Forces and motion · GCSE Physics
Force and acceleration required practical
Plan the GCSE Physics force–acceleration investigation: trolley on a ramp or linear air track, F = ma setup, varying mass while controlling force, measuring acceleration, graph and friction controls.
Keep resultant force constant (same hanging mass). Change trolley mass. Measure acceleration from v–t graph or light gates. Plot a against 1/m; friction and ramp angle are controls.
The important bits
What you need to know
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Aim: investigate how acceleration depends on mass when the resultant force is constant, testing F = ma (a = F/m for constant F).
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Typical setup: trolley on a smooth ramp or track; string over a pulley to a hanging mass that provides the accelerating force F = mg (in newtons, using m in kg and g = 9.8 N/kg).
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Independent variable: total mass of the trolley (add slotted masses to the trolley while reducing the hanging mass so resultant force stays constant — or keep the same hanging mass and vary trolley mass only if the specification method allows).
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Dependent variable: acceleration a in m/s², found from a velocity–time graph gradient, from a = 2s/t² for a trolley starting from rest over measured distance s, or from light gates spaced distance d apart: a ≈ (v₂² − v₁²)/(2d).
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Control variables: same ramp angle, same surface (low friction track), same starting position, same hanging mass if that is the chosen constant force, and no extra pushes at release.
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Friction controls: use a compensated ramp (tilt until trolley moves at constant speed with no driving mass — then add the hanging mass), or use a linear air track to reduce friction. Lubricate wheels; repeat and mean.
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Predict inverse relationship: if F is constant, a ∝ 1/m. Plot acceleration (y) against 1/mass (x) — expect a straight line through the origin if friction is small and F is truly constant.
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Safety: keep feet clear of the ramp; slotted masses can fall; secure the pulley and clamp the track. Do not stand where the trolley runs off the end.
Quotations worth analysing
Short evidence. Real method.
“resultant force F = mass × acceleration (F = ma)”
For this practical, F is the resultant. If friction is small, the hanging weight provides F. Use kg and m/s² so F is in newtons.
“For constant resultant force, acceleration is inversely proportional to mass.”
Double the trolley mass, halve the acceleration if friction is negligible. The graph of a against 1/m tests that prediction.
“Compensate for friction by inclining the track until the trolley moves at steady speed without a driving force.”
If you skip this, measured acceleration is lower than theory and the line may not pass through the origin.
Go deeper
How do I measure acceleration without guessing?
Release the trolley from rest each time. Use two light gates a distance d apart: record speeds v₁ and v₂. Then a = (v₂² − v₁²)/(2d). Or film with a ruler and timer, plot v–t, take the gradient. Or measure time t to travel distance s from rest: s = ½at² so a = 2s/t². Whichever method you use, keep it identical for every mass. The hanging mass m_h gives driving force F ≈ m_h g only if friction is compensated and the string is parallel to the track. Convert grams to kilograms before F = ma. If you add 0.5 kg to the trolley, total mass is trolley plus added masses plus hanging mass if it moves with the system — check your teacher’s exact model (some treat hanging mass separately). Consistency matters more than fancy kit.
Go deeper
Why must friction be controlled on a ramp?
Real tracks have friction and air resistance. Without compensation, the resultant force is less than m_h g, and acceleration is systematically low. Tilting the ramp until the trolley rolls at constant speed with no hanging mass means friction down the slope balances the component of weight — then when you add a small hanging mass, that extra force is closer to the true resultant. A linear air track lifts the trolley on a cushion of air and cuts friction dramatically. In evaluation, say that friction causes the graph of a against 1/m to curve or miss the origin. Improvements: compensate the ramp, use a track, repeat releases, and use light gates instead of stopwatch timing over short distances (human reaction time distorts t).
Go deeper
What graph proves F = ma at GCSE level?
With constant F, plot a (y) against 1/m_total (x). Expect a straight line through the origin: steeper gradient means larger F. Alternatively plot F (y) against a (x) with fixed mass — gradient equals mass. Students sometimes plot a against m and expect a straight line — that is wrong; it should be a curve decreasing hyperbolically. Label axes with units. Calculate gradient: on a vs 1/m, gradient ≈ F. Compare to the hanging weight used. If gradient is much smaller, friction absorbed part of F. Anomalies: bump at release, string rubbing on edge, trolley hitting the pulley. Mean two or three runs per mass.
See the idea in action
Hanging mass 0.050 kg → F ≈ 0.050 × 9.8 = 0.49 N. Trolley mass 0.80 kg → total moving mass 0.85 kg if the string mass is ignored. Light gates 0.50 m apart give v₁ = 0.40 m/s and v₂ = 1.20 m/s. a = (1.20² − 0.40²)/(2 × 0.50) = (1.44 − 0.16)/1.0 = 1.28 m/s². Check F/m = 0.49/0.85 ≈ 0.58 m/s² — close, but friction lowers the measured a. With trolley mass 1.30 kg total, a falls to about 0.84 m/s². Plot a against 1/m: points should lie on a line through the origin if friction is compensated. IV: mass. DV: acceleration. Controls: same F, same ramp, same method.
Exam technique
Turn knowledge into marks
State how you keep resultant force constant and how you measure acceleration with units. Mention friction compensation or a low-friction track. Plot a against 1/m for constant F, or explain F = ma from the gradient. Repeat and mean.
Common mistakes
Do not give these marks away
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Plotting acceleration against mass and expecting a straight line, instead of a against 1/m for constant force.
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Using mass in grams in F = ma, or forgetting that the hanging mass provides the driving force in newtons.
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Ignoring friction and then claiming results “prove F = ma” when measured acceleration is much lower than F/m.
With constant resultant force, how should acceleration change if the total mass of the trolley is doubled?
AAcceleration doubles
BAcceleration stays the same
CAcceleration halves
DAcceleration quadruples
Show the answer
Acceleration halves. From a = F/m, if F is constant and mass doubles, acceleration is halved. A graph of a against 1/m should be a straight line through the origin.
Quick questions
If this is the bit you searched
What is the independent variable in the force–acceleration practical?
The mass of the trolley (or total moving mass), while keeping the resultant force constant — usually the same hanging mass on the string.
How do you reduce friction in this investigation?
Use a compensated ramp, a linear air track, smooth wheels, and a clean track. Tilt the ramp until the trolley moves at constant speed without a driving force before adding the hanging mass.
What graph should you draw for constant force?
Acceleration on the y-axis and 1/mass on the x-axis. Expect a straight line through the origin if friction is small.
How is acceleration measured on the ramp?
From light gates and v² = u² + 2as, from the gradient of a velocity–time graph, or from a = 2s/t² for motion from rest over distance s.