
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: 19981116   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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