# Homework Help: Imple physics question dealing with conseration of energy!

1. Nov 24, 2008

### davidelete

1. The problem statement, all variables and given/known data
One end of a board rests on some books lying on a table. A ball is released from the raised end of the board and it rolls down the incline and along the horizontal table for 10.0 cm before falling onto the floor. How far from the edge of the table does the ball land if the table is 70.0 cm high and the end of the board is raised 30.0 cm?

2. Relevant equations
I tested the experiment instead, and got 67 cm. I truly cannot figure out the correct way to do this.

3. The attempt at a solution

2. Nov 24, 2008

### LowlyPion

How long does it take something to fall from rest 70 cm?

Knowing that time, and knowing how fast it was going horizontally will give you the distance along the floor from the table.

So what was the velocity horizontally on the table?

Recall anything about Potential energy and kinetic energy that might be useful like conservation of energy?

3. Nov 24, 2008

### davidelete

t=$$\sqrt{2y/g}$$
Therefore, time would be 4.518 s in the downward direction, but this does not seem to play well. The velocity is not given, but if I multiply 4.518 s by 9.8 m/s2, I get that the velocity is 44.27 m/s. I can get this a plethora of different ways, so I am expecting that it is correct.

If this is the case, potential energy is m*g*h, and kinetic energy is 1/2mv2. Mass can be neglected in this, so g*h and 1/2v2 are all that matter. Now if I plug 9.8 in for gravity, 1 m in for height, and 44.27 in for velocity, I find that through gh=1/2v2, I get 1 m as the distance. I still do not know if this is correct, or if I must incorporate the 10 cm in some way or another. Should I subtract the 10 from 100 and get 90 cm?

4. Nov 25, 2008

### LowlyPion

I think that's a little off.

Gravity is 9.8 m/s2

70 cm is only .7 m.

5. Nov 25, 2008

### davidelete

I know gravity. For the .7m, wouldn't you have to contain the rolling 30 as well?

6. Nov 25, 2008

### LowlyPion

Mind your units is my point. If gravity is in SI units of meters then your other units need to be compatible. 44m/s is 158 km/hr. or 4 sec to drop less than a meter? Now I'm sure the ball wasn't traveling at that speed, or that it took that long to fall off your table .7 m table. So yes, you also need to treat the height of the incline as .3m as well.

Units matter. They must match.

7. Nov 25, 2008

### D H

Staff Emeritus
There's not enough information to answer the question. What kind of ball is it? A solid ball such as a super ball will hit the floor at a different location than will a hollow ball such as a tennis ball.

In other words, you need to know something about the ball's moment of inertia to be able to answer the question.

8. Nov 25, 2008

### davidelete

Mass, volume, and density are insignificant. These are apparently taking place in a vacuum on earth. I turned the paper in, I didn't have all the required information at that point, but it should suffice to let me keep a high A. Things like this are really bumps in the road. I need to know how to do it, and my notes and book are not clear to the point on topics such as these.

9. Nov 25, 2008

### D H

Staff Emeritus
There's another important piece of missing information: The length of the board. Why are these two pieces of information important?

You cannot solve the problem without knowing the length of the board. Suppose the board is 30 cm long. That means the board must be pitched straight up if the top is to be 30 cm above the table. The ball will have no horizontal velocity when it leaves this minimal length board, and thus the ball will have zero horizontal distance between leaving the edge of the board and hitting the floor.

Now suppose the board is longer than 30 cm. When the ball leaves the board, some of its rolling velocity will be directed downward and some horizontally. The ball will travel. How far depends on the magnitude of the exit velocity and the direction. While the direction is solely a function of the setup (i.e., the angle of the ramp), the exit velocity is a function of both the setup (i.e., the height of the board above the table) and the nature of the ball. Try your experiment with a super ball, a tennis ball, and something like a Hot Wheels car.

The key word here is "rolling". Ignoring friction, the super ball, a tennis ball, and model car will all gain the same amount of kinetic energy as they roll from the top of the ramp to the bottom, and in an amount equal to the decrease in potential energy. (i.e., mgh). However, how this kinetic energy is distributed into rotational and translational energy depends a lot on the makeup of the object. The model car's wheels don't have much rotational kinetic energy, so almost of the kinetic energy will be in the form of translation. This is not the case for the tennis ball or the super ball. A good chunk of the kinetic energy goes into rotation, and that rotational kinetic energy does not contribute to the motion of the ball after it leaves the track. The car will go considerably further from the table than will either ball, and the super ball will go somewhat further than the tennis ball. If you have studied moment of inertia you will know why. If you haven't studied that yet, the instructor had no business talking about "rolling".