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.

  • 1
    $\begingroup$ First rule of experimenting: Include measurement errors $\endgroup$
    – ACuriousMind
    Oct 7, 2014 at 22:39
  • $\begingroup$ @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. $\endgroup$
    – user43617
    Oct 7, 2014 at 22:41
  • 1
    $\begingroup$ 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. $\endgroup$
    – ACuriousMind
    Oct 7, 2014 at 22:47
  • $\begingroup$ 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). $\endgroup$
    – SJuan76
    Oct 7, 2014 at 22:51
  • $\begingroup$ @ACuriousMind Thank you for the very good advice. The calculated standard deviation came out to be 0.09403. $\endgroup$
    – user43617
    Oct 7, 2014 at 23:01

1 Answer 1


When you run into a problem like this, there are several things to consider:

  1. Systematic error. For example, when you say "three, two, one, go" you never release exactly on "go" (plus, there is a time delay between when the person says "go" and when you hear it...). Other example: the floor is not level (as suggested by ACuriousMind). When you measure the height of the table, did you make sure you measured the vertical distance? Did you mean to use the arc sin, rather than the arc tan, in order to compute the angle (small effect - but nonzero). A final example is calibration: is the stopwatch calibrated? Is the tape measure calibrated? Is your coefficient of gravity calibrated to your latitude (it changes by 7% from North Pole to equator, and also depends on altitude). Actually you seem to be using a value of 10, which is clearly erroneous at the accuracy you are aiming for.
  2. Random error. Although you may have no systematic errors, you may find that the variation in results is sufficient to explain the difference. I note you wrote you discarded differences in time "greater than 1 meter" between you and your partnet which worries me... what units did you mean to use? When you compute the error on your measurements, then combining these measurements leads to a compound error. You want to confirm what that error is, and whether the "expected" answer is consistent with your measurement. So if you expect 1.08 m/s and you measure 1.11 ± 0.05 m/s, then while the answers are different they are not inconsistent.
  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).

You want to make sure that you think carefully about potential sources of systematic error, and design the experiment to minimize them.

I think that the biggest error is likely to come from the release method. It's really easy to give a small systematic "push" to the cart when you let go, and that is going to have a big effect (since on a small slope the initial velocity is very low).

One thing to consider: time the cart at three different points along the table, at well marked points (maybe even a tiny bump so you get an audible "click"). With three time points you can eliminate the error in start time and start velocity.


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