Investigating Newton's second law

Practical method

Aim

Investigate how the acceleration of a trolley depends on resultant force at constant mass and on mass at constant force, testing a ∝ F and a ∝ \(\dfrac{1}{m}\).

Variables

Independent variable: accelerating force (or total mass)
Dependent variable: acceleration
Control variables: total accelerated mass when force is varied; force when mass is varied; friction-compensated runway.

Apparatus

• Trolley on a friction-compensated runway
• Pulley, string and hanging masses
• Two light gates or a motion sensor with data logger
• Card of known length
• Balance

Method

1. Tilt the runway until the trolley moves at constant speed when pushed (friction compensation).
2. Attach a hanger over the pulley; keep spare masses on the trolley.
3. Release the trolley and record the speeds at two light gates and the time between them; a = \(\dfrac{v_{2} - v_{1}}{t}\).
4. Move masses from the trolley to the hanger to increase the force while keeping the total mass constant. Repeat for five forces.
5. Then keep the hanging mass fixed and add mass to the trolley.
6. Plot a against F (straight line through the origin, gradient \(\dfrac{1}{m}\)) and a against \(\dfrac{1}{m}\) (straight line, gradient F).

Sample results

Total mass (trolley + hanger) 1.00 kg for the force series.

Analysis

a is proportional to F: the gradient is 0.99 kg⁻¹, so the total mass is 1.01 kg, agreeing with the measured 1.00 kg. For the mass series, a against \(\dfrac{1}{m}\) is a straight line with gradient equal to the accelerating force.

Uncertainty

With light gates, speeds are uncertain by about 1%, so each acceleration is uncertain by about 3%. The weights of the slotted masses are typically good to 1%.

Errors and improvements

Risk assessment

Common mistakes

• Forgetting that the hanger accelerates too, so it is part of the total mass.
• Adding masses to the hanger without removing them from the trolley.
• Using a stopwatch over short times.

Worked analysis

Speeds at two gates are 0.30 and 0.79 m s⁻¹, 0.62 s apart: a = \(\dfrac{0.49}{0.62}\) = 0.79 m s⁻². With F = 0.784 N, m = \(\dfrac{F}{a}\) = 0.99 kg.

Common misconception

The accelerating force is the weight of the hanging mass, but the mass being accelerated is the trolley plus the hanger.