# Frictionless Cart on a Ramp (Experimental Design Question)

Question:

Why is the calculated value for our final velocity higher than our predicted value?

Since our prediction neglected air resistance and friction, shouldn't the velocity for the actual cart be lower than that of the predicted cart?

This lab had three parts.

Part 1:

We had to design our system and setup. We were given a low friction cart, our ramp (the table), a meter stick, and stopwatch.

The table was elevated by two textbooks under the legs of one side. The length of the table was measured several times to be 152.3 cm. To calculate theta, we measured the heights of the table from both ends and took the difference. Height 1 was 84.6 cm, height 2 was 76.5 cm, thus, the difference is 8.9 cm.

Taking the arctan of 8.9/152.3, we found the angle of the ramp to be 3.34 degrees.

Part 2:

The forces acting on the cart are the force of gravity in the x-direction, the normal force is canceled by the y-component of gravity, and we neglect friction since we are assuming that we have a frictionless cart.

$$F_{gx} = \boldsymbol{F_{g}}\sin{\theta}$$ $$\boldsymbol{F_{g}} = m \cdot \boldsymbol{a}$$

$$F_{gx} = \boldsymbol{F_{g}}\sin{\theta}$$ $$m \cdot {a_{x}} = m \cdot \boldsymbol{a}\sin{\theta}$$ $$a_{x} = \boldsymbol{a}\sin{\theta}$$ $$a_{x} = -10\frac{\text{m}}{\text{s}^2} \cdot \sin{3.34^\circ} = -.583 \frac{\text{m}}{\text{s}^2}$$

$$v_{x}^2 = v_{x_0}^2 + 2a_{x}(x_{f} - x_{0})$$ $$v_{x}^2 = 2(-.583 \frac{\text{m}}{\text{s}^2})(\text{1 m})$$ $$v_{x}^2 = -1.17 \frac{\text{m}^2}{\text{s}^2}$$ $$v_{x} = 1.08 \frac{\text{m}}{\text{s}}$$

Part 3:

We took the cart and calculated the time it took to be displaced exactly 1 meter. We took 15 trials in order to mitigate the effects of error of reaction times. Thus, partner 1 had 1 stopwatch and the cart. Partner 1 would tell partner 2 when he would release the cart, and both partners would note the time to take it to be displaced 1 m. If our times differed by more than $\pm 1$ meter, we would redo the trial. Average time was 1.8 seconds.

Using the kinematic equation

$$x - x_{0} = v_{x0}t + \frac{1}{2}a_{x}t^2$$

the calculated acceleration comes out to -.62 m/s2. And the velocity comes to be 1.11 m/s.

Thus, back to the question. Why is our calculated velocity, 1.11 m/s higher than our predicted value for velocity 1.08 m/s, even though our predictions neglects friction as well as air resistance.

• First rule of experimenting: Include measurement errors – ACuriousMind Oct 7 '14 at 22:39
• @ACuriousMind What do you mean by measurement errors? While the actual measuring tools and measurements may be off, I don't think it would cause such a large significant difference in the actual value of the velocity. We measured all lengths several times, and took 15 trials for the time to mitigate such errors. – user43617 Oct 7 '14 at 22:41
• Yes, but nevertheless, no measurement is precise. All measured quantities possess an uncertainty, and proper experimenting should include them. For example, when measuring the time 15 times and averaging the results, you should give the standard deviation of the values you averaged over as error. If you measured the times yourself instead of using a sensor, there's also the human reaction time to consider as a systematic error. When measuring lengths, you should think about how precise you can actually read your measurement device. – ACuriousMind Oct 7 '14 at 22:47
• A possible experiment would be taking the same table and elevate it in the opposite sense, to verify that the floor you are putting it on is, indeed, level (you measured the angle by measuring the heights from the floor). – SJuan76 Oct 7 '14 at 22:51
• @ACuriousMind Thank you for the very good advice. The calculated standard deviation came out to be 0.09403. – user43617 Oct 7 '14 at 23:01

3. Errors in physics. You already stated you ignored friction; you also ignored the inertia of the wheels of your cart. When a wheel rolls down a slope its inertia is greater than the mass - because it has to rotate, not just translate. This would tend to make the final velocity lower than the naive calculation, so in this case it doesn't help. You said you computed $\theta$ from the arctan, but it seems to me you should have used the sin. It's a difference of 0.1% for these small angles... The biggest error in physics is your use of $g=10 m/s^2$ - that alone comprises a 2% error (but again, in the wrong direction).