# Homework Help: Mousetrap Experimental Investigation

1. Jul 27, 2013

### DelToro

For an investigation, I have built a mousetrap car. The two variables in this investigation is the mass of the mousetrap car and the distance it travels. When there is no added mass on the cart, it does not travel far due to massive wheel slippage. As I increased the mass it carried, it travelled further up to a certain point (around 700grams). From that point onwards, adding extra mass decreased the distance it travelled. The resulting graph of mass v distance gave a parabola shape for the initial masses (a sharp increase with the initial masses) and then a linear decrease past the optimal mass.

After doing some research I have found that there is an amount of usable traction before the wheels start to slip. This is given by the equation: usable traction = μ (coefficient of static friction) x Fn (Normal force).

I have been unable to explain the shape of this graph. Theoretically, if the usable traction is a linear line (as there are no exponentials in the equation), then the initial distance increase as mass increase should be linear, but it is not. Is there some other variable I am missing? Why would the distance increase at such a large rate initially, but then decrease after the optimal mass?

2. Jul 27, 2013

### Staff: Mentor

With more mass, the length of the acceleration part increases, so you get more energy per force.

3. Jul 27, 2013

### DelToro

That makes sense, thank you for responding.
So the mousetrap would be applying an impulse for a longer period of time resulting in the larger increase in the distance travelled?

4. Jul 27, 2013

### Staff: Mentor

The momentum view (instead of energy) is possible, too, right.

5. Jul 27, 2013

### DelToro

In any case, such as mass increasing the acceleration or with impulse, wouldn't that only produce a linear line as mass increases? I'm finding it difficult to explain the shape of the curve in the graph. Only after it reaches it peak does it produce a linear line downward (which makes sense). The only conclusion i can make is that wheel slippage decrease exponentially as mass increase? But then that would be linear as well, as usable traction is equivalent to the product of static friction and the normal force. I've attached the graph i ended up getting. The mass is on the x-axis and the distance travelled is on the y-axis. I don't know if i'm making myself look stupid or not, I have no idea how to justify the curve.

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6. Jul 28, 2013

### Staff: Mentor

You reduce the slipping time, and acceleration without slipping is much more effective. I would expect that ~1 is the mass where slipping does not occur any more.

I doubt that it is linear (especially for large masses - the distance won't get negative), the data is just not sufficient to see how the curve looks like.

7. Jul 28, 2013

### DelToro

I know, there definitely is not enough data to know for sure that the decrease is linear. However, theoretically there might be a chance that it is (although i'm almost certain that there will be a flaw in my logic). So, past the point at which there is an optimal mass for a peak distance, there is no more wheel slippage and the energy available to the cart is constant. Hence, these factors are now of little relevance apart from the fact there is constant amount of energy. So, the remaining forces acting in the system are mostly the opposing forces, like rolling resistance and friction in the bearings, both of which are given by the equation f=μ*Fn. So, when these are plotted on a graph, they produce a linear line increasing as mass increases. Because the opposing force is increasing in a linear fashion, it would follow that the net force would decrease in a linear fashion. In turn, the total traveled distance would decrease in a linear fashion. Have a missed something completely here or have a completely mess up a concept?

8. Jul 28, 2013

### Staff: Mentor

A constant energy with a linear friction force leads to a 1/x-curve, that is not linear. The inverse distance would be linear with mass.

9. Jul 29, 2013

### DelToro

Okay, i see what you mean, i think.
Is it then reasonable to say that the 1/x curve is due to acceleration being inversely proportional to mass and directly proportional to displacement? Hence, as mass increases the acceleration decreases in an inverse fashion thus the traveled distance decreases in an inverse fashion.