# Mousetrap car problem

1. Mar 28, 2004

### impulsive

I am working on a mousetrap-powered car for my physics class and ran into a problem. i'm not sure if I am doing this right, but I ran the car from rest and it traveled 7.16 meteres in 4.63 seconds. Using the equation x= V initial(t) + (1/2)at^2, a determined that the car's acceleration is 0.668 m/s^2. Trying to determine the coefficient of friction, I used F-f = m*a. The mass of my car is .100 kg, and I measured the force from the mousetrap to be 10.4 N. As a result of using this equation, i determined that f, the frictional force, is 10.3 N. Using the relationship between the frictional force and the normal force, which is 0.98 N, I determined the coefficient of friction to be 10.5, which seems way too large. I don't know if that all makes sense, but can anyone help Me!?!

2. Mar 28, 2004

### Staff: Mentor

Your equations assume that the mousetrap exerts a constant force throughout the motion. That doesn't seem realistic to me. (Why would it stop?) Describe how this car works.

3. Mar 28, 2004

### jdavel

impulsive,

"....it traveled 7.16 meteres in 4.63 seconds."
Are you sure it was accelerating over the whole time, or was it coasting for part of it? (see my next comment)

"Using the equation x= V initial(t) + (1/2)at^2, a determined that the car's acceleration is 0.668 m/s^2."
This only works if a is constant, which means you have to be sure the mousetrap was applying force to the wheels over the whole time interval t, and (as Doc Al said) that the force the mousetrap was applying didn't change over that interval. Both of those seem unlikely.

"I measured the force from the mousetrap to be 10.4 N."
How did you do this? If you used a spring or balance attached to the bail (I think that's what the part of the trap that snaps on the mouse's neck is called) then you need to account for whatever linkaged and leverage exists between the bail and the rim of the car's wheels, since that's where the force relevant to the car's mass and acceleration is being applied.

"I determined the coefficient of friction to be 10.5, which seems way too large."
Unless your car can stay put on an incline that rises 10.5 units for every horizontal unit (which would be nearly stright up!) I agree, that number is way to high.

4. Mar 29, 2004

### impulsive

Turns out I stupidly made these calculations missing some major factors, which I was reminded of shortly after I made this post! Thank you for your help anyhow. I will post again when I have more details.

5. Mar 29, 2004

### gta-maloy]

I, not too long ago, had to do the same type of exercise. I did not manage to get that amount of speed out of my car (probably a bigger trap). I measured the N's at the start point and again at the end point to find the true force produced by the spring of the trap. Because it is a spring there will be a difference between the two points, a spring looses alot of force as the two metals aproach each other. You then need to figure out aproxametly how far the car will travel with force aplied to it, this is where it got tricky for me, in the end I did farely well on the project but that was merely because it was for my 12th grade physics class, and well, my calculation were IMO greater than they should have been for the time of the exercises. I hope this helps you out any.

6. Mar 29, 2004

### impulsive

Okay, I measured the distance over which the force is being applied(about 3.8 m), but I still don't understand what's going on. Am I relating force or work? Or torque? I assume the distance of the trap from the drive axle(where the force is being applied) has something to do with it, but I can't figure out how that factors in. This is supposed to be a research project, but none of my research is yielding any help.

7. Mar 29, 2004

### NateTG

Well, what are you trying to figure out?

One of the things you might consider doing is running the car past some yardsticks while videotaping it. That way you can get velocity measurements.

8. Mar 30, 2004

### impulsive

Basically, right now I need to know the force that the mousetrap is supplying to the car. I first tried to use the 10.4 N force which was too large, then I tried to use work (10.4 N)(3.80M) which was of course also not right, so I know the force must be a LOT smaller than 10.4 N, but I can't figure out any way to determine this. Maybe I could measure the velocity and work backwards. gta-maloy, exactly how did you measure the force at the start and at the end???

9. Mar 30, 2004

### jdavel

impulsive,

You're close!

Suppose you have a lever; you push down through some distance with a force of 10.4N at one end, the fulcrum is somewhere in the middle, and a weight (unknown) on the other end goes up 3.8m. You need one more thing to figure out what the unknown weight is. What is it? What does it correspond to on your car?

Here's another idea. If you wind up the mousetrap, and put the car on an incline, it'll accelerate more slowly, right? Why? Now suppose you raise the incline until the car won't accelerate at all; it just stays where it is. What does this tell you about the force that the mousetrap must be applying in this situation?

10. Mar 30, 2004

### impulsive

Aaaaahhhh....

Now I am seeing how I can figure this out...I did some more runs today... Thanks jdavel, and everyone else.

11. Mar 30, 2004

### NateTG

Really, the mouse trap supplies torque. You can get more force and less distance by hooking up closer to the spring on the mouse trap, or less force and more distance by hooking up farther away. I suggest that you do not worry about the interactions between the mouse trap and the car for now.

Perhaps it would be helpful to split the time that the car travels into a piece where it is accelerating, and a piece where it is decelerating?