Coaching Centre
Dynamics review
A summary of Module 2 Dynamics, how forces, energy and momentum connect, the key equations, key words, and a mixed HSC-style test covering Newton's laws, equilibrium, inclined planes, friction, connected bodies, work, energy, power, momentum, impulse and collisions.
Use this page to check the whole module. Read the summary, make sure you know when to use each tool, then do the test without looking back.
Module summary
- 2A Forces and free-body diagrams. A force is an interaction between two objects; contact vs field forces. Newton's three laws. Third-law pairs: equal, opposite, same type, on different objects. Free-body diagrams show only forces on one object. Apparent weight $N = m(g + a)$.
- 2B Net force and components. Algebraic addition along a line; tip-to-tail in 2D; $F_x = F\cos\theta$, $F_y = F\sin\theta$. Resultant and equilibrant.
- 2C Equilibrium. $\Sigma F_x = 0$, $\Sigma F_y = 0$. Static and dynamic. Closed triangle of forces. $T = mg/(2\sin\theta)$ for a symmetric hang.
- 2D Inclined planes. $mg\sin\theta$ along, $N = mg\cos\theta$. Frictionless $a = g\sin\theta$. With friction $a = g(\sin\theta - \mu_k\cos\theta)$. Slips when $\tan\theta = \mu_s$. Practical with light gates: $a = 2s/t^2$.
- 2E Friction and resistance. $f_s \le \mu_sN$, $f_k = \mu_kN$, $\mu_k \lt \mu_s$. Friction can drive motion (walking, tyres). Rolling resistance roughly constant; air resistance about $\propto v^2$; terminal velocity.
- 2F Constant net force and connected bodies. Constant $F_{net}$ gives constant $a$. System method: $a$ = net external force ÷ total mass, then isolate a body for tension. F = ma practical with constant total mass.
- 2G Work and kinetic energy. $W = Fs\cos\theta$; area under F–s graph; $K = \tfrac12mv^2$; $W_{net} = F_{net}s = \Delta K$.
- 2H Conservation of mechanical energy. $\Delta U = mg\Delta h$. If only gravity does work, $K + U$ is constant. Friction turns mechanical energy into heat, $Q = fd$.
- 2I Power. $P = \Delta E/t$, $P = Fv$. At constant speed, driving force = resistance. Efficiency.
- 2J Conservation of momentum. $\vec p = m\vec v$. $\vec F_{net} = \Delta\vec p/\Delta t$. Closed system: $\Sigma m\vec v_{before} = \Sigma m\vec v_{after}$, separately in $x$ and $y$.
- 2K Impulse and car safety. $F\Delta t = \Delta p$ = area under F–t graph. Longer stopping time, smaller force. Crumple zones, seatbelts, airbags.
- 2L Elastic and inelastic collisions. Elastic: $K$ conserved. Inelastic: $K$ lost to heat, sound, deformation. Completely inelastic: objects stick. Ballistic pendulum: momentum for the collision, energy for the swing.
Putting it together
Module 2 gives you three tools, and choosing the right one is half the battle.
| Tool | Use it when | Watch out for |
|---|---|---|
| Forces: $\Sigma\vec F = m\vec a$ | you need acceleration, time, a tension or a normal force | drawing every force on one object; components on slopes |
| Energy: $W_{net} = \Delta K$, $K + U$ conserved | you need a speed at a position, the path is curved, or time doesn't matter | friction converts energy to heat; vertical height, not path length |
| Momentum: $\Sigma m\vec v$ conserved | objects collide, stick together or push apart | direction and signs; kinetic energy is usually not conserved |
Many exam questions chain them together. A block slides down a ramp (energy, with friction), hits another block (momentum), then the pair slides to a stop (forces or work–energy). Split the problem into stages and pick the right law for each stage, exactly as in the ballistic pendulum.
A 0.80 kg block slides from rest down a ramp 1.5 m high and inclined at 30°, with $\mu_k = 0.15$. At the bottom it moves onto a frictionless floor and sticks to a stationary 1.2 kg block. Find the speed of the pair.
Key equations
| Equation | Meaning |
|---|---|
| $\vec F_{net} = m\vec a$ | Newton's second law |
| $\vec F_{AB} = -\vec F_{BA}$ | Newton's third law |
| $F_x = F\cos\theta$, $F_y = F\sin\theta$ | components of a force |
| $f = \mu N$ | friction (maximum static, or kinetic) |
| $W = Fs\cos\theta$, $W_{net} = F_{net}s = \Delta K$ | work; work–energy theorem |
| $K = \tfrac12mv^2$, $\Delta U = mg\Delta h$ | kinetic and gravitational potential energy |
| $P = \dfrac{\Delta E}{t}$, $P = Fv$ | power |
| $\vec p = m\vec v$, $\Sigma m\vec v_{before} = \Sigma m\vec v_{after}$ | momentum and its conservation |
| $\Delta\vec p = \vec F\Delta t$ | impulse |
Key words
air resistance · apparent weight · coefficient of friction · completely inelastic · connected bodies · contact force · dynamic equilibrium · efficiency · elastic collision · equilibrant · field force · free-body diagram · gravitational potential energy · impulse · inclined plane · inelastic collision · kinetic energy · kinetic friction · mechanical energy · momentum · net force · normal force · power · resultant · rolling resistance · static equilibrium · static friction · tension · terminal velocity · third-law pair · work
Test yourself
A 1.2 kg book is pressed against a vertical wall by a horizontal 20 N push and doesn't slip. The friction force on the book is
Vertically the book is in equilibrium, so friction balances its weight: $1.2 \times 9.8 = 11.8$ N upward.
A 900 kg car accelerates from rest to 20 m s⁻¹ in 8.0 s on a level road. The average power delivered to the car's kinetic energy is
$\Delta K = \tfrac12(900)(20)^2 = 1.8 \times 10^5$ J; $P = 1.8 \times 10^5/8.0 = 2.25 \times 10^4$ W.
Two carts are pushed apart by a spring from rest. Cart X has twice the mass of cart Y. Afterwards, compared with Y, cart X has
Equal and opposite momenta: $2m v_X = m v_Y$, so $v_X = v_Y/2$. $K_X = \tfrac12(2m)(v/2)^2 = \tfrac14mv^2$, half of $K_Y = \tfrac12mv^2$.
A block slides down a slope at constant velocity. Which statement is correct?
Constant velocity: $mg\sin\theta = \mu_k mg\cos\theta$, so $\mu_k = \tan\theta$. Potential energy is being turned into heat, so mechanical energy is not conserved.
A force–time graph for a collision is a rectangle 300 N high and 0.050 s wide. The change in momentum of the object is
Area $= 300 \times 0.050 = 15$ N s.
A 2.0 kg mass and a 3.0 kg mass hang over a frictionless pulley. The tension in the string is
$T = \dfrac{2m_1m_2g}{m_1 + m_2} = \dfrac{2(2.0)(3.0)(9.8)}{5.0} = 23.5$ N, between the two weights.
Label the forces on the block on the slope.
Which tool is the best first choice for each question?
Order these by size of average force on the same falling egg (smallest first).
The forces in a Newton's third-law pair act on objects. On a slope, the normal force equals $mg$. Work is the area under a –displacement graph, and impulse is the area under a force– graph. In every collision in a closed system, is conserved, but kinetic energy is conserved only if the collision is .
A 60 kg skateboarder rolls down a ramp from rest, dropping a vertical height of 2.5 m. Assume friction on the ramp is negligible. At the bottom, still on level ground, she jumps onto a stationary 3.0 kg skateboard that was waiting for her (she leaves the first board behind, which stops). (a) Calculate her speed at the bottom of the ramp. (2 marks) (b) Calculate her speed just after landing on the second board. (2 marks) (c) Analyse the energy changes in the whole process, including the landing. (3 marks)
(a) No friction, so mechanical energy is conserved: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2.5} = 7.0$ m s⁻¹.
(b) Landing on the board is a collision in which momentum is conserved (horizontal external forces are negligible): $60 \times 7.0 = (60 + 3.0)V$, so $V = 6.7$ m s⁻¹.
(c) At the top she has $U = 60 \times 9.8 \times 2.5 = 1470$ J. Rolling down the frictionless ramp, all of this becomes kinetic energy: $\tfrac12(60)(7.0)^2 = 1470$ J. Landing on the board is a completely inelastic collision (she and the board move together), so momentum is conserved but kinetic energy is not. Afterwards $K = \tfrac12(63)(6.67)^2 = 1400$ J, so about 70 J (about 5%) is transformed into sound, heat and the work done flexing the board and her legs. The loss is small because the board's mass is small compared with hers. Total energy is conserved throughout.