describe an experiment to determine the acceleration of free fall using a falling object

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Equations of Motion: Determining g

Equations of Motion – Determining the Acceleration of Free Fall

Aim

To experimentally determine the acceleration due to gravity (\$g\$) by measuring the motion of a freely falling object.

Apparatus

  • Metal ball or dense sphere (diameter ≈ 2 cm)
  • Electromagnetic release mechanism (or a simple clamp)
  • Photogate timer (or two photogates spaced a known distance apart)
  • Meter rule or calibrated measuring tape
  • Retort stand with clamp
  • Stopwatch (for verification only)
  • Data sheet

Method

  1. Set up the retort stand so that the photogate(s) are aligned vertically.
  2. Measure and record the vertical distance \$s\$ between the release point and the centre of the first photogate.
  3. If using two photogates, measure the separation \$d\$ between them.
  4. Attach the ball to the release mechanism at the measured height.
  5. Release the ball without imparting any initial velocity (initial speed \$u = 0\$).
  6. Record the time \$t\$ taken for the ball to pass each photogate (or the time between the two photogates).
  7. Repeat the measurement at least five times and calculate the average time.
  8. Vary the release height and repeat steps 2–7 to obtain a set of \$s\$\$t\$ data.

Theoretical Background

For an object released from rest and moving under constant acceleration \$a\$ (here \$a = g\$), the equation of motion is

\$s = ut + \frac{1}{2} a t^{2}\$

Since \$u = 0\$, this simplifies to

\$s = \frac{1}{2} g t^{2} \quad\Longrightarrow\quad g = \frac{2s}{t^{2}}\$

If two photogates are used, the velocity \$v\$ at the first gate can be found from

\$v = \frac{d}{t_{12}}\$

and then \$g\$ can be obtained from the relation

\$v^{2} = u^{2} + 2gs \quad\Rightarrow\quad g = \frac{v^{2}}{2s}\$

Both approaches should give consistent values of \$g\$ within experimental uncertainties.

Data Collection

TrialHeight \$s\$ (m)Time \$t\$ (s)\$t^{2}\$ (s²)Calculated \$g\$ (m s⁻²)
10.500.3190.1029.80
20.600.3500.1239.76
30.700.3780.1439.79
40.800.4040.1639.81
50.900.4280.1839.84

Data Analysis

  • Plot \$s\$ (vertical axis) against \$t^{2}\$ (horizontal axis). The graph should be a straight line passing through the origin.
  • The gradient \$m\$ of the line equals \$\frac{1}{2}g\$. Hence \$g = 2m\$.
  • Calculate the gradient using linear regression or by drawing a best‑fit line.
  • Determine the experimental value of \$g\$ and compare it with the accepted value \$9.81\ \text{m s}^{-2}\$.

Uncertainty and Error Analysis

  • Random errors: reaction time of the release mechanism, timing resolution of the photogate.
  • Systematic errors: mis‑alignment of photogates, air resistance (negligible for dense sphere), incorrect measurement of height.
  • Calculate the percentage uncertainty in \$g\$ using propagation of uncertainties:

    \$\frac{\Delta g}{g} = \sqrt{\left(\frac{\Delta s}{s}\right)^{2} + \left(2\frac{\Delta t}{t}\right)^{2}}\$

  • Discuss how each source of error could be minimised (e.g., use a longer fall distance to reduce relative timing error).

Sample Calculation

For a height \$s = 0.70\ \text{m}\$ and an average time \$t = 0.378\ \text{s}\$:

\$t^{2} = (0.378)^{2} = 0.143\ \text{s}^{2}\$

\$g = \frac{2s}{t^{2}} = \frac{2 \times 0.70}{0.143} = 9.79\ \text{m s}^{-2}\$

Safety Considerations

  • Ensure the falling object cannot strike anyone; keep the area beneath the experiment clear.
  • Secure the stand and clamps to prevent them from toppling.
  • Do not use glass or fragile objects as the falling mass.

Suggested diagram: Schematic of the set‑up showing the retort stand, photogate(s), release mechanism, and measured height \$s\$.