HomeLearning HubIB DP PhysicsA.2 Forces and momentum
A.2

Forces and momentum

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

🎯What you need to be able to do

  • Draw free-body diagrams and find the resultant force on a body.
  • Apply Newton’s three laws, and identify third-law pairs correctly.
  • Work with friction, drag, normal, tension, buoyancy and elastic restoring forces.
  • Treat circular motion as a force problem: centripetal acceleration and the forces that supply it.
  • Use impulse and the impulse–momentum relationship, including from a force–time graph.
  • Apply conservation of momentum, and distinguish elastic from inelastic collisions and explosions.

📚The physics

Free-body diagrams come first. Draw the body as a dot, then one arrow for every force acting on it, and nothing else. No velocity arrows, no acceleration arrows, no forces the body exerts on other things. If you cannot name what is exerting a force, that force does not exist. Most force questions become arithmetic once the diagram is right.

Newton’s first law says a body keeps constant velocity unless a resultant force acts. Constant velocity includes being at rest, and it includes moving at a steady speed in a straight line — but not moving in a circle at steady speed, because velocity is a vector and its direction is changing.

Newton’s second law is \( F = ma \) when the mass is constant, but the more general and more useful form is \( F = \dfrac{\Delta p}{\Delta t} \): the resultant force equals the rate of change of momentum. Use the second form whenever mass is changing or whenever the question is about impulse.

Newton’s third law is the one most often stated correctly and used wrongly. A third-law pair acts on two different bodies, is the same type of force, and is equal in magnitude and opposite in direction. The weight of a book on a table and the normal force from the table are not a third-law pair — they act on the same body, and one is gravitational while the other is electromagnetic. The partner of the book’s weight is the gravitational pull the book exerts on the Earth.

Friction. Static friction adjusts itself up to a maximum of \( \mu_s N \); it is whatever it needs to be to prevent sliding, which is why you cannot simply write \( f = \mu_s N \) for a stationary block. Once sliding begins, kinetic friction is \( \mu_k N \) and \( \mu_k < \mu_s \).

Circular motion is not a new force. Centripetal force is a role, not a source. Something real always plays it: tension for a ball on a string, gravity for a satellite, friction for a car cornering, the normal force for a wall-of-death rider. Never add a separate centripetal force to a free-body diagram — identify which of the forces already there is doing the job. The magnitude required is

\[ F_c = \frac{mv^{2}}{r} = m\omega^{2}r \qquad \text{directed towards the centre} \]

There is no centrifugal force in an inertial frame. What passengers feel in a cornering car is their own inertia and the seat pushing them inwards.

Momentum and impulse. Momentum \( p = mv \) is a vector, so signs matter and you must choose a positive direction before you start. Impulse is the change in momentum, \( J = \Delta p = F\Delta t \) for a constant force, and for a varying force it is the area under the force–time graph. This is why crumple zones, airbags and bending your knees all work: the change in momentum is fixed by the collision, so extending the time reduces the force.

Conservation of momentum holds for any system with no external resultant force, and it holds in collisions and explosions alike. Kinetic energy is a different matter: it is conserved only in an elastic collision. In an inelastic collision momentum is still conserved while kinetic energy is not — it goes into deformation, heat and sound. In an explosion the total momentum stays zero if it started at zero, while kinetic energy increases from stored energy.

✏️Worked example

A 1200 kg car travelling east at 15 m s\(^{-1}\) collides head-on with a 2400 kg van travelling west at 8.0 m s\(^{-1}\). They lock together. Find their common velocity after the collision, and show that the collision is inelastic.

Choose a direction. Take east as positive. The car’s momentum is \( +1200 \times 15 = +18000 \) kg m s\(^{-1}\). The van’s is \( 2400 \times (-8.0) = -19200 \) kg m s\(^{-1}\).

Conserve momentum. Total before \( = +18000 - 19200 = -1200 \) kg m s\(^{-1}\). After the collision the combined mass is 3600 kg, so

\[ v = \frac{-1200}{3600} = -0.33\ \text{m s}^{-1} \]

The wreck moves west at 0.33 m s\(^{-1}\). The negative sign is the answer, not an error — the van wins because its momentum was larger.

Check the energy. Before: \( \tfrac{1}{2}(1200)(15^{2}) + \tfrac{1}{2}(2400)(8.0^{2}) = 135000 + 76800 = 212 \) kJ. After: \( \tfrac{1}{2}(3600)(0.33^{2}) = 0.2 \) kJ. Essentially all the kinetic energy has gone, so the collision is inelastic — as any collision in which the objects stick together must be.

The trap. Adding 18000 and 19200 because both vehicles are “moving fast” gives 37200 and a nonsense answer of 10.3 m s\(^{-1}\). Momentum is a vector. Fix the positive direction on the page before substituting anything.

🔭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.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. A 5.0 kg box slides down a frictionless slope inclined at 30°. Find its acceleration.
The component of weight along the slope is \( mg\sin\theta \), so \( a = g\sin 30^\circ = 9.81 \times 0.50 = 4.9 \) m s\(^{-2}\). The mass cancels — every object slides down a frictionless slope at the same rate.
2. A 70 kg person stands in a lift accelerating upwards at 1.5 m s\(^{-2}\). Find the normal force from the floor.
Taking up as positive: \( N - mg = ma \), so \( N = m(g + a) = 70 \times (9.81 + 1.5) = 792 \) N. That is more than their weight of 687 N, which is the heavy feeling as a lift starts upwards.
3. An 8.0 kg block is pulled along level ground by a horizontal force of 40 N. The coefficient of kinetic friction is 0.35. Find the acceleration.
Friction: \( f = \mu_k mg = 0.35 \times 8.0 \times 9.81 = 27.5 \) N. Resultant: \( 40 - 27.5 = 12.5 \) N. So \( a = 12.5/8.0 = 1.6 \) m s\(^{-2}\).
4. A 0.20 kg ball on the end of a 0.80 m string is whirled in a horizontal circle at 3.0 revolutions per second. Find the tension in the string.
\( \omega = 2\pi f = 2\pi \times 3.0 = 18.8 \) rad s\(^{-1}\). The tension supplies the centripetal force: \( T = m\omega^{2}r = 0.20 \times 18.8^{2} \times 0.80 = 57 \) N. Note there is no extra "centripetal force" — the tension is it.
5. A 0.058 kg tennis ball is served from rest to 45 m s\(^{-1}\); the racket is in contact for 5.0 ms. Find the average force on the ball.
\( \Delta p = m\Delta v = 0.058 \times 45 = 2.61 \) kg m s\(^{-1}\). Then \( F = \Delta p/\Delta t = 2.61/(5.0 \times 10^{-3}) = 5.2 \times 10^{2} \) N — roughly a thousand times the ball’s weight, which is why the contact time matters so much.
6. Two skaters stand at rest facing each other on frictionless ice. They push apart; the 40 kg skater moves off at 3.0 m s\(^{-1}\). Find the velocity of the 60 kg skater.
Total momentum before is zero, and no external horizontal force acts, so it stays zero: \( 0 = 60v + 40 \times 3.0 \), giving \( v = -2.0 \) m s\(^{-1}\). The 60 kg skater moves at 2.0 m s\(^{-1}\) in the opposite direction. Kinetic energy has increased from zero — supplied by their muscles — which is exactly what an explosion is.

🔗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 Hypertextbook — momentum and impulse