop) then it's true the lighter one is easier to break. this would also be ignoring air resistance, which would complexify things even more.

9. Nov 5, 2006

### QuantumCrash

Yes, that does make sense. My mistake. Larger normal force, larger friction. So mass does not affect stopping distance in a very simple model like two blocks of different masses.

However, in the case of the truck and the car, friction from the ground contributes only a small proportion of the stopping force. Most of it is provided by air resistance, friction within the system, etc.