Force, mass and acceleration

Practical method

Aim

Investigate how the acceleration of a trolley depends on the resultant force (constant mass) and on its mass (constant force), testing F = ma.

Variables

Independent variable: resultant force (or total mass)
Dependent variable: acceleration
Control variables: total accelerated mass when force is varied (move masses from trolley to hanger); force when mass is varied; the same track, slope and starting position.

Apparatus

• Dynamics trolley on a bench or friction-compensated runway
• Pulley, string and slotted masses on a hanger
• Two light gates with a card on the trolley (or ticker timer/motion sensor)
• Metre rule
• Balance

Method

1. Tilt the runway slightly so the trolley just rolls at constant speed when pushed; this compensates for friction.
2. Attach the trolley to a 1.0 N hanger load over a pulley. Keep all the other slotted masses on the trolley.
3. Release the trolley from the same point and record its speed at each light gate and the time between them; calculate acceleration a = \(\dfrac{v - u}{t}\).
4. Move a mass from the trolley to the hanger to increase the force while keeping the total mass constant. Repeat for five forces.
5. Repeat each run three times and calculate mean accelerations.
6. To vary mass, keep the hanger load fixed and add masses to the trolley.
7. Plot acceleration against force; a straight line through the origin shows a ∝ F, with gradient \(\dfrac{1}{m}\).

Sample results

Total accelerated mass 1.0 kg, friction-compensated runway.

Analysis

Acceleration is directly proportional to resultant force: the graph is a straight line through the origin with gradient ≈ 0.98 kg⁻¹, so m = \(\dfrac{1}{0.98}\) ≈ 1.0 kg, as expected. With the force fixed at 0.50 N, doubling the mass from 0.50 kg to 1.00 kg roughly halves the acceleration (0.98 to 0.49 m/s²), showing a ∝ \(\dfrac{1}{m}\).

Errors and improvements

Risk assessment

Common mistakes

• Adding masses to the hanger without removing them from the trolley.
• Forgetting the hanger's own mass in the force.
• Using speed instead of change in velocity per second.

Worked analysis

Two light gates give speeds of 0.20 m/s and 0.62 m/s, 0.70 s apart. a = (0.62 − 0.20) ÷ 0.70 = 0.60 m/s². With a total mass of 1.0 kg, F = ma = 1.0 × 0.60 = 0.60 N.

Common misconception

A heavier hanging mass does not only increase the force: it also increases the total mass being accelerated unless masses are moved from the trolley.

Variations required by some exam boards

On a ramp: a trolley can be released down a fixed ramp with different masses added. Measure its acceleration with light gates each time and compare with a = \(\dfrac{F}{m}\).