Cambridge A-Level Physics 9702 – Equations of Motion: Determining gEquations 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
- Set up the retort stand so that the photogate(s) are aligned vertically.
- Measure and record the vertical distance \$s\$ between the release point and the centre of the first photogate.
- If using two photogates, measure the separation \$d\$ between them.
- Attach the ball to the release mechanism at the measured height.
- Release the ball without imparting any initial velocity (initial speed \$u = 0\$).
- Record the time \$t\$ taken for the ball to pass each photogate (or the time between the two photogates).
- Repeat the measurement at least five times and calculate the average time.
- 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
| Trial | Height \$s\$ (m) | Time \$t\$ (s) | \$t^{2}\$ (s²) | Calculated \$g\$ (m s⁻²) |
|---|
| 1 | 0.50 | 0.319 | 0.102 | 9.80 |
| 2 | 0.60 | 0.350 | 0.123 | 9.76 |
| 3 | 0.70 | 0.378 | 0.143 | 9.79 |
| 4 | 0.80 | 0.404 | 0.163 | 9.81 |
| 5 | 0.90 | 0.428 | 0.183 | 9.84 |
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
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\$.