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A.2

Forces and momentum

Theme A · Space, time and motion · SL and HL

A.1 described motion. A.2 explains it. Everything here comes from three laws and one conserved quantity, and almost every question begins the same way: draw one object, draw every force acting on it, and add them as vectors. Get that diagram right and the physics is usually one line of algebra.

🎯What you need to be able to do

  • Describe a force properly — its size, direction, the object it acts on and the object exerting it — and name the common types.
  • Add forces as vectors and resolve a force into components, especially on a slope.
  • Draw free-body diagrams, and use translational equilibrium (\( \sum F = 0 \)).
  • Apply Newton’s three laws, including \( F = ma \) and \( F = \dfrac{\Delta p}{\Delta t} \), and identify third-law pairs correctly.
  • Distinguish mass from weight, and use Hooke’s law to measure a force.
  • Work with static and dynamic friction, \( F_f \le \mu_s R \) and \( F_f = \mu_d R \).
  • Use viscous drag (Stokes’ law) and buoyancy (\( F_b = \rho V g \)), including terminal speed in a fluid.
  • Use momentum and impulse, including the area under a force–time graph.
  • Apply conservation of momentum to collisions and explosions, and tell elastic from inelastic.
  • Treat circular motion as a force problem: \( a = \dfrac{v^{2}}{r} \), and know which real force is doing the job.

🧲What a force is

A force is a push or a pull, measured in newtons, and you can recognise one by what it does: it deforms something, or it changes something’s velocity. That second half matters more than it looks. A change in velocity is an acceleration, so a resultant force causes an acceleration — and no resultant force is needed to keep an object moving at constant velocity.

A force is an interaction between two objects, never a property of one. To describe a force completely you need five things:

  • its magnitude and its direction (it is a vector);
  • the object it acts on;
  • the object exerting it;
  • the nature of the force.

So a proper description reads: “a 50 N push, at 20° to the horizontal, acting ON the football FROM the boot”. If you cannot name the object exerting a force, that force does not exist — which is the single most useful test in this whole topic.

The types you will meet. Every one of them is ultimately gravitational or electromagnetic in origin; almost everything in daily life is electromagnetic.

  • Normal force, \( F_N \) — the part of the contact force between two surfaces that acts at right angles to them. Between smooth surfaces it is the only contact force there is.
  • Surface friction, \( F_f \) — opposes relative motion of the surfaces, and acts along them.
  • Tension and the elastic restoring force, \( F_H \) — a stretched spring pulls equally on the objects at both ends, in proportion to its extension (Hooke’s law).
  • Viscous drag, \( F_d \) — opposes motion through a fluid, and grows with speed.
  • Buoyancy, \( F_b \) — the upward force on anything immersed in a fluid.
  • Gravitational force, \( F_g \) — between masses; near the Earth we call it weight.
  • Electric, \( F_e \), and magnetic, \( F_m \) — between charges, and between currents or magnets.

One useful split is contact forces (normal, friction, tension, drag, buoyancy) against forces that act at a distance (gravitational, electric, magnetic). The first group needs surfaces touching; the second does not.

📐Forces as vectors, and free-body diagrams

Because forces are vectors, two of them at an angle do not add as numbers. Add them tip to tail, or resolve them into perpendicular components and add each direction separately — whichever makes the geometry easier.

Two panels. On the left, a 4.0 newton force east and a 3.0 newton force north are redrawn tip to tail and the closing vector is the 5.0 newton resultant at 37 degrees. On the right, a block rests on a slope of angle theta: its weight is drawn straight down and resolved by dashed arrows into a component W cos theta into the slope, balanced by the normal reaction R, and a component W sin theta down the slope which nothing balances.
Adding two forces, and splitting one force into components. On a slope, resolve along and perpendicular to the surface — never horizontally and vertically.

Free-body diagrams are the discipline that makes all of this work. Choose one object, then draw every force acting on that object and nothing else.

A book resting on a table, then two separate free-body diagrams. For the book: the contact force from the table upwards and the gravitational pull of the Earth downwards, equal because it is at rest. For the table: the push from the book downwards, the weight of the table downwards, and the reaction from the floor upwards, so that the floor's reaction equals the weight plus the push.
One situation, two objects, two different diagrams. The book’s push on the table appears only on the table’s diagram.
What does not belong on a free-body diagram. No velocity or acceleration arrows — they are not forces. No forces the object exerts on other things. No “centripetal force” added alongside the tension or friction that is already producing it. And no force you cannot attribute to a named object.

⚖️Newton’s first law and equilibrium

An object continues at rest, or moving at constant velocity in a straight line, unless a resultant external force acts on it. Read carefully, that adds nothing to the definition of a force — it simply insists on the converse: no resultant force means no acceleration, and observing no acceleration tells you the forces balance.

Four cases with their force arrows. A book at rest with equal up and down arrows; a car at a steady speed with equal engine and drag arrows; a parachutist at terminal velocity with equal air-resistance and weight arrows; and a person in a lift accelerating upwards, where the upward normal force arrow is visibly longer than the weight arrow.
Only the last case has a resultant force — and it is the only one where the two arrows are different lengths.

An object with zero resultant force is in translational equilibrium:

\[ \sum F = 0 \]

Because force is a vector, that means zero resultant in every direction. In two dimensions it is enough to show the forces balance in any two non-parallel directions — usually the two you chose when you resolved them. For a mass hanging on a string and pulled sideways by a force \(P\), equilibrium gives \( T\sin\theta = P \) horizontally and \( T\cos\theta = W \) vertically, and those two equations solve the whole problem.

Equilibrium does not mean “at rest”. A car at a steady 90 km h\(^{-1}\) is in equilibrium; a child at the very end of a swing’s arc is momentarily at rest but is not in equilibrium, because there is a resultant force about to accelerate them back. Being instantaneously stationary and having zero resultant force are different statements.

🧮Newton’s second law

The first law says a resultant force causes acceleration. The second says how much. In its most general form, the resultant force equals the rate of change of momentum:

\[ F = \frac{\Delta p}{\Delta t} \qquad \text{and, when the mass is constant,} \qquad F = ma \]

Four things are worth pinning down. \(F\) is always the resultant force, so work that out first. The mass must be in kilograms for \(F = ma\) to give newtons. The law is experimental — it was not derived from anything. And when the mass changes (a rocket burning fuel, coal landing on a conveyor belt) or the force varies, go back to \( F = \Delta p / \Delta t \), which still holds.

✏️Worked example 1 — a block on a rough slope

A 3.0 kg block is placed on a slope at 30° to the horizontal. The maximum friction force between the block and the slope is 8.0 N. Take \( g = 9.81 \) m s\(^{-2}\). Does the block slide, and if so with what acceleration?

Resolve along and perpendicular to the slope. The weight is \( W = 3.0 \times 9.81 = 29.4 \) N, so

down the slope: \( W\sin 30^\circ = 14.7 \) N
into the slope: \( W\cos 30^\circ = 25.5 \) N

Perpendicular to the slope nothing accelerates, so the normal reaction must be \( R = 25.5 \) N — not 29.4 N, which is the mistake waiting to be made here.

Along the slope, the 14.7 N pulling it down exceeds the 8.0 N of friction available, so it does slide. Friction then acts up the slope at its maximum value:

\[ F_{\text{resultant}} = 14.7 - 8.0 = 6.7\ \text{N} \qquad a = \frac{6.7}{3.0} = 2.2\ \text{m s}^{-2}\ \text{down the slope} \]
Check it. The answer must be smaller than \( g\sin 30^\circ = 4.9 \) m s\(^{-2}\), the acceleration on a frictionless slope of the same angle — and 2.2 is. If friction ever appears to make something accelerate faster, a sign is wrong.

🤝Newton’s third law

When two bodies A and B interact, the force A exerts on B is equal in magnitude and opposite in direction to the force B exerts on A. In symbols, \( F_{AB} = -F_{BA} \). Forces always come in pairs, which makes the third law a checklist: if you have drawn a force, something somewhere is feeling its partner.

On the left, marked with a red cross, the weight of a book and the table's upward push on it are shown as a supposed third-law pair, with the explanation that they act on the same object and are different types of force. On the right, marked with a green tick, the two genuine pairs: the Earth pulls the book down while the book pulls the Earth up, and the table pushes the book up while the book pushes the table down.
The classic wrong answer, and the two right ones. A pair never appears on a single free-body diagram.

Two properties settle every third-law question. The pair acts on two different objects — so two forces on the same free-body diagram are never a pair. And the pair is the same type of force: if A pulls B gravitationally, then B pulls A gravitationally too.

The law also explains things that look one-sided. Two skaters push apart and feel equal forces, but the lighter one accelerates more, because \( a = F/m \). A person pushing off a wall is accelerated by the wall’s push on them; the wall (attached to the Earth) has far too much mass for its acceleration to be noticed. And a car accelerates because the road pushes it forwards — the engine only turns the wheels, which push backwards on the road.

⚖️Mass, weight and Hooke’s law

Mass is the amount of matter in an object, measured in kilograms, and it does not change when you move the object. Weight is a force, measured in newtons. Take an object to the Moon and its mass is unchanged while its weight falls to about a sixth, because \( g \approx 1.6 \) N kg\(^{-1}\) there against 9.81 N kg\(^{-1}\) here.

The word “weight” is genuinely ambiguous in physics: some people mean the gravitational force \( mg \), others mean the reading on a supporting scale. In equilibrium the two agree, and in an accelerating lift they do not — which is exactly why you feel heavy as a lift starts upwards, while the gravitational force on you has not changed at all. Say “gravitational force” when you mean \( mg \) and the ambiguity disappears.

Hooke’s law turns a spring into a force meter. Up to its elastic limit, the extension \(x\) of a spring is proportional to the force stretching it:

\[ F = kx \qquad k \text{ in N m}^{-1} \]

so a graph of force against extension is a straight line through the origin whose gradient is the spring constant. Measure the extension and you have measured the force — a newtonmeter is nothing more than this.

🧱Solid friction

Friction opposes the relative motion of two surfaces, and it arises because no surface is smooth at the microscopic scale. It comes in two varieties, and the difference matters.

On the left, a block pushed with 0, 5, 10 and 15 newtons: friction matches the push exactly at 0, 5 and 10 newtons while the block stays still, then drops to 9 newtons once the block slides. On the right, the same story as a graph of friction against applied push: a straight line of gradient one up to a maximum, then a constant, slightly lower horizontal line once sliding starts.
Static friction is self-adjusting; dynamic friction is roughly constant. The break point is \( F_{max} = \mu_s R \).

Static friction acts between surfaces that are not sliding, and it takes whatever value is needed to prevent sliding — up to a maximum. Dynamic (or kinetic) friction acts once they are sliding, and is roughly constant:

\( F_f \le \mu_s R \)
\( F_f = \mu_d R \)
\( \mu_d < \mu_s \)

where \(R\) is the normal reaction. The coefficients are ratios of two forces, so they have no units, and they are usually less than 1. Two consequences surprise people: the maximum friction does not depend on the area of contact when the surfaces are held together by gravity, and friction drops at the moment the object starts to move — which is why a heavy box lurches forward as it finally gives way.

Never write \( F_f = \mu_s R \) for something at rest. That is the maximum static friction, not the actual value. A block sitting untouched on a table has zero friction acting on it, however large \( \mu_s R \) may be. Use the inequality, and get the actual value from the other forces.

Tilting a slope until an object just slides is the neatest way to measure \( \mu_s \), because everything else cancels.

A block on a slope of angle theta with its weight resolved into components along and perpendicular to the surface, the normal reaction, and friction acting up the slope. The algebra beside it shows F equals W sin theta, R equals W cos theta, and that at the angle where the block just slides the coefficient of static friction equals tan theta.
At the angle where sliding just begins, \( \mu_s = \tan\theta \) — the weight cancels, so the result is the same for any block.

💧Viscosity, drag and buoyancy

An ideal fluid slides past itself freely. A real fluid does not: its layers rub, and that internal friction is viscosity, \( \eta \). It is defined as the ratio of the tangential stress applied to a fluid to the velocity gradient it produces,

\[ \eta = \frac{\text{tangential stress}}{\text{velocity gradient}} = \frac{F/A}{\Delta v / \Delta y} \]

which gives units of Pa s (equivalently N s m\(^{-2}\) or kg m\(^{-1}\) s\(^{-1}\)). Water is about \( 1.0 \times 10^{-3} \) Pa s at room temperature and thick syrup about \( 1.0 \times 10^{2} \) Pa s — five orders of magnitude apart. Viscosity is extremely sensitive to temperature, which is why engine oil is graded for it and why syrup pours so much better warm.

For a sphere moving slowly through a fluid, the drag force is given by Stokes’ law, and an immersed object also feels an upthrust given by Archimedes’ principle.

A sphere of density rho falling through a fluid of density sigma and viscosity eta, with three force arrows: weight down, upthrust up and viscous drag up, balancing at terminal velocity. Beside it, Stokes' law with each symbol labelled and its conditions listed, and two spring balances showing a block weighing 22 newtons in air but 17 newtons in water, the missing 5 newtons being the weight of water displaced.
Three forces, not two. The upthrust is there whether the object is moving or not; the drag appears only when it moves.
\( F_d = 6\pi\eta r v \)
\( F_b = \rho V g \)

In the buoyancy equation \( \rho \) is the density of the fluid and \(V\) the volume displaced, so the upthrust equals the weight of fluid pushed aside. A floating object displaces exactly its own weight of fluid, which is the whole of why ships float.

Stokes’ law is not universal. It assumes the sphere moves slowly enough for the flow around it to stay streamlined, that the fluid is effectively infinite (no nearby container walls), and that the object really is a sphere. Push any of those and it fails.

✏️Worked example 2 — the size of an oil drop

An oil drop of density \( 8.9 \times 10^{2} \) kg m\(^{-3}\) falls through a gas of viscosity 14 µPa s and reaches a terminal velocity of 18 cm s\(^{-1}\). Estimate the radius of the drop. The density of the gas is negligible compared with that of the oil.

At terminal velocity the forces balance: weight down, upthrust and drag up. With the gas density neglected the upthrust drops out, leaving weight = drag:

\[ \tfrac{4}{3}\pi r^{3}\rho g = 6\pi\eta r v \qquad \Longrightarrow \qquad r^{2} = \frac{9\eta v}{2\rho g} \]

Substituting \( \eta = 14 \times 10^{-6} \) Pa s, \( v = 0.18 \) m s\(^{-1}\), \( \rho = 890 \) kg m\(^{-3}\) and \( g = 9.81 \) m s\(^{-2}\):

\[ r^{2} = \frac{9(14 \times 10^{-6})(0.18)}{2(890)(9.81)} = 1.30 \times 10^{-9}\ \text{m}^{2} \qquad r = 3.6 \times 10^{-5}\ \text{m} \]
Check it. That is 36 µm, about half the width of a human hair — the right order of magnitude for a drop of mist, and small enough that the assumption of slow, streamlined flow is safe. An answer in millimetres would have meant an arithmetic slip, because a drop that big would fall far faster than 18 cm s\(^{-1}\).

🎯Momentum and impulse

Linear momentum is mass times velocity,

\[ p = mv \qquad \text{in kg m s}^{-1} \text{ (or N s)} \]

and because velocity is a vector, momentum is a vector: signs matter, and you must fix a positive direction before substituting anything. The impulse of a force is the change of momentum it produces,

\[ J = \Delta p = F\Delta t \]

which is just Newton’s second law rearranged. For a force that varies during an impact — and every real impact does — the impulse is the area under the force–time graph.

Two force-time graphs. On the left a single bell-shaped pulse with the area under it shaded and labelled as the impulse, equal to the change in momentum. On the right, two pulses of equal area on the same axes: a tall narrow one for hitting something rigid and a low wide one for hitting an airbag, showing that stretching the time reduces the peak force.
Same area, same change in momentum. Only the peak force differs — which is the entire design principle behind airbags, crumple zones and bending your knees when you land.

✏️Worked example 3 — a jet of water on a wall

A hose of cross-sectional area 25 cm\(^{2}\) delivers water at 50 m s\(^{-1}\) horizontally at a wall, where it is brought to rest. The density of water is 1000 kg m\(^{-3}\). Find the force on the wall.

Work out what arrives in one second. A 50 m length of the jet hits the wall each second, so the volume arriving per second is \( 2.5 \times 10^{-3} \times 50 = 0.125 \) m\(^{3}\), and its mass is \( 0.125 \times 1000 = 125 \) kg.

Then use the momentum form of the second law. That water arrives at 50 m s\(^{-1}\) and stops, so the momentum destroyed each second is

\[ \Delta p = 125 \times 50 = 6250\ \text{kg m s}^{-1} \qquad F = \frac{\Delta p}{\Delta t} = \frac{6250}{1} = 6.3 \times 10^{3}\ \text{N} \]
Check it. \( F = ma \) is no use here — the mass involved is growing every second, which is exactly the situation \( F = \Delta p/\Delta t \) exists for. By Newton’s third law the wall pushes back on the water equally hard; that is what stops it.

🔒Conservation of momentum

The total momentum of a system of interacting particles is constant provided no resultant external force acts. This is not a new law — it follows from the second and third laws together. When A and B collide, the force on B from A is equal and opposite to the force on A from B, and they act for exactly the same time. So the impulses are equal and opposite, the momentum changes are equal and opposite, and the total is unchanged.

Kinetic energy is a different question entirely, and this is where marks are lost.

A truck of mass m and speed v hitting an identical stationary truck, in three outcomes. Elastic: the first stops and the second leaves at v, with kinetic energy unchanged. Totally inelastic: they couple and move off at v over 2, with half the kinetic energy gone. Inelastic: they separate at v over 4 and 3v over 4, with five eighths of the kinetic energy remaining. Momentum is conserved in all three.
Momentum survives every collision. Kinetic energy survives only the elastic one.
  • Elastic: no kinetic energy is lost at all. Real everyday objects never quite manage it — collisions between molecules are the honest example. A useful test: in an elastic collision the relative velocity of approach equals the relative velocity of separation.
  • Totally inelastic: the objects stick together, so their relative velocity of separation is zero. This loses the most kinetic energy of any collision with that momentum — to deformation, heat and sound.
  • Inelastic: everything in between, which is almost every collision you will ever meet.
  • Explosions run the same law backwards: the total momentum stays whatever it was (usually zero), while kinetic energy increases, released from stored chemical or elastic energy.

✏️Worked example 4 — two masses on a pulley

A 4.0 kg mass and a 6.0 kg mass are joined by a light string over a frictionless pulley and released from rest. Find the acceleration of the masses and the tension in the string.

Two free-body diagrams, two equations. Take the direction of motion as positive for each mass: the 6.0 kg falls, the 4.0 kg rises. For the rising mass, \( T - 4.0g = 4.0a \). For the falling mass, \( 6.0g - T = 6.0a \). Adding them eliminates \(T\):

\[ (6.0 - 4.0)g = (6.0 + 4.0)a \qquad a = \frac{2.0 \times 9.81}{10.0} = 1.96\ \text{m s}^{-2} \]

Then substitute back into either equation: \( T = 4.0(9.81 + 1.96) = 47\ \text{N} \).

Check it. Use the other equation: \( T = 6.0(9.81 - 1.96) = 47 \) N — the same. And notice the tension lies between the two weights (39 N and 59 N), as it must: it is too small to hold the heavy mass up and more than enough to lift the light one.
The trap. The tension is not the weight of either mass, and it is not their difference. It has one value throughout a light string over a frictionless pulley — but that value only comes out of solving both equations together.

🎡Uniform circular motion

An object going round a circle at constant speed is accelerating, because its velocity is constantly changing direction. That is the whole idea, and everything else follows from it.

An object at two points A and B of a circle with tangential velocity arrows of equal length but different directions. Beside it the two velocities are redrawn from a common point and the vector joining their tips, the change in velocity, points towards the centre of the circle. A panel gives the centripetal acceleration and force formulas, and a row of examples names the real force doing the job in each case: friction for a cornering car, tension for a ball on a string, gravity for a satellite and the normal force for a wall-of-death rider.
Subtracting the two velocities shows the change — and therefore the acceleration and the resultant force — points at the centre.

Angles in radians make the algebra work. An angle in radians is the arc length divided by the radius, \( \theta = s/r \), so a full circle is \( 2\pi \) rad. The angular velocity \( \omega \) is the angle turned per second, in rad s\(^{-1}\):

\( \omega = \dfrac{\Delta\theta}{\Delta t} = \dfrac{2\pi}{T} \)
\( v = \omega r \)

and the centripetal acceleration and force follow, each written three equivalent ways:

\[ a = \frac{v^{2}}{r} = \omega^{2}r = \frac{4\pi^{2}r}{T^{2}} \qquad F = \frac{mv^{2}}{r} = m\omega^{2}r = \frac{4\pi^{2}mr}{T^{2}} \]
Centripetal force is a job, not a source. Never add a “centripetal force” arrow to a free-body diagram. Identify which real force — tension, gravity, friction, the normal force, or a component of one of them — is pointing at the centre, and set that equal to \( mv^{2}/r \). There is also no centrifugal force in an inertial frame: what passengers feel in a cornering car is their own inertia and the door pushing them inwards.

Because the centripetal force is always perpendicular to the velocity, it does no work and cannot change the speed. That is why a conical pendulum keeps a steady speed while its direction changes continuously, and why the horizontal component of the normal force on a banked track can turn a car with no help from friction at all.

A vertical circle is the interesting case, because now gravity has a component along the path for most of the journey and the speed is no longer constant.

A mass swung on a string in a vertical circle, slowest at the top and fastest at the bottom. Two free-body diagrams: at the top both the tension and the weight point down towards the centre, giving T plus mg equals mv squared over r, and setting T to zero gives the minimum speed as the square root of gr; at the bottom the tension points up against the weight, giving T minus mg equals mv squared over r.
At the top gravity helps; at the bottom it fights. The string is nearest to breaking at the lowest point, where the speed is highest too.

🔭See it happen

PhET, Collision Lab. Set the elasticity slider to 100% and then to 0% with the same starting velocities, and watch the momentum readout stay put while the kinetic energy readout collapses. That single comparison settles the difference between the two conservation statements better than any paragraph.

PhET, Forces and Motion: Basics. Turn on the force arrows and push a crate. The applied-force and friction arrows are drawn to scale, so you can watch friction match your push exactly, then break away and settle at a lower value — the graph on this page, happening live.

📝Practise

Work through these, then reveal the answer. Between them they cover every objective at the top of the page. Take \( g = 9.81 \) m s\(^{-2}\) throughout.

1. A 1.0 kg mass rests on a frictionless surface. A force of 1.0 N acts on it for exactly 1.0 s. How far does it move (a) in the first second, (b) in the first ten seconds?

(a) While the force acts, \( a = F/m = 1.0 \) m s\(^{-2}\), so \( s = \tfrac{1}{2}(1.0)(1.0)^{2} = 0.50 \) m.

(b) After 1.0 s the force is removed and the surface is frictionless, so by Newton’s first law the mass carries on at the speed it had reached, \( v = at = 1.0 \) m s\(^{-1}\). In the remaining 9.0 s it travels a further 9.0 m, giving 9.5 m in total.

The point of the question is part (b): nothing slows it down, so no force is needed to keep it going.

2. An insect of mass 2.8 g jumps vertically from rest, leaving the ground at 3.5 m s\(^{-1}\). Its body is accelerated over a distance of 42 mm during take-off. Calculate (a) the average force the ground exerts on it, and (b) the maximum height it reaches.

(a) During take-off, \( a = \dfrac{v^{2}}{2s} = \dfrac{3.5^{2}}{2(0.042)} = 146 \) m s\(^{-2}\). The ground’s push \(R\) must both accelerate the insect and support its weight: \( R - mg = ma \), so \( R = m(a + g) = 0.0028(146 + 9.81) = 0.44 \) N.

(b) After take-off it is in free fall, so \( h = \dfrac{v^{2}}{2g} = \dfrac{3.5^{2}}{2(9.81)} = 0.62 \) m — more than 200 times its own body length.

Forgetting the weight in part (a) gives 0.41 N, which is the resultant force, not the force from the ground.

3. A 1200 kg car travels at a constant 50 km h\(^{-1}\) and makes a 90° turn on a circular path of radius 22 m, keeping the same speed. Find (a) the time the turn takes, (b) the magnitude of the change in momentum, (c) the minimum coefficient of static friction needed for the car not to skid.

First convert: \( 50 \) km h\(^{-1} = 13.9 \) m s\(^{-1}\).

(a) A quarter of the circumference is \( \tfrac{1}{4}(2\pi \times 22) = 34.6 \) m, so \( t = 34.6/13.9 = 2.5 \) s.

(b) The momentum has the same magnitude before and after, \( p = 1200 \times 13.9 = 1.67 \times 10^{4} \) kg m s\(^{-1}\), but its direction has turned through 90°. The change is the vector difference, so \( |\Delta p| = \sqrt{2}\,p = 2.4 \times 10^{4} \) kg m s\(^{-1}\), directed at 45° to both — not zero, and not \(p\).

(c) The centripetal acceleration is \( a = v^{2}/r = 13.9^{2}/22 = 8.8 \) m s\(^{-2}\), supplied by friction: \( \mu mg \ge ma \), so \( \mu \ge a/g = 8.8/9.81 = 0.90 \). That is a high value — this is a fast turn for that radius.

4. A railway truck of mass 800 kg moving at 3.0 m s\(^{-1}\) collides with a stationary truck of mass 1200 kg. The two couple together. (a) Is the collision elastic? (b) Find their common velocity. (c) Find the impulse received by the 800 kg truck. (d) Estimate the energy lost.

(a) Inelastic — they stick together, so kinetic energy cannot be conserved.

(b) \( 800 \times 3.0 = 2000v \), so \( v = 1.2 \) m s\(^{-1}\).

(c) For the 800 kg truck, \( \Delta p = 800(1.2 - 3.0) = -1.4 \times 10^{3} \) N s: an impulse of 1440 N s opposing its original motion. (The other truck receives \(+1440\) N s — equal and opposite, as the third law requires.)

(d) Before: \( \tfrac{1}{2}(800)(3.0)^{2} = 3600 \) J. After: \( \tfrac{1}{2}(2000)(1.2)^{2} = 1440 \) J. So 2160 J — 60% of it — has gone into deformation, heat and sound.

5. Coal falls vertically onto a horizontal conveyor belt at a rate of 4.5 kg s\(^{-1}\). The belt runs at 2.5 m s\(^{-1}\). (a) Estimate the minimum force needed to keep the belt moving at that speed. (b) Explain why the real force is larger.

(a) Each second, 4.5 kg of coal has to be brought from rest to 2.5 m s\(^{-1}\) horizontally, so \( F = \dfrac{\Delta p}{\Delta t} = 4.5 \times 2.5 = 11 \) N. Note that \( F = ma \) is useless here: the mass on the belt is changing, so the momentum form of the second law is the one that works.

(b) Because the belt also has to overcome friction in its rollers and bearings, and coal that lands and slips before matching the belt’s speed wastes further energy. 11 N is a floor, not a prediction.

6. A block of mass \(M\) rests on a horizontal surface which is slowly tilted. At an angle \(\theta\) the block just begins to slide. (a) Derive an expression for the coefficient of static friction. (b) Once it is moving, does it slide at constant velocity or accelerate?

(a) While it is still at rest, resolving along and perpendicular to the slope gives \( F = Mg\sin\theta \) and \( R = Mg\cos\theta \). At the angle where it just slides, friction is at its maximum, \( F = \mu_s R \). Dividing one by the other, the mass and \(g\) cancel: \( \mu_s = \tan\theta \).

(b) It accelerates. Once sliding starts the friction available drops from \( \mu_s R \) to \( \mu_d R \), and \( \mu_d < \mu_s \), so the component of weight down the slope now exceeds the friction and there is a resultant force. This is why a block that finally gives way tends to lurch rather than creep.

7. A fairground ride is a hollow cylindrical room of radius 2.5 m that spins about its vertical axis. People stand against the wall. When the wall is moving at 9.0 m s\(^{-1}\) the floor is lowered and they stay put. (a) What provides the horizontal force on a person? (b) Estimate the minimum coefficient of friction between a person and the wall.

(a) The normal force from the wall, pushing them horizontally inwards towards the axis. It is the centripetal force — not an extra force, just the wall pushing.

(b) That normal force is \( N = \dfrac{mv^{2}}{r} \). Vertically, friction must hold the person’s whole weight: \( \mu N \ge mg \). Substituting, \( \mu \ge \dfrac{gr}{v^{2}} = \dfrac{9.81 \times 2.5}{9.0^{2}} = 0.30 \). The mass cancels — the ride works the same for everyone, which is just as well.

8. A small mass on a string moves in a vertical circle of radius \(r\) where the gravitational field strength is \(g\). Deduce (a) the minimum speed it must have at the top of the circle, and (b) the minimum speed at the bottom.

(a) At the top, both the tension and the weight point downwards, towards the centre: \( T + mg = \dfrac{mv^{2}}{r} \). The slowest possible pass is when the string goes slack, \( T = 0 \), leaving \( mg = \dfrac{mv^{2}}{r} \), so \( v_{top} = \sqrt{gr} \). Gravity alone is then providing exactly the centripetal force needed.

(b) The string does no work (it is always perpendicular to the velocity), so mechanical energy is conserved between the top and the bottom, a height \(2r\) below: \( \tfrac{1}{2}mv_{bot}^{2} = \tfrac{1}{2}mv_{top}^{2} + mg(2r) \). With \( v_{top}^{2} = gr \) this gives \( v_{bot}^{2} = gr + 4gr = 5gr \), so \( v_{bot} = \sqrt{5gr} \) — more than twice the speed at the top.

9. A book rests on a table. Name the partner of each of the two forces acting on the book, saying what type each is and what it acts on.

The book feels two forces: the Earth’s gravitational pull downwards, and the table’s contact push upwards.

The partner of the gravitational pull is the book’s gravitational pull on the Earth, upwards, and equally large.

The partner of the table’s push is the book’s push down on the table, a contact force.

The two forces on the book are not a third-law pair, even though they happen to be equal here: they act on the same object, and they are different types of force. Tilt the table and they stop being equal, while both third-law pairs stay equal exactly as before.

10. A 2.0 kg gun fires a 4.0 g bullet at 100 m s\(^{-1}\). Find the recoil velocity of the gun, and explain what happens to the kinetic energy.

Total momentum before is zero, and no external horizontal force acts, so it stays zero: \( 0 = (0.0040)(100) + 2.0v \), giving \( v = -0.20 \) m s\(^{-1}\) — 0.20 m s\(^{-1}\) backwards.

Kinetic energy has increased from zero to \( \tfrac{1}{2}(0.0040)(100)^{2} + \tfrac{1}{2}(2.0)(0.20)^{2} = 20.0 + 0.04 = 20 \) J, released by the propellant. That is what an explosion is: momentum conserved, kinetic energy created. Notice how lopsidedly it is shared — the light bullet takes essentially all of it, because for a given momentum the kinetic energy \( p^{2}/2m \) is larger for the smaller mass.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Classroom — Newton’s Laws, and Momentum and Collisions
  • The Physics Classroom — Circular Motion and Satellite Motion
  • The Physics Hypertextbook — momentum and impulse
  • PhET — Collision Lab, and Forces and Motion: Basics