Mathematical models in mechanics: the modelling words and their assumptions

Inextensible means equal accelerations. Suppose a string of fixed length $L$ joins particle $A$ on a table to particle $B$ hanging over the edge.

If $A$ is a distance $x$ from the edge and $B$ hangs a distance $y$ below it, then $x + y = L$ at every instant.

If $x$ goes down by $0.3$ m, $y$ must go up by $0.3$ m; distances changed by the same amount in the same time mean equal speeds, and equal changes in speed mean equal accelerations.

Light means equal tensions. Take a short piece of string with tension $T_1$ pulling one end and $T_2$ pulling the other.

Newton's second law for that piece gives $T_1 - T_2 = m_s a$, where $m_s$ is its mass.

If the string is light, $m_s = 0$, so $T_1 = T_2$ whatever the acceleration.

A heavy rope does not have this property: the top of a hanging rope holds up the rope below it as well as the load.

• M1: a method mark for a correct method applied to the question. In mechanics it is usually an equation from a principle: resolving, Newton's second law, moments, a constant acceleration formula. A1 is an accuracy mark for a correct equation or answer and needs its M1. B1 is an independent mark for a stated fact (a reaction, a zero tension, a graph feature).

• DM1 (capital D in M1 schemes): a method mark that counts only if the earlier M1 it depends on was earned. A1*: the answer is printed in the question, so every step to it must be seen with no slip. A1ft or B1ft: follow-through on your own earlier wrong value.

• cso means correct solution only; isw means a correct answer followed by extra wrong working keeps its marks; oe means any equivalent form; cao means correct answer only. M1 schemes do not use awrt; they write "or better" or "2 or 3 s.f. only". A misread that does not make the question easier loses two of the A or B marks in that part.

• An equation earns its M1 only if it passes three tests: the right number of terms, every term with the same dimensions (all forces, or all mass times acceleration), and every force that needs resolving resolved with its sine or cosine. A missing or extra $g$ is an accuracy slip; a missing mass is a method error.

• The M1 cover tells you to take $g = 9.8\ \text{m s}^{-2}$. No question repeats it, so use $9.8$ every time; using $9.81$ costs a mark once per question.

• An answer that used $g = 9.8$ must be given to 2 or 3 significant figures. $23.52$ is over-accurate and scores A0; so do a fraction like $\tfrac{7}{5}$ or a surd once $9.8$ has been used.

• An exact multiple of $g$, such as $\tfrac{4g}{5}$ or $2.4mg$, is accepted as it stands. Keep $g$ as a letter until the last line, then substitute once.

• Over-accuracy is penalised once per question; rounding too early in the working can cost a mark every time it happens.

A small box of mass $3$ kg rests on a rough horizontal table.

It is attached to one end of a light inextensible string.

The string passes over a small smooth pulley fixed at the edge of the table, and a ball of mass $2$ kg hangs from the other end.

Write down what each of the following lets you assume in your working: (a) "small box" modelled as a particle; (b) "light"; (c) "inextensible"; (d) "smooth pulley"; (e) "rough table".

1. (a) The box's size is negligible: every force on it acts through one point and it does not tip or rotate. B1

2. (b) The string has no mass, so the tension is the same along each straight part of it. B1

3. (c) The string does not stretch, so while it is taut the box and the ball have the same speed and the same magnitude of acceleration. B1

4. (d) There is no friction at the pulley, so the tension is the same on both sides of it: one letter $T$ serves for the box and the ball. B1

5. (e) Friction acts on the box, opposing its motion along the table, so the horizontal equation for the box needs a friction term. B1

Answer: five assumptions, each of which becomes part of an equation: one $T$, one $a$, forces through a point, and a friction force on the box only.

Notice that no assumption is made about the ball's air resistance: a ball "hanging freely" is also a particle with air resistance ignored.

Items like this did not appear in the 2024-25 papers, so treat this as understanding, not a common exam question.

A lamp of mass $0.6$ kg hangs at rest from a ceiling on a string.

The lamp is modelled as a particle.

(a) The string is modelled as light. Find the tension in the string. (2)

(b) A more realistic model gives the string a mass of $0.05$ kg. Find the tension at the top of the string in this model. (2)

1. (a) Resolve vertically for the lamp: $T = 0.6g$ M1

2. $T = 0.6 \times 9.8 = 5.88$ N, so $5.9$ N or $5.88$ N A1

3. (b) Resolve vertically for the lamp and string together; the top of the string holds both: $T_{\text{top}} = (0.6 + 0.05)g$ M1

4. $T_{\text{top}} = 0.65 \times 9.8 = 6.37$ N, so $6.4$ N A1

Answer: (a) $5.9$ N (or $5.88$ N); (b) $6.4$ N (or $6.37$ N).

Both used $g = 9.8$, so both are given to 2 or 3 s.f.

In the refined model the tension is $5.88$ N at the bottom of the string and $6.37$ N at the top, so "the tension in the string" no longer has a single value.

That is exactly what the word "light" removes.

The difference is under 10%, which is why the light-string model is usually good enough.

Particle $A$, of mass $5$ kg, lies on a smooth horizontal table.

It is attached to one end of a light inextensible string, which passes over a small smooth pulley fixed at the edge of the table.

Particle $B$, of mass $2$ kg, hangs freely from the other end of the string, as shown in Figure 1.

The system is released from rest with the string taut.

(a) Show that the acceleration of $A$ is $2.8\ \text{m s}^{-2}$. (4)

(b) Find the tension in the string. (2)

Take the direction of motion as positive: towards the pulley for $A$, downwards for $B$.

1. (a) Inextensible string: $A$ and $B$ have the same acceleration $a$. Light string, smooth pulley: the same tension $T$ acts on both. Smooth table: no friction on $A$.

2. N2L for $B$: $2g - T = 2a$ M1 A1

3. N2L for $A$: $T = 5a$ M1

4. Add the equations: $2g = 7a$, so $a = \dfrac{2g}{7} = 2.8\ \text{m s}^{-2}$ A1*

5. (b) $T = 5a = \dfrac{10g}{7}$ M1

6. $T = 14$ N A1

Answer: (a) shown; (b) $14$ N (or $\tfrac{10g}{7}$ N).

The exact multiple $\tfrac{10g}{7}$ is accepted as it stands; once you substitute $9.8$, give 2 or 3 s.f., so $14$ N or $14.0$ N. Check the meaning: $T$ is less than $B$'s weight ($2g = 19.6$ N), so $B$ accelerates downwards, as it must, and $a$ is less than $g$ because the weight of $B$ has to accelerate $A$ as well.

Every line used a modelling word.

With a rough table you would need a friction term in line 3, and with a heavy string the two tensions would differ.

For each situation, write down a suitable model for the object named, and one assumption you would add back if the predictions turned out poor.

(a) A parcel sliding down a conveyor ramp. (b) A plank used as a see-saw. (c) A ring threaded on a taut washing line.

1. (a) The parcel as a particle on a rough inclined plane. B1 Refine by including air resistance or by letting friction vary along the ramp. B1

2. (b) The plank as a uniform rod (a rigid body), because turning about the pivot is the point of the question; a particle cannot turn. B1 Refine by making it a non-uniform rod if it is thicker at one end. B1

3. (c) The ring as a bead on a smooth wire. B1 Refine by allowing friction between the ring and the line (a rough wire). B1

Answer: a particle on a rough plane, a uniform rod and a bead on a smooth wire.

The test for each choice is the same: does the ignored feature change the answer to the question being asked?

Like Worked example 1, this is for understanding: no 2024-25 item asked you to choose or refine a model.