| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Elastic string equilibrium |
| Difficulty | Standard +0.3 This is a straightforward M3 equilibrium problem requiring resolution of forces (weight, tension, applied force) and application of Hooke's law. The calculation involves standard trigonometry with given tan θ = 3/4, finding extension, and solving for k. While it requires multiple steps, all techniques are routine for M3 students with no novel insight needed. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = mg\cos\theta\) | M1A1 | Resolve parallel to string; correct no. of terms; condone sign errors and sin/cos confusion; or resolve in two directions and eliminate unknown force or use trig on right-angled triangle |
| \(T = \dfrac{2mg\!\left(\frac{21}{10}a - ka\right)}{ka}\) | M1A1 | Use Hooke's Law with correct structure; correct equation |
| \(\dfrac{4}{5}mg = \dfrac{2mg\!\left(\frac{21}{10}a - ka\right)}{ka}\) | dM1 | Substitute trig and eliminate \(T\) to produce equation in \(k\) only; dependent on previous M marks; if \(x\) used for extension: \(x = \frac{2k}{5}a \rightarrow ka + \frac{2k}{5}a = \frac{21}{10}a\) |
| \(k = \dfrac{3}{2}\) or \(1.5\) | A1 (6) | cao |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = mg\cos\theta$ | M1A1 | Resolve parallel to string; correct no. of terms; condone sign errors and sin/cos confusion; or resolve in two directions and eliminate unknown force or use trig on right-angled triangle |
| $T = \dfrac{2mg\!\left(\frac{21}{10}a - ka\right)}{ka}$ | M1A1 | Use Hooke's Law with correct structure; correct equation |
| $\dfrac{4}{5}mg = \dfrac{2mg\!\left(\frac{21}{10}a - ka\right)}{ka}$ | dM1 | Substitute trig and eliminate $T$ to produce equation in $k$ only; dependent on previous M marks; if $x$ used for extension: $x = \frac{2k}{5}a \rightarrow ka + \frac{2k}{5}a = \frac{21}{10}a$ |
| $k = \dfrac{3}{2}$ or $1.5$ | A1 (6) | cao |
---2.
\begin{figure}[h]
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\caption{Figure 2}
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A light elastic string $A B$ has modulus of elasticity $2 m g$ and natural length $k a$, where $k$ is a constant.\\
The end $A$ of the elastic string is attached to a fixed point. The other end $B$ is attached to a particle of mass $m$. The particle is held in equilibrium, with the elastic string taut, by a force that acts in a direction that is perpendicular to the string. The line of action of the force and the elastic string lie in the same vertical plane. The string makes an angle $\theta$ with the downward vertical at $A$, as shown in Figure 2.
Given that the length $A B = \frac { 21 } { 10 } a$ and $\tan \theta = \frac { 3 } { 4 }$, find the value of $k$.
\hfill \mbox{\textit{Edexcel M3 2023 Q2 [6]}}