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A teacher at this school did a science experiment where he had two balls of different masses go down an incline which was raised about 20cm above the table. The balls were released from the same point on the incline and he assumes they accelerated at the same rate and so when they left the incline they were going at the same speed (I'm not sure this is true). In any case, the ball with the larger mass hit the table at a farther distance than the ball with the smaller mass. Could this be true if they were going at the same speed when they left the incline? It seems to me that they should hit at the same place. I don't think the friction of the air would cause enough resistance to cause any difference. The larger ball would have more momentum but this shouldn't affect the rate at which a ball falls. Am I missing something or what accounts for the different distances the balls go through the air?
Question Date: 1998-11-16
Answer 1:

It is true that if the balls are going at the same speed when they leave the incline, and if the effect of air resistance is small, then they should land at the same spot. In this experiment, since the two balls didn't land at the same spot, they must have left the incline at different speeds.

It may seem strange that the two balls left the incline at different speeds even though they were released from the same point on the incline. After all, when we just drop two balls from the same height and let them fall freely to the floor, they accelerate at the same rate and are going at the same speed when they hit the floor. When we send them down an incline, however, they don't necessarily have the same acceleration anymore. To demonstrate this fact, I would try the following experiment:

Find an incline wide enough to allow you to send the two balls down the incline at the same time, right next to each other (although not contacting each other, of course). Release them from the same height, at exactly the same time, and then you will easily be able to see whether or not they accelerate at the same rate and leave the incline at the same speed.

The Physics Learning Center here at UCSB has just such an experiment set up, actually. It has two cylinders of the same mass and same shape, but when you let them roll down an inclined plane, they accelerate at dramatically different rates. For an explanation of why this is, see the next answer.

Answer 2:

As I understand the experiment, the ramp finishes some twenty centimeters above the table and the balls roll down the incline, and then fall the final twenty centimeters onto the table moving with a horizontal motion as well as a vertical one, so that they fall some distance away from the end of the ramp. Your experiment shows that apparently different mass balls have a different horizontal velocity and so hit the table at different distances.

Well, your physical intuition that the balls have different velocities at the end of the ramp is correct; in a nutshell some of the energy the ball gains goes into rotating the ball (which it has to do if it is rolling).

The different balls require more or less of the energy to be used up in getting the ball to rotate and so there is less energy left over to go into straight line speed, the balls have differing velocities at the end of the ramp.

It turns out that it is the distribution of the mass throughout the ball which determines how much energy it takes to spin a ball at a particular rotational speed. If I have three balls of equal mass, one with most of its mass concentrated at the center (imagine a small heavy ball embedded in a light round plastic case), one with a uniform distribution ( a regular rubber ball), and one with its mass concentrated near the outside (imagine a hollow ball) then, on rolling them down the ramp the first ball ends up going faster than the second, which goes faster than the third. (In fact it doesn't even matter if they have different masses, the result will be the same, the full explanation is below). Why don't you build three balls as I have just described and try it out ?

To explain any further I need to delve into some math, this is the full work for the teacher rather than an explanation for the junior high student. Ignore the rest of this if it gets too complicated, I got carried away figuring it all out and you don't need to know these details...

Both balls gain the same energy in coming down the ramp

E = mgh


(m is the mass of the ball, g is the acceleration due to gravity and h is the vertical height the balls have descended). If the balls were just allowed to drop vertically (remove the ramp) the situation is clear:both balls accelerate at the same rate, since the mass m drops out of the equations; the velocity just before hitting the table can be computed by equating the kinetic energy gained to the potential energy lost.

2 (1/2)mv = mgh


therefore,
v = Squareroot (2gh)


The balls take the same amount of time to hit the table and gain no horizontal velocity and so hit the table directly beneath the point they are released.

Now put the ramp back in. If the ramp is perfectly smooth (i.e. there is no friction between the ball and the ramp) the ball just slides down the ramp, note that it does not turn at all, it just slides. In this case the situation is similar to the case with no ramp, the ball gains a kinetic energy and the velocity can be computed as before. The difference is that this velocity is now at an angle theta to the vertical (the angle of the ramp) when the ball gets to the end the ramp. This is independent of the mass, as above, and so each ball would have the same velocity at the end of the ramp. The horizontal component of the balls velocity is unchanged during the final fall to the table and you would expect the balls to hit the table at the same distance. This is almost certainly what you were thinking of when you looked into the problem.

OK, now for the bit that makes it different. You have a ramp with friction and I guess that your ball rolls down the ramp rather than slides. As well as the translational kinetic energy (the (1/2) m v we were discussing above) an object can have rotational kinetic energy given by

(1/2) I w


where I is the moment of inertia and w is the angular velocity (radians per second) of the object. The gravitational potential energy the ball loses as it comes down the ramp is converted into translational and rotational kinetic energy i.e. we have to modify our equations to read


(1/2)mv + (1/2)I w = mgh


The angular velocity w and the regular translational velocity v are linked if the ball is rolling


v = r w
, where r is the radius of the ball

The larger the moment of inertia I, the smaller the final translational velocity v that the ball is going to attain at the end of the ramp.

The moment of inertia is given by the distribution of the mass in the ball away from the axis of rotation, a ball with a lot of mass concentrated at the center is easier to spin than one with al

Answer 3:

I was talking to another physicists and we agree that the two balls should travel the same distance. They have the same speed when they leave the ramp because they both accelerate at the same rate. Once they leave the ramp, the distance they travel only depends on their initial velocity leaving the ramp. If they have the same diameter, the air resistance will be the same on both of them and effect them both the same way.

Maybe the difference is that the heavier ball deformed the ramp more than the lighter ball causing them to travel differently.

Answer 4:

The reason the bigger ball went farther is because of it's greater diameter. It's nothing to do with the mass.

Yes, they will accelerate at the same rate, assuming identical mass distributions. But, because the larger ball must travel further to leave the ramp, it will gain more velocity prior to leaving the ramp, and thus fly further.



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