Newton's Second Law Virtual Lab

Newton's second law links three ideas: the resultant force acting on an object, the object's mass, and its acceleration. For an object of constant mass it is written F = ma. A push makes a heavy trolley speed up slowly and a light one speed up quickly, and the law says exactly how much.

This laboratory tests the law with the standard dynamics-trolley practical. A trolley sits on a track. A string runs from the trolley over a pulley to a small hanging mass. The hanging weight pulls the string, the trolley accelerates along the track, and light gates measure how fast its speed changes. You can change the trolley mass, the hanging mass, the friction, the tilt of the track and the air drag, then check every result against theory.

The notes work in three layers. Foundation material needs basic algebra and the idea of acceleration as change of speed. Core material matches typical A-level, IB Diploma, AP Physics 1, WASSCE and Cambridge AS/A Level courses and uses free-body diagrams, simultaneous equations and graph gradients. Extension material goes towards IB HL, AP Physics C and first-year university and uses quadratic drag, terminal velocity, energy methods and the effect of the pulley.

The simulator opens with m₁ = 0.50 kg, m₂ = 0.10 kg, friction coefficient 0.05 and a 2.0 m track. Before you press Release, predict the acceleration. The default run gives a = 1.226 m/s².

Learning outcomes

Foundation

• State that a resultant force on an object causes it to accelerate in the direction of the force.
• Describe how acceleration changes when the force changes and when the mass changes.
• Distinguish mass (kilograms) from weight (newtons).
• Use F = ma in simple one-step problems.
• Explain how a light gate and a card give the speed of a trolley.

Core (A-level, IB, AP, WAEC)

• Draw free-body diagrams for the trolley and for the hanging mass.
• Identify the correct mass to use (trolley plus hanging mass) and explain why.
• Derive a = (m₂g − μm₁g) / (m₁ + m₂) and the string tension T.
• Explain why the tension is less than m₂g while the hanging mass accelerates.
• Explain friction compensation by tilting the track.
• Plot acceleration against force and against 1/mass, draw lines of best fit, and find the mass from a gradient with its uncertainty.

Extension (HL, AP C, university)

• Solve the motion with quadratic air drag and find the terminal velocity v_t = √(F / k).
• Use the work-energy theorem and the impulse-momentum theorem to check results.
• Explain how the rotational inertia of a real pulley adds to the effective mass.
• Describe how the intercepts of the two graphs reveal the friction coefficient.

Force, mass and acceleration

A force is a push or a pull, measured in newtons (N). One newton is the resultant force that gives a mass of 1 kg an acceleration of 1 m/s². Acceleration is the rate of change of velocity, in m/s². Mass is a measure of how strongly an object resists changes in its motion (its inertia), in kilograms.

Newton's second law states that the resultant force on an object equals its rate of change of momentum. For constant mass this becomes

ΣF = ma

Three points matter in every problem. The force is the resultant (vector sum) of all forces on one chosen object. The acceleration has the same direction as that resultant. The mass is the mass of that same object, or of the whole system when the objects move together.

Mass and weight

Weight is the gravitational force on an object, W = mg, with g = 9.81 N/kg near the Earth's surface. A 0.20 kg mass weighs 0.20 × 9.81 = 1.962 N. Mass does not change from place to place, but weight does. On the Moon the same mass weighs about one sixth as much.

The trolley and hanging mass system

A trolley of mass m₁ rests on a level track. A string passes over a light, frictionless pulley and carries a hanging mass m₂. The string is light and does not stretch, so both masses always have the same speed and the same size of acceleration. When released, the hanging mass falls and the trolley moves towards the pulley.

Free-body diagrams

A free-body diagram shows one object alone, with an arrow for every force acting on it, drawn from the object. Forces the object exerts on other things are left out. The simulator draws live free-body diagrams for both masses.

• Trolley: tension T along the track towards the pulley, friction F opposing the motion, normal reaction N upwards and weight m₁g downwards. N and m₁g cancel on a level track, because there is no vertical acceleration.
• Hanging mass: weight m₂g downwards and tension T upwards. These do not cancel, because the hanging mass accelerates downwards.

Why the whole system is used

The string forces the two masses to move together, so the moving system has mass M = m₁ + m₂. The string tension is an internal force of the system: it pulls the trolley forward and pulls the hanging mass back by the same amount, so it cancels when the system is treated as one body. The only external forces along the line of motion are the weight m₂g (driving) and friction (opposing). That gives F_net = m₂g − F, and the law is applied with mass M.

Why the tension is less than m₂g

If the tension equalled m₂g, the resultant force on the hanging mass would be zero and it would fall at constant speed, so it could never accelerate the trolley from rest. The hanging mass accelerates downwards, so the resultant force on it is downwards: m₂g − T = m₂a. Hence T = m₂(g − a), which is smaller than m₂g by m₂a. The tension equals the weight only when a = 0, for example when the trolley is held still or has reached terminal velocity.

Newton's third law

The pull of the string on the trolley and the pull of the trolley on the string are a third-law pair: equal in size, opposite in direction, acting on different bodies. They do not cancel in a free-body diagram, because each appears on a different diagram. In the same way, the string pulls the hanging mass up with the same tension that the string pulls the trolley forward (for a light string over a frictionless pulley).

Friction and its compensation

Kinetic friction opposes motion and has size F = μN, where N is the normal reaction and μ the coefficient of friction. On a level track N = m₁g. A trolley will not start moving unless the driving weight exceeds the friction limit, m₂g > μm₁g. To compensate for friction, raise the end of the track away from the pulley until a trolley given a gentle push runs at steady speed. The slope component m₁g sin α then cancels the friction μm₁g cos α, so tan α = μ.

Acceleration without friction (Core)

Take the direction of motion as positive. For the hanging mass, with the resultant force downwards:

m₂g − T = m₂a

For the trolley on a level, frictionless track, the only horizontal force is T:

T = m₁a

Add the two equations. T cancels:

m₂g = (m₁ + m₂)a   ⇒   a = m₂g / (m₁ + m₂)

Substituting back gives the tension:

T = m₁a = m₁m₂g / (m₁ + m₂)

Two limits are a useful check. If m₁ is huge, a tends to zero and T tends to m₂g. If m₁ is tiny, a tends to g and T tends to zero, which is free fall of the hanging mass.

With friction (Core)

With kinetic friction F = μm₁g on a level track, the trolley equation becomes T − μm₁g = m₁a. Adding it to the hanging-mass equation:

a = (m₂g − μm₁g) / (m₁ + m₂)     T = m₁m₂g(1 + μ) / (m₁ + m₂)

This is the theory formula used by the simulator and by the comparison tool in the Analyse step. The trolley moves only if m₂ > μm₁.

With a tilted track (Core)

If the track is tilted by α with the pulley end lower, the weight of the trolley has a component m₁g sin α down the slope (helping the motion) and the normal reaction becomes m₁g cos α. Then

a = [m₂g + m₁g(sin α − μ cos α)] / (m₁ + m₂)

When tan α = μ, the bracket is zero and a = m₂g / (m₁ + m₂), as if there were no friction.

Constant acceleration from rest

Because the force is constant, so is a, and the equations of uniform acceleration apply: v = at, s = ½at², v² = 2as. A velocity-time graph is a straight line through the origin with gradient a.

Light-gate acceleration (Core)

A card of width w interrupts a light gate for a time Δt, so the mean speed over the card is v = w / Δt. For constant acceleration this mean speed equals the speed at the middle of the interrupt. Two gates give speeds v₁ and v₂ at times t₁ and t₂ (the middle times), and

a = (v₂ − v₁) / (t₂ − t₁)

The two graph methods (Core)

Acceleration against force, fixed total mass. Move masses between the trolley and the hanger so that M = m₁ + m₂ stays constant, and let F = m₂g. Since m₁ = M − m₂, the friction formula gives

a = [(1 + μ)F − μMg] / M   ⇒   gradient = (1 + μ) / M,   intercept = −μg

With friction compensated or absent (μ = 0 in the formula), the line passes through the origin with gradient 1/M, so M = 1 / gradient.

Acceleration against 1/mass, fixed force. Keep m₂ fixed and add mass to the trolley. With M = m₁ + m₂:

a = (1 + μ)m₂g × (1 / M) − μg   ⇒   gradient = (1 + μ)m₂g,   intercept = −μg

With no net friction the gradient is the driving force m₂g, so a is proportional to 1/M.

Quadratic drag and terminal velocity (Extension)

A sail on the trolley gives a drag force kv² opposing the motion, where k has units kg/m. Applying the law to the system:

(m₁ + m₂) dv/dt = F_net − kv²

where F_net = m₂g − μm₁g on a level track. The acceleration falls as v rises and reaches zero when kv² = F_net, giving the terminal velocity

v_t = √(F_net / k)

Separating variables gives the exact solution v = v_t tanh(t / τ), where τ = M / (kv_t) is the time constant, and the distance travelled is s = v_tτ ln cosh(t / τ). The simulator integrates the equation numerically with a fourth-order Runge-Kutta method and checks it against these formulae.

Effect of the pulley (Extension)

A real pulley of moment of inertia I and radius r must also be spun up. For a string that does not slip, its kinetic energy adds I/r² to the effective mass: a = m₂g / (m₁ + m₂ + I/r²), and the tension is different on the two sides of the pulley.

Every example uses g = 9.81 m/s². Intermediate values are kept at full calculator precision and only final answers are rounded. The pulley and string are light and the pulley is frictionless unless stated.

Motion graphs for one run (Foundation and Core)

• Position-time: a curve that gets steeper. For constant acceleration from rest it is a parabola, s = ½at². The gradient at any instant is the velocity.
• Velocity-time: a straight line through the origin when the resultant force is constant. The gradient is the acceleration and the area under the line is the distance travelled.
• Acceleration-time: a horizontal line at the constant value a. With air drag the line falls towards zero.

The velocity-time tab in the simulator also marks the two light-gate readings. They lie on the line, which shows that the gate method is measuring the same acceleration as the whole curve.

Force graphs (Core)

Newton's second law predicts two straight-line graphs. Each one changes a single variable while the other is held fixed.

Why is the friction line below the other? Friction removes μm₁g from the net force, and m₁ = M − m₂ shrinks as more mass moves to the hanger, so the loss falls as F rises. The two lines meet only when all the mass hangs on the string. Where the friction line crosses the force axis, the trolley just begins to move.

Reading a straight-line graph

• Straight line through the origin: direct proportionality.
• Gradient: use a large triangle that covers most of the line, and read points that are not data points.
• Intercept: a non-zero intercept is information. Here it tells you about friction. A line that bends at low force suggests that static friction or a sticking pulley is affecting the first points.
• Curve instead of a line: if you changed the total mass at the same time as the force, the graph of a against m₂ curves, because a = m₂g / (m₁ + m₂) is not linear in m₂ when m₁ is fixed.

Friction in a real trolley (Core)

Wheel bearings and the contact with the track produce a small friction force that opposes motion. A real trolley has rolling resistance rather than the sliding friction of the simple model, and its size changes a little with load and speed. The model F = μN with a constant μ is a useful approximation. Static friction is usually somewhat larger than kinetic friction, so the first few millimetres of motion can be uneven. The simulator uses one coefficient for both.

Compensation procedure. Raise the end of the track that is furthest from the pulley, with no hanging mass attached. Give the trolley a gentle push and adjust the height until it keeps a steady speed (equal gaps between points on a ticker tape, or constant light-gate speeds). At that tilt the component of weight down the slope equals friction. Check the result with a different load on the trolley, because tan α = μ is independent of the mass. In the simulator, press Compensate friction and release with the hanger removed to see the steady speed, then add it.

Air drag (Core and Extension)

At low speeds, drag on a smooth object is roughly proportional to speed. At the speeds of a lab trolley with a large card or sail, quadratic drag F = kv² is a better model, with k = ½ρC_dA (air density ρ, drag coefficient C_d and frontal area A). Drag is zero when the trolley starts, so the initial acceleration is unchanged, but it grows as the speed rises. A plain trolley has very small drag over a metre or two, which is why a sail is fitted to make the effect easy to see.

Terminal velocity (Extension)

As v rises, kv² increases until it equals the net driving force. The resultant force is then zero, so by the second law (and the first) the acceleration is zero and the speed stays constant at v_t = √(F_net / k). Notice that the terminal speed depends on the driving force and on k, but not on the trolley mass: mass only changes how long it takes to get there, through the time constant τ = M / (kv_t).

Other real-world effects (Core)

• Pulley: a real pulley has mass and bearing friction. Its rotational inertia acts like extra mass, and its friction lowers the net force. This is one reason measured accelerations are a little smaller than the formula predicts.
• String: a real string has mass and stretches slightly, and the string must run parallel to the track. If it slopes, part of the tension is wasted on lifting the trolley.
• Track: a track that is not level adds or removes a force of m₁g sin α. A tilt of only 0.5° changes the force on a 0.80 kg trolley by 0.07 N, which is not small when m₂g is about 0.5 N.
• Air track: a track with an air cushion reduces friction almost to zero, so no compensation is needed.
• Hanger height: the hanging mass accelerates only while the string is taut and above the floor. The measured run must finish before it lands.

Research questions (Core)

1. How does the acceleration of a trolley depend on the resultant force when the mass is fixed?
2. How does it depend on the mass when the force is fixed?
3. Does the measured acceleration agree with a = (m₂g − μm₁g) / (m₁ + m₂)?

Variables

Method for acceleration against force

1. Set the track level and compensate for friction. In the simulator, press Compensate friction.
2. Put all the slotted masses on the trolley, so the hanger carries only its own small mass. Record the total, for example M = 0.80 kg.
3. Tick Keep total mass constant. Move a 0.05 kg slotted mass from the trolley to the hanger.
4. Release the trolley from the same start line, and record the gate acceleration. Repeat steps 3 and 4 for m₂ = 0.05, 0.10, 0.15 and 0.20 kg. The ideal values are 0.613, 1.226, 1.839 and 2.453 m/s².
5. Plot a against F = m₂g and fit a line.

Moving masses rather than adding them keeps M constant. If you only added mass to the hanger, the total mass would change as well, and the graph would curve.

Method for acceleration against 1/mass

1. Keep one hanging mass, for example m₂ = 0.10 kg, throughout.
2. Add mass to the trolley in steps: m₁ = 0.30, 0.60, 0.90 and 1.20 kg. Calculate M = m₁ + m₂ each time.
3. Record a for each. Plot a against 1/M.

Choosing gate positions

Place gate 1 a little way from the start so that the trolley is moving and the first interrupt is short. Place gate 2 well downstream so that the speeds differ enough to give a precise difference. Leave a margin before the end of the track so the trolley does not strike the buffer inside the gate.

Repeats, range and risk

Use at least five different values of the independent variable over a wide range, repeat each run two or three times and average the accelerations. Keep the hanging mass small enough that the acceleration is well below g, so the run lasts long enough to time. Real hanging masses are safe, but a falling slotted-mass stack can hurt feet, so put a soft catch below the hanger.

Calculating a from the light gates (Core)

Take the default run, with m₁ = 0.50 kg, m₂ = 0.10 kg and no friction, and a 0.050 m card. Gate 1 at 0.30 m is interrupted for 0.0485 s, with its middle time at 0.630 s. Gate 2 at 1.40 m is interrupted for 0.0232 s, with its middle time at 1.320 s.

• v₁ = 0.050 / 0.0485 = 1.031 m/s and v₂ = 0.050 / 0.0232 = 2.155 m/s.
• a = (2.155 − 1.031) / (1.320 − 0.630) = 1.124 / 0.690 = 1.63 m/s².

The formula gives 1.635 m/s², so the agreement is within the rounding of the readings.

Uncertainty in a measurement (Core)

• Absolute uncertainty: the size of the possible error in the same unit, such as 0.0002 s.
• Percentage uncertainty: (absolute uncertainty / value) × 100. A timing uncertainty of 0.0002 s on an interrupt of 0.0485 s is 0.41%.
• Combining: for products and quotients add the percentage uncertainties. For a difference, such as v₂ − v₁, add the absolute uncertainties, which is why a small difference gives a poor percentage uncertainty. This is why gate 2 should be far from gate 1.
• Repeat readings: the spread of repeated values (half the range) estimates the random uncertainty. The mean is more reliable than any single value.

The simulator's Realistic timer scatter adds about 0.2 ms of random error to each gate time. Run the same release five times with scatter switched on to see the spread in a.

Line of best fit and gradient uncertainty (Core)

Draw the line so that points lie evenly on both sides. The Analyse tool uses least squares, which minimises the sum of the squared vertical distances. It reports the gradient and intercept with standard errors, and r², which is close to 1 for a good straight line.

By hand, the worst-line method draws the steepest and shallowest lines that still pass through the uncertainty bars of the points. Half the difference between their gradients is the uncertainty in the gradient. The dotted lines on the Analyse graph show the same idea using the standard error.

Turning gradient into mass

For the force graph with friction compensated, the gradient is 1/M, so M = 1/gradient. The fractional uncertainty in M equals the fractional uncertainty in the gradient. Example 6 in section D gives M = 0.80 ± 0.02 kg. State the answer to the same number of decimal places as its uncertainty.

Comparing with theory

A result agrees with theory when the theory value lies within the experimental uncertainty (often within two standard errors). Compute the percentage difference, (measured − theory) / theory × 100, and say whether random scatter, a systematic error or a model limit is the likeliest cause.

• Random errors scatter points on both sides of the line and are reduced by repeating and averaging.
• Systematic errors shift every point the same way: an un-levelled track, a hidden friction force, a wrong card width or a mass that was not measured. They show up as an unexpected intercept or a gradient that is consistently off.

What the intercept tells you

Both force graphs have the intercept −μg. An intercept of −0.49 m/s² means μ = 0.49 / 9.81 = 0.05. If the intercept is zero within uncertainty, friction was successfully compensated.

The equipment (Foundation and Core)

• Dynamics trolley and track: a low-friction trolley with slots or pegs for extra masses, on a straight, level track.
• Pulley: light and well oiled, mounted at the end of the track, with the string level with the trolley's hook.
• Hanger and slotted masses: usually 10 g slotted masses on a 10 g hanger. Mass is transferred between hanger and trolley.
• Light gates and timer: two gates connected to a timer or data logger, with an interrupt card of known width on the trolley.
• Balance: to measure the trolley and each mass. A trolley is often 0.5 to 1 kg, with a precision of 1 g.
• Metre rule: to measure the gate separation and the card width.

Alternative ways to measure acceleration

• Ticker timer: dots on a tape every 0.02 s. Gaps grow as the trolley accelerates, and the change of speed per interval gives a.
• Motion sensor or video: a data logger plots velocity-time directly and the gradient is a.
• One gate and a long card: a card with two interrupt segments gives two speeds and the time between them from a single gate.
• Distance and time: for a fixed distance s, time with a stopwatch and use a = 2s/t². Reaction time of 0.2 s is large next to a run of 2 s, so this is the least accurate method.

Sources of error and good practice (Core)

• Check the track is level, with a spirit level, before compensating.
• Release the trolley without a push, from the same start line each time.
• Measure the card width with callipers or a rule that reads to 0.5 mm. A 0.050 m card measured to 0.5 mm has a 1% uncertainty.
• Do not forget the mass of the hanger itself, and keep m₂ as the total hanging mass.

What the simulation includes

• A trolley of adjustable mass on a track of adjustable length, pulled by a hanging mass over an ideal pulley.
• Kinetic friction with a constant coefficient, and an adjustable track tilt.
• Optional quadratic drag, integrated numerically and checked against the exact formulae.
• Two light gates and an interrupt card, with optional timing scatter.

What it leaves out

• The pulley has no mass and no friction, and the string is light and does not stretch.
• The friction coefficient is constant, and starting friction is the same as moving friction.
• The track is perfectly straight and the string is perfectly parallel to it.
• The hanging mass never reaches the floor before the trolley reaches the end of the track, and the drop shown on screen is not drawn to scale.
• The trolley stops being simulated when it reaches the end of the track, so no collision with the buffer is modelled.
• Rolling resistance, wheel spin and string vibration are ignored.

Safety in a real laboratory

• Stop the trolley with a buffer or a soft barrier before it reaches the end of the bench.
• Place a cushion on the floor under the hanger, and keep feet clear.
• Keep the string short enough that the hanger cannot swing into people.
• Follow your own school or institution's practical safety rules and risk assessment.

• Using only the trolley mass. The hanging mass is accelerated too. Always divide by m₁ + m₂ when the whole system is the object.
• Saying T = m₂g. This is true only when the hanging mass is not accelerating. While it accelerates, T = m₂(g − a) is smaller.
• Thinking the acceleration is g. The hanging mass is not in free fall, because the string pulls it up. a = g only in the limit of a trolley with no mass.
• Adding mass to the hanger in the force experiment. That changes both the force and the total mass. Move the mass from trolley to hanger instead.
• Mixing up mass and weight. The F in F = ma is a force in newtons. Convert a hanging mass in kilograms to a weight with g before comparing forces.
• Forgetting friction is present. A line that misses the origin is not a mistake in the graph. It shows friction, which can be removed by compensation.
• Believing a force is needed to keep an object moving. A resultant force is needed to change the velocity. At terminal velocity the resultant force is zero but the object still moves.
• Believing the heavier object accelerates faster. With the same force, the heavier trolley accelerates less. In free fall, mass cancels because weight is proportional to mass.
• Treating action and reaction as one pair of forces on one body. Newton's third-law pairs act on different bodies, so they never cancel in a single free-body diagram.
• Drawing the normal reaction as the weight's third-law partner. N and weight cancel because of Newton's second law (no vertical acceleration). The third-law partner of the weight is the trolley's own gravitational pull on the Earth.
• Using the gradient wrongly. For a against F the gradient is 1/M, not M. For a against 1/M the gradient is F.
• Reading a very short interrupt time as exact. Rounding a time of 0.0232 s to 0.02 s changes the speed by about 16%. Keep three significant figures in timings.

Command words

• State, name: a short fact, no explanation needed.
• Calculate: show the equation, substitute numbers with units, then give the answer to a sensible number of significant figures.
• Explain: give a reason with a link, such as 'the hanging mass accelerates, so the resultant force on it is not zero, so T is less than its weight'.
• Determine: use data, a graph or a given formula to find the value.
• Suggest: apply your physics to an unfamiliar situation.

A method for connected-body problems

1. Sketch the situation and choose a positive direction for the motion.
2. Draw a free-body diagram for each body.
3. Write ΣF = ma for each body. The accelerations have the same size because the string is taut.
4. Either solve the two equations together, or use the whole-system shortcut, a = (driving force − opposing forces) / total mass, then find T from one body.
5. Check by substituting into the other body's equation. Check the answer is less than g, and that T lies between 0 and m₂g.

Graph questions

• Write the theory equation, compare it with y = mx + c, and say what the gradient and intercept stand for.
• Use a gradient triangle that spans more than half the line, and show the coordinates you read.
• For uncertainty, give the percentage uncertainty in the gradient, and carry it into the final answer.

Habits that earn marks

• Give units on every numerical answer. Mass in kg, force in N, acceleration in m/s².
• Keep extra digits until the last step, then round.
• Say 'resultant force', not just 'force', when the force is the one in F = ma.
• Marks in this laboratory are for learning, not an official grade.

This map is a guide to topic areas where Newton's second law and the dynamics-trolley practical are commonly taught. It is not an official statement from any exam board, and it gives topic names only. Always check your own current specification, syllabus or course framework for the exact content, the required practicals and the equations you are given.

The three level tags used in these notes (Foundation, Core, Extension) help you match the material to your course. This page supports, and does not replace, the specification you are following.

• Lifts and cranes: the cable tension is greater than the weight when the lift accelerates upwards and less when it accelerates downwards. The tension-less-than-weight idea from this laboratory applies to the downward case.
• Atwood machines and counterweights: a counterweight lift joins two masses over a pulley and accelerates the difference. Engineers choose the counterweight so that the motor handles only a small net force.
• Road vehicles: braking distance depends on friction and mass. A fully loaded vehicle needs more force to stop in the same distance.
• Skydiving and parachutes: air drag grows with speed until it equals weight, giving terminal velocity. Opening a parachute increases k, so the terminal speed falls.
• Rockets and launch systems: thrust minus weight is the resultant force, and the mass falls as fuel burns, so the acceleration rises. The momentum form of the law is needed.
• Sports science: sprinters accelerate against their own mass; a heavier athlete needs a greater push for the same acceleration. Force plates measure those forces directly.
• Conveyor and cable systems: a loaded cart pulled by a cable or a hanging counterweight is the same mechanism as this experiment.

The Atwood machine (Extension)

Two masses m₁ and m₂ (m₁ > m₂) hang on either side of a light, frictionless pulley. The two equations m₁g − T = m₁a and T − m₂g = m₂a give

a = (m₁ − m₂)g / (m₁ + m₂)     T = 2m₁m₂g / (m₁ + m₂)

For 0.30 kg and 0.20 kg, a = 0.10 × 9.81 / 0.50 = 1.962 m/s² and T = 2 × 0.30 × 0.20 × 9.81 / 0.50 = 2.354 N. The tension lies between the two weights (1.962 N and 2.943 N).

A pulley with inertia (Extension)

If the pulley has moment of inertia I and radius r, then I/r² adds to the effective mass. For m₁ = 0.50 kg, m₂ = 0.10 kg and I/r² = 0.10 kg, a = 0.981 / 0.70 = 1.401 m/s², against 1.635 m/s² for an ideal pulley. The tension is then different on the two sides: it is 0.701 N on the trolley side and 0.841 N on the hanging side. Experimenters can estimate I/r² by plotting 1/a against m₁ + m₂ for a fixed hanging mass: the line cuts the mass axis at −I/r².

Momentum form of the law

The general statement is F = dp/dt. For a rocket or a conveyor belt that gains or loses mass, F = ma is not enough. The momentum form also remains valid at speeds close to the speed of light, where p = γmv.

Drag regimes

Drag is roughly linear in v at low Reynolds number (small, slow objects in a viscous fluid) and roughly quadratic at high Reynolds number (larger, faster objects in air). Linear drag gives an exponential approach to terminal velocity, v = v_t(1 − e^−t/τ), while quadratic drag gives the tanh form used in this laboratory.

Numerical integration

The Euler method updates v and s using the acceleration at the start of each step and has an error that shrinks in proportion to the step size. The fourth-order Runge-Kutta method averages four slopes per step and its error shrinks with the fourth power of the step. The simulator uses Runge-Kutta, and its self-test compares the results with the exact tanh solution.

• The resultant force on an object equals its mass times its acceleration: F = ma. The acceleration is in the direction of the resultant force.
• In the trolley experiment the accelerating mass is m₁ + m₂. The tension is internal to that system.
• a = (m₂g − μm₁g) / (m₁ + m₂), and without friction a = m₂g / (m₁ + m₂), which is always less than g.
• T = m₂(g − a), which is less than m₂g by m₂a. T equals m₂g only if a = 0.
• Free-body diagrams show every force on one body. N and m₁g cancel on a level track. T and m₂g do not cancel on the hanger.
• Friction is compensated by tilting the track so that tan α = μ. The net force is then m₂g only.
• Graph of a against F at constant total mass: straight line, gradient (1 + μ)/M, which is 1/M if friction is compensated; M = 1/gradient.
• Graph of a against 1/M at constant hanging mass: straight line, gradient (1 + μ)m₂g, which is the driving force if friction is compensated.
• Both graphs have intercept −μg. Zero intercept means friction is absent or compensated.
• With quadratic drag, terminal velocity v_t = √(F_net / k). The resultant force is zero there and the speed is constant. Mass affects only how quickly v_t is reached.
• Compare with theory using percentage difference and uncertainty, not by eye.

Acceleration

The rate of change of velocity with time; a vector, in m/s².

Air drag

A force opposing motion through air, increasing with speed.

Compensation (of friction)

Tilting a track so that the weight component down the slope cancels friction.

Coefficient of friction (μ)

The ratio of the friction force to the normal reaction between two surfaces; it has no unit.

Drag constant (k)

The constant in F = kv², in kg/m. It equals ½ρC_dA.

Free-body diagram

A diagram showing one body alone with all the forces acting on it.

Friction

A force between touching surfaces that opposes relative motion.

Gradient

The slope of a graph; change in y divided by change in x.

Inertia

The tendency of an object to resist a change in its motion; measured by mass.

Internal force

A force between two parts of a system, which cancels when the system is treated as one body.

Kinetic friction

Friction on surfaces that slide over each other.

Light gate

A sensor with a light beam that records the time for which an object interrupts it.

Mass

The amount of matter in an object and a measure of its inertia, in kg.

Newton (N)

The force that gives a 1 kg mass an acceleration of 1 m/s².

Normal reaction (N)

The contact force perpendicular to a surface.

Resultant force

The single force equal to the vector sum of all the forces on an object.

Runge-Kutta method

A numerical method that advances a solution in small steps using several slope estimates per step.

Standard error

An estimate of the uncertainty in a fitted quantity such as a gradient.

System

The collection of bodies chosen for analysis; here the trolley and the hanging mass.

Tension

The pulling force transmitted along a string or rope.

Terminal velocity

The constant speed reached when drag balances the driving force, so that the resultant force is zero.

Time constant (τ)

M / (kv_t); the time scale on which the speed approaches its terminal value.

Weight

The gravitational force on an object, W = mg, in newtons.

Only titles and publishers are given, together with stable home pages. Search each site for the title to reach the exact page.

• OpenStax, University Physics Volume 1, chapters on Newton's laws of motion and their applications. Free and open textbook. https://openstax.org/
• OpenStax, College Physics. Free algebra-based textbook with chapters on dynamics, force and Newton's laws. https://openstax.org/
• Bureau International des Poids et Mesures (BIPM), The International System of Units (SI Brochure), for the definition of the newton and the conventional value of standard gravity, 9.80665 m/s². https://www.bipm.org/
• Feynman, R. P., Leighton, R. B. and Sands, M., The Feynman Lectures on Physics, Volume I, on Newton's laws of dynamics. The California Institute of Technology hosts a free online edition. https://www.feynmanlectures.caltech.edu/
• Taylor, J. R., Classical Mechanics, University Science Books. Contains Newton's laws, drag forces and numerical methods (extension reading).
• Kleppner, D. and Kolenkow, R., An Introduction to Mechanics, Cambridge University Press (extension reading).

The simulation uses g = 9.81 m/s², a rounded value near the conventional 9.80665 m/s². Marks given by the examination are for learning, not an official grade.