Friction and Resistance - Inertia

[keith m]

Two identical cars are at the top of a hill, one has a driver and three passengers, the other just the driver. Both drivers release the hand break at the same time, which would reach the bottom first.

I have used two cars as the example as this will reduce the possible variables such as wind resistance; in this example every condition is the same for both cars other than the weight.

I believe they would reach the bottom at the same time as the only force acting on them is gravity. I am far from being an expert but seem to remember this from my days at college, my colleague argues that the heavier car will reach the bottom first due to momentum gains, please could someone put us out of our misery?

 

[AJ Bentley]

This very experiment was the birth of scientific study.

http://galileo.rice.edu/lib/student_work/experiment95/inclined_plane.html"

It would seem your friend quite literally doesn't know the first thing about science.

 

[torquil]

Two identical cars are at the top of a hill, one has a driver and three passengers, the other just the driver. Both drivers release the hand break at the same time, which would reach the bottom first.

I have used two cars as the example as this will reduce the possible variables such as wind resistance; in this example every condition is the same for both cars other than the weight.

I believe they would reach the bottom at the same time as the only force acting on them is gravity. I am far from being an expert but seem to remember this from my days at college, my colleague argues that the heavier car will reach the bottom first due to momentum gains, please could someone put us out of our misery?

Neglecting wind resistance and differences in rolling resistance due to different car masses, both will reach the bottom at the same time.

More realistically, including wind resistance (but still neglecting rolling resistance), the heavier car will reach the bottom a bit before the other one. Consider two geometrically identical cars of masses m and 2m at speed v down an inclined plane. Both experience the same drag force Fd. The force due to gravity is Fg and 2Fg respectively (this includes trig. factors due to inclined plane). Newtons law gives in the two cases:

Car of mass m:

m*a = Fg - Fd -> a = Fg/m - Fd/m

Car of mass 2m:

2m*a = 2Fg - Fd -> a = Fg/m - Fd/2m

So you see, in the second case a is larger, i.e. the acceleration is larger since the negative effect of wind resistance on acceleration is halved. Also, the terminal speed at which a=0 is larger.

This illustrates the point somewhat crudely. Rolling friction would typically have the opposite effect since the "rolling coefficient of friction" is typically larger for the heavier car, but the wind resistance will dominate in most cases so the point still stands.