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Educerie · IB Diploma · Physics
Theme A Space, time and motion · A.2 Forces and momentum
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| State and apply Newton's three laws, including translational equilibrium | SL, HL | "State Newton's first law" (1 mark); "Explain why the lift moves at constant velocity" (2 marks) |
| Treat forces as interactions and identify Newton's third-law pairs | SL, HL | Paper 1A: "Which force forms a third-law pair with…" (1 mark) |
| Draw and interpret free-body diagrams, and find the resultant force in one or two dimensions | SL, HL | "Draw the free-body diagram for the block" (2 marks), then a calculation |
| Use the contact forces: normal force, static and dynamic friction, tension, Hooke's law, viscous drag, buoyancy | SL, HL | A slope or terminal-speed calculation, 3–4 marks |
| Use the field forces: weight, and electric and magnetic forces as field forces | SL, HL | Labelling the forces on a charged drop or a falling ball |
| Use p = mv, conservation of momentum, impulse J = FΔt and J = Δp | SL, HL | Force–time graph: "Determine the impulse" (2 marks) |
| Explain when F = ma applies and when F = Δp ÷ Δt is needed | SL, HL | A jet, rocket or conveyor calculation (2–3 marks) |
| Analyse elastic and inelastic collisions and explosions, with the energy changes, in one dimension | SL, HL | "Determine whether the collision is elastic" (3 marks) |
| Analyse collisions and explosions in two dimensions | HL only | A glancing collision resolved into components (4–5 marks) |
| Use centripetal acceleration and force, angular velocity and period, for horizontal and vertical circles | SL, HL | Tension at the top and bottom of a vertical circle (3–4 marks) |
Before you start
You need A.1: velocity and acceleration as vectors, the equations of motion, and resolving a vector into two perpendicular components with sine and cosine. The only new unit is the newton (N), the force that gives 1 kg an acceleration of 1 m s⁻², so 1 N = 1 kg m s⁻².
1The idea in one paragraph
A force is one body pushing or pulling on another, and forces always come in pairs, one on each body. Add up every force on a single body and you get the resultant force. If it is zero, the body keeps doing exactly what it was doing; if it is not, the body's momentum changes, at a rate equal to the resultant force. Because the paired forces between two bodies are equal and opposite, whatever momentum one body gains the other loses, so the total momentum of bodies that only push on each other never changes. That one idea predicts collisions and explosions. A body moving in a circle is changing direction, so it too needs a resultant force, and that force points to the centre.
2Forces as interactions, and Newton's three laws
A force is an interaction between two bodies: a push or a pull by one body on another. It is a vector, measured in newtons. Some forces need contact (a hand on a door); others act at a distance through a field (the Earth's gravity on a falling apple). Either way there are always two bodies involved, and that is the whole of the third law.
Newton's first law. A body stays at rest, or keeps moving in a straight line at constant speed, unless a resultant force acts on it. Constant velocity needs no resultant force. A car cruising at a steady 30 m s⁻¹ on a level road still has its engine working, but only to cancel the air resistance and friction; the forces balance. When the resultant force on a body is zero, it is in translational equilibrium, and it is either at rest or moving at constant velocity.
Newton's second law. The resultant force on a body equals the rate of change of its momentum. When the mass does not change, this becomes the familiar form.
Resultant force = rate of change of momentum. For constant mass: F = ma, with F and a in the same direction.
Newton's third law. If body A exerts a force on body B, then B exerts a force on A that is equal in size, opposite in direction and of the same type. The two forces act on different bodies. Figure 1 is the example that trips people up.
A book rests on a table. Two forces act on the book: the table pushes it up (the normal force) and the Earth pulls it down (its weight). They are equal and opposite, but they are not a third-law pair. They act on the same body, they are different types of force, and they are equal only because the book is in equilibrium: put the table in a lift that is accelerating upwards and they are no longer equal. Panel (b) shows the real pairs. The table pushes up on the book and the book pushes down on the table: a contact pair. The Earth pulls down on the book and the book pulls up on the Earth: a gravitational pair.
A test that never fails: write each force as "X on Y". Its third-law partner is "Y on X". If you cannot swap the two names, you have not found a pair.
3Free-body diagrams and the resultant force
A free-body diagram shows one body on its own, with every force acting on it drawn as an arrow starting at the body, in the right direction, with a length that suggests its size and a label, either the symbol the data booklet uses or the force's name. Forces the body exerts on other things do not belong on it. Most mechanics questions become routine once this diagram is right.
To find the resultant force, add the forces as vectors. In one dimension, choose a positive direction and add with signs. In two dimensions, resolve every force into two perpendicular directions, add each direction separately, and apply F = ma in each.
Worked example 1 · equilibrium in two dimensions. A 4.0 kg shop sign hangs from two cables, each at 30° to the horizontal, one on each side. Find the tension in each cable. Take g = 9.8 m s⁻².
The sign is at rest, so the resultant force is zero in both directions. Horizontally, the two cables' horizontal parts cancel, which is why the tensions are equal. Vertically:
Each cable carries 39 N, as much as the whole weight of the sign, because only half of each tension points upwards. Hang the sign from flatter cables and the tension grows without limit, which is why a washing line can never be pulled perfectly straight.
On a slope, resolve along the slope and perpendicular to it, because the motion is along the slope. Figure 2 shows the free-body diagram of a block resting on a rough slope.
The weight splits into mg sin θ down the slope and mg cos θ into the slope. Check the trig with a limit: on a flat surface θ = 0, sin θ = 0, and there should be no pull down the slope. There isn't.
Worked example 2 · will it slide? A 2.0 kg block is placed on a slope at 35°. The coefficient of static friction is 0.60 and of dynamic friction 0.40. Does it slide, and if so what is its acceleration?
4The contact forces
The guide names six contact forces. Each has a direction you must get right and, for four of them, an equation from the data booklet.
Normal force. When a body presses on a surface, the surface pushes back. The normal force is the part of that contact force perpendicular to the surface ("normal" means at right angles). It is only as big as it needs to be to stop the body sinking into the surface, so it is not always equal to the weight: on a slope it is mg cos θ, and in a lift accelerating upwards it is more than mg.
Friction. The surface frictional force acts parallel to the surface, opposing the sliding or the tendency to slide. Figure 3 shows how it behaves as you push harder on a heavy box.
While the box is at rest, static friction matches your push exactly, so the box stays in equilibrium. It can only grow to a limit. Once your push exceeds that limit, the box slides and dynamic friction takes over, which is usually a little smaller and stays roughly constant. The data booklet gives both. The coefficient of static friction (μ with subscript s) and the coefficient of dynamic friction (subscript d) are plain numbers with no unit, and they depend on the two surfaces.
The ≤ sign is the thing to understand. A 2.0 kg block on a level floor with a coefficient of static friction of 0.60 can have up to 0.60 × 19.6 = 11.8 N of static friction. Push it with 5 N and the friction is 5 N, not 11.8 N.
Tension. The pulling force in a string, rope or cable. It always pulls, along the string, away from the body it is attached to, and in a light string it has the same size all the way along.
Elastic restoring force. A stretched or compressed spring pulls or pushes back towards its natural length. Figure 4 shows it. For a spring obeying Hooke's law,
where x is the extension (or compression) from the natural length and k is the spring constant in N m⁻¹, the force per metre of extension. The minus sign says the force points opposite to the extension; its size is kx. Hooke's law holds up to the spring's limit of proportionality.
Viscous drag. A body moving through a fluid is slowed by a drag force opposite to its velocity. For a small sphere moving slowly and smoothly through a fluid, the data booklet gives Stokes' law:
where η (eta) is the fluid's viscosity in Pa s (N s m⁻²), a measure of how thick and sticky it is, r is the sphere's radius and v its speed. Drag is proportional to speed, which is exactly why a falling body reaches a terminal speed.
Buoyancy. A body in a fluid pushes some of the fluid out of the way, and the fluid pushes back upwards with a force equal to the weight of the fluid displaced:
where ρ is the density of the fluid (not the body) and V is the volume of fluid displaced. Figure 5 puts weight, buoyancy and drag on one sphere.
Worked example 3 · terminal speed in oil. A steel ball of radius 1.0 mm and density 7800 kg m⁻³ falls through an oil of density 1260 kg m⁻³ and viscosity 1.4 Pa s. Find its terminal speed.
At terminal speed the resultant force is zero, so the two upward forces balance the weight. Both the weight and the buoyancy depend on the ball's volume, V = (4/3)πr³.
A centimetre per second. Measuring that speed and working backwards is how the viscosity of a liquid is found in the laboratory.
5The field forces
Three forces act through a field, with no contact. The gravitational force on a body near the Earth is its weight, mg, acting towards the centre of the Earth; g is 9.8 N kg⁻¹ at the surface, the same number as the acceleration of free fall. The electric force acts on a charged body in an electric field, and the magnetic force acts on a moving charge or a current in a magnetic field. You need to recognise these two here and draw them on free-body diagrams; how to calculate them is Theme D.
Mass and weight are different things. Mass is the amount of matter, in kilograms, the same everywhere. Weight is a force, in newtons, and on the Moon it is about a sixth of its value on Earth.
6Momentum and impulse
The linear momentum of a body is its mass times its velocity:
Newton's second law in its full form says the resultant force equals the rate of change of momentum. Multiply both sides by the time the force acts and you get the impulse:
where F is the average resultant force and Δt the time of contact. Impulse is the change in momentum, whatever the force and however long it acts. On a force–time graph the impulse is the area under the curve, which is how you find it when the force is not constant. Figure 6 shows two impacts with the same area.
The same change in momentum can be delivered by a large force for a short time or a small force for a long time. Every safety device uses this: crumple zones, airbags, helmets and bending your knees on landing all lengthen the contact time so that the peak force is smaller.
Worked example 4 · a tennis ball. A 58 g tennis ball arrives at 20 m s⁻¹ and leaves the racket at 30 m s⁻¹ in the opposite direction. The contact lasts 5.0 ms. Find the impulse and the average force.
Take the direction the ball leaves in as positive, so the arriving velocity is −20 m s⁻¹.
Getting 0.58 N s (from 30 − 20) is the classic mistake: reversing direction means the velocity changes by 50 m s⁻¹, not 10.
When the mass changes. F = ma assumes the mass is constant. When mass is being gained or lost, as in a rocket burning fuel, a hose spraying water or sand falling onto a conveyor belt, use the form F = Δp ÷ Δt, which allows for it. A garden hose sends 0.50 kg of water per second at a wall at 12 m s⁻¹, and the water stops at the wall. Each second, 0.50 kg loses 12 m s⁻¹ of velocity, so Δp ÷ Δt = 0.50 × 12 = 6.0 N, and that is the force the water exerts on the wall. The acceleration of any single piece of water never enters the calculation.
7Collisions, explosions and conservation of momentum
When two bodies collide, the force of A on B and of B on A are equal and opposite (third law) and last for exactly the same time. So their impulses are equal and opposite, and whatever momentum A gains, B loses. That gives the conservation law.
The total momentum of a system stays constant unless a resultant external force acts on it.
"External" means from outside the system: the forces between the colliding bodies are internal and cancel. In a car crash, friction from the road is external, but it is tiny compared with the collision forces over the few milliseconds of the impact, so momentum is conserved to a good approximation.
Figure 7 shows the two types of event you must handle in one dimension.
Worked example 5 · the cars in panel (a). A 1200 kg car at 15 m s⁻¹ runs into the back of a stationary 800 kg car and the two lock together. Find their common velocity and the kinetic energy lost.
Momentum is conserved in every collision. Kinetic energy is a different matter, and it sorts collisions into two types.
- In an elastic collision the total kinetic energy after equals the total before. Collisions between gas molecules and between hard steel balls come close.
- In an inelastic collision some kinetic energy is transferred to other forms, mainly internal energy (the bodies warm up and deform) and sound. When the bodies stick together, as in panel (a), the collision is called perfectly inelastic and the loss is as large as conservation of momentum allows.
To decide which type a collision is, compare the total kinetic energy before and after. Checking momentum tells you nothing, because momentum is conserved either way. For example, a 0.20 kg ball at 3.0 m s⁻¹ hits a stationary 0.10 kg ball; afterwards the first moves at 1.0 m s⁻¹ and the second at 4.0 m s⁻¹, both forwards. Momentum: 0.60 before, 0.20 + 0.40 = 0.60 after. Kinetic energy: ½ × 0.20 × 3.0² = 0.90 J before, 0.10 + 0.80 = 0.90 J after. The collision is elastic. You will not have to solve the momentum and energy equations together to find unknown velocities; the guide leaves that out.
Explosions. In an explosion, a body or a group of bodies at rest pushes itself apart, as in panel (b), where a compressed spring pushes two trolleys away from each other. The total momentum was zero before, so it is zero after: the two momenta are equal and opposite. Kinetic energy increases from zero, and it comes from a store inside the system, here elastic energy in the spring, in a firework chemical energy.
Notice that the lighter trolley carries two-thirds of the energy. Writing kinetic energy in terms of momentum shows why:
Both trolleys have the same size of momentum, 0.60 N s, so the kinetic energy is inversely proportional to mass. The same rule explains why a rifle's recoil is survivable while the bullet is not.
8Circular motion
A body moving in a circle at constant speed is accelerating, because its velocity keeps changing direction. Figure 8 shows the velocity along the tangent at each point and the acceleration pointing towards the centre.
This centripetal acceleration is directed radially towards the centre, and the data booklet gives three forms of it:
Here T is the period, the time for one revolution, and ω (omega) is the angular velocity, the angle swept per second in radians per second. One revolution is 2π radians, so
A bicycle wheel of radius 0.30 m turning twice a second has T = 0.50 s, so ω = 2π ÷ 0.50 = 12.6 rad s⁻¹, the rim moves at v = 12.6 × 0.30 = 3.8 m s⁻¹, and a point on the rim accelerates at ω²r = 47 m s⁻² towards the axle.
By the second law, a centripetal acceleration needs a resultant force towards the centre, the centripetal force, F = mv² ÷ r. It acts perpendicular to the velocity, so it changes the direction of motion but not the speed. It is not an extra force to add to the free-body diagram. It is the name for the resultant of the real forces, and one of them always supplies it: tension for a stone on a string, friction for a car on a flat bend, gravity for the Moon.
Worked example 6 · a car on a flat bend. A 1200 kg car rounds a flat bend of radius 50 m at 15 m s⁻¹. What friction force is needed, and what is the least coefficient of static friction that will do it?
The friction here is static: the tyres grip the road sideways. On ice the coefficient falls far below 0.46, the needed force cannot be supplied, and the car carries on in a straighter line than the road.
Vertical circles, and non-uniform circular motion. When a ball on a string is whirled in a vertical circle, its speed changes: it slows as it rises and speeds up as it falls, so the circular motion is non-uniform. The guide asks you to analyse the forces only at the top and the bottom, where they are all vertical. Figure 9 shows both.
At each point, the resultant force towards the centre equals mv² ÷ r.
- At the top, the tension and the weight both point down, towards the centre: T + mg = mv² ÷ r.
- At the bottom, the tension points up, towards the centre, and the weight points away: T − mg = mv² ÷ r.
Worked example 7. A 0.30 kg ball on a string of length 0.80 m moves in a vertical circle. At the top its speed is 4.0 m s⁻¹, and at the bottom 6.9 m s⁻¹ (the energy method of A.3 gives this). Find the tension at each point.
The tension is greatest at the bottom, where the ball is fastest and the string must also hold up the weight; that is where a string snaps. At the top, if the ball is slow enough, the weight alone can supply all of mv² ÷ r and the tension falls to zero. The least speed at the top is where T = 0, mg = mv² ÷ r, so v = √(gr) = √(9.8 × 0.80) = 2.8 m s⁻¹. Any slower and the string goes slack and the ball leaves the circle.
A body can also move in a horizontal circle at changing speed, a car speeding up round a roundabout, for example. The part of the resultant force towards the centre still equals mv² ÷ r; an extra part along the tangent changes the speed.
Because the centripetal force is always perpendicular to the velocity, it does no work on the body, and in uniform circular motion the kinetic energy stays the same. A.3 uses that.
9HLCollisions and explosions in two dimensions
SL students can skip to section 10.
Momentum is a vector, so conservation of momentum holds separately in every direction. In a glancing collision, choose two perpendicular axes, usually along and across the original motion, and write one conservation equation for each.
Worked example 8. A puck A moving at 2.0 m s⁻¹ strikes an identical stationary puck B off-centre. Afterwards A moves at 1.0 m s⁻¹ at 60° to its original direction. Find the velocity of B.
The masses are equal, so every term has the same m in it and it cancels.
Figure 10 shows the result both as motion and as a vector triangle.
Panel (b) is the method in picture form: the momentum vectors after the collision, placed nose to tail, add up to the momentum before. Check the energy too: ½m(2.0²) = 2.0m before and ½m(1.0² + 1.73²) = 2.0m after, so this collision is elastic. When one body strikes an equal mass at rest, an elastic glancing collision always sends the two off at 90° to each other, which is what the right angle in panel (b) shows. A snooker player uses this every frame.
An explosion in two dimensions works the same way. A stationary firework shell that bursts into three fragments has zero momentum before, so the three momentum vectors afterwards must form a closed triangle. Knowing two of them fixes the third.
10Where marks are lost
Calling the normal force and weight a third-law pair. They act on the same body. A third-law pair acts on two different bodies and is always the same type of force.
Thinking motion needs a force. Constant velocity needs zero resultant force (first law). A force is needed to change the velocity.
Drawing forces the body exerts on other things, or inventing a "centripetal force" arrow. A free-body diagram shows only forces acting on the body. The centripetal force is the resultant of forces already on the diagram, never an extra arrow.
Using the maximum static friction when the body is not about to slip. Static friction is only as large as it needs to be. The coefficient of static friction times the normal force is its largest possible value, reached only on the point of slipping.
Losing the sign when a body bounces. A ball that arrives at 20 m s⁻¹ and leaves at 30 m s⁻¹ the other way has changed velocity by 50 m s⁻¹. Define positive before writing any momentum.
Testing for an elastic collision with momentum. Momentum is conserved in every collision. Only kinetic energy separates elastic from inelastic.
Saying energy is "lost" in an inelastic collision and stopping there. Say where it went: to internal energy of the bodies, and to sound.
Writing the weight into the vertical circle with the wrong sign. At the top, T and mg both point to the centre; at the bottom they point opposite ways. Draw the free-body diagram at each point before writing an equation.
11Draw it right
Free-body diagrams are marked on a short list of things. Check every one.
- Draw one body, as a dot or a simple box, on its own.
- Every arrow starts on the body and points the way the force acts on it.
- Every arrow has a label: the symbol the data booklet uses or the force's name (normal force, friction, weight, tension, drag, buoyancy). A symbol with no arrow is not enough.
- Arrow lengths reflect the sizes. If the body is in equilibrium, opposite arrows are equal in length; if it accelerates, the arrows on that side are longer.
- The normal force is perpendicular to the surface and friction is parallel to it, opposing the (tendency to) slide.
- No arrow for a "centripetal force", a "resultant force" or a "force of motion". The resultant is found from the diagram, not drawn on it.
- For a collision, draw a before and an after sketch with masses, velocities and a stated positive direction.
- HL: in two dimensions, draw the momentum vector triangle or state both components.
12Try it
Marks in brackets. Take g = 9.8 m s⁻². Answers and marker's notes are at the end.
Q1. A person stands at rest on bathroom scales on a floor. Which force forms a Newton's third-law pair with the person's weight? 1 mark
A. the normal force of the scales on the person · B. the force of the person on the scales · C. the gravitational force of the person on the Earth · D. the force of the floor on the scales
Q2. A 5.0 kg crate rests on a ramp inclined at 20° to the horizontal.
(a) Draw a labelled free-body diagram for the crate. 2 marks
(b) Calculate the frictional force on the crate. 2 marks
(c) The ramp is slowly made steeper. The crate starts to slide when the angle reaches 31°. Determine the coefficient of static friction. 2 marks
Q3. (data-based, Paper 1B style) A 0.60 kg trolley moving at 1.5 m s⁻¹ runs into a force sensor fixed to a wall. The sensor's record shows the force rising steadily from 0 to 12 N in 0.10 s, then falling steadily back to 0 at 0.20 s.
(a) Sketch the force–time graph and determine the impulse on the trolley. 2 marks
(b) Determine the velocity of the trolley after the collision. 2 marks
(c) Deduce whether the collision is elastic. 2 marks
Q4. Two ice skaters, of mass 60 kg and 45 kg, stand at rest facing each other and push apart. The 45 kg skater moves off at 2.0 m s⁻¹.
(a) Calculate the velocity of the 60 kg skater. 2 marks
(b) Calculate the total kinetic energy after the push, and state where it came from. 2 marks
Q5. A 0.20 kg stone on a string of length 0.60 m is whirled in a vertical circle.
(a) Calculate the least speed the stone can have at the top if the string is to stay taut. 2 marks
(b) At the bottom its speed is 5.5 m s⁻¹. Calculate the tension in the string there. 2 marks
(c) Explain why the string is most likely to break at the bottom of the circle. 2 marks
Q6 (HL). A 2.0 kg puck moving east at 3.0 m s⁻¹ strikes a stationary 3.0 kg puck. Afterwards the 2.0 kg puck moves north at 1.0 m s⁻¹. Determine the speed and direction of the 3.0 kg puck, and deduce whether the collision is elastic. 5 marks
13In one breath
A force is an interaction between two bodies. First law: zero resultant force means constant velocity, rest included, which is translational equilibrium. Second law: resultant force equals rate of change of momentum, F = ma for constant mass and F = Δp ÷ Δt when the mass changes. Third law: A on B is equal and opposite to B on A, same type, different bodies, so the normal force and the weight on a book are not a pair. Draw one body with every force on it, resolve, and add. The normal force is perpendicular to the surface; friction is parallel, only as large as needed at rest up to μ times the normal force, and equal to the dynamic μ times the normal force when sliding; a spring pulls back with kx; a small sphere in a fluid feels drag 6πηrv and buoyancy ρVg. Momentum p = mv is conserved unless an external resultant force acts; impulse FΔt = Δp is the area under a force–time graph. Every collision conserves momentum, only elastic ones conserve kinetic energy, and explosions release it from a store, with Eₖ = p² ÷ 2m. Circular motion needs a resultant force mv² ÷ r towards the centre: T + mg at the top of a vertical circle, T − mg at the bottom. HL: in two dimensions, conserve momentum along two perpendicular axes.
Answers
Q1. C. The weight is the Earth pulling on the person, so its partner is the person pulling on the Earth: same type (gravitational), opposite direction, on the other body. C only. A is the most common wrong answer: it is equal and opposite to the weight here, but it acts on the same body and is a contact force.
Q2. (a) The crate drawn alone with three labelled arrows from it: weight (mg) vertically down, normal force perpendicular to the ramp surface and away from it, friction parallel to the surface, pointing up the slope. 1 for all three forces present and labelled, 1 for all three in correct directions. An extra "force down the slope" arrow loses the second mark.
(b) The crate is in equilibrium, so friction balances the component of the weight down the slope: friction = mg sin 20° = 5.0 × 9.8 × 0.342 = 17 N, up the slope. M1 for mg sin θ with equilibrium stated or implied, A1 for 17 N. Using μ FN scores 0, because μ is not given and the crate is not about to slip.
(c) On the point of slipping, friction reaches its maximum, μ × normal force, where μ is the coefficient of static friction. So μmg cos θ = mg sin θ, and μ = tan θ = tan 31° = 0.60. M1 for equating μmg cos θ to mg sin θ, A1 for 0.60 with no unit.
Q3. (a) A triangle with its peak of 12 N at 0.10 s and base from 0 to 0.20 s. Impulse = area = ½ × 0.20 × 12 = 1.2 N s (away from the wall). M1 for area under the graph, A1 for 1.2 N s.
(b) Take the trolley's initial direction as positive; the impulse from the wall is −1.2 N s. Δv = −1.2 ÷ 0.60 = −2.0 m s⁻¹, so v = 1.5 − 2.0 = −0.50 m s⁻¹: 0.50 m s⁻¹ back from the wall. M1 for Δp = J with the sign handled, A1 for 0.50 m s⁻¹ with the direction. An answer of 3.5 m s⁻¹ comes from adding the impulse the wrong way and scores 0.
(c) Eₖ before = ½ × 0.60 × 1.5² = 0.675 J; after = ½ × 0.60 × 0.50² = 0.075 J. Kinetic energy has decreased, so the collision is inelastic. M1 for both kinetic energies, A1 for the conclusion drawn from them. A conclusion from momentum alone scores 0.
Q4. (a) Total momentum is zero before, so 0 = 45 × 2.0 + 60v, giving v = −1.5 m s⁻¹: 1.5 m s⁻¹ in the opposite direction to the lighter skater. M1 for conservation of momentum from zero, A1 for 1.5 m s⁻¹ with the direction.
(b) Eₖ = ½ × 45 × 2.0² + ½ × 60 × 1.5² = 90 + 67.5 = 158 J (157.5 J). It came from chemical energy in the skaters' muscles, transferred by the work their arms did in the push. A1 for the total, A1 for the source. "From momentum" scores 0: momentum is not a form of energy.
Q5. (a) At the least speed the tension is zero and the weight alone provides the centripetal force: mg = mv² ÷ r, so v = √(gr) = √(9.8 × 0.60) = 2.4 m s⁻¹. M1 for mg = mv²/r with T = 0, A1 for 2.4 m s⁻¹.
(b) T − mg = mv² ÷ r, so T = 0.20 × 5.5² ÷ 0.60 + 0.20 × 9.8 = 10.1 + 1.96 = 12 N. M1 for T − mg = mv²/r, A1 for 12 N. Writing T = mv²/r and forgetting the weight gives 10 N and scores M0 A0.
(c) At the bottom the stone is moving fastest, so the centripetal force mv² ÷ r needed is greatest; and the weight acts away from the centre there, so the tension must provide mv² ÷ r plus mg. The tension is therefore at its maximum at the bottom. 1 for the greatest speed, so the greatest mv²/r, 1 for the weight opposing so T = mv²/r + mg.
Q6 (HL). Take east as x and north as y. East: 2.0 × 3.0 = 3.0vₓ, so the east component vₓ = 2.0 m s⁻¹. North: 0 = 2.0 × 1.0 + 3.0 × (north component), so the north component = −0.667 m s⁻¹, that is 0.667 m s⁻¹ south. Speed = √(2.0² + 0.667²) = 2.1 m s⁻¹, direction tan⁻¹(0.667 ÷ 2.0) = 18° south of east. Kinetic energy before = ½ × 2.0 × 3.0² = 9.0 J; after = ½ × 2.0 × 1.0² + ½ × 3.0 × 2.11² = 1.0 + 6.7 = 7.7 J. Kinetic energy decreased, so the collision is inelastic. M1 for conserving momentum in the x-direction, M1 in the y-direction, A1 for 2.1 m s⁻¹, A1 for 18° south of east, A1 for the energy comparison and conclusion. Adding speeds instead of components scores only the energy mark.
Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section A.2 Forces and momentum. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.