| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Horizontal elastic string on rough surface |
| Difficulty | Standard +0.8 This M3 question requires understanding of elastic strings, energy methods, and friction in a multi-stage problem. Part (a) involves equilibrium analysis with friction (relatively standard), but part (b) requires careful energy accounting with work done against friction over an unknown distance, then proving the particle stops before reaching natural length—this demands systematic problem-solving and inequality manipulation beyond routine exercises. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(F \leq \mu\cdot 4mg\) | B1 | Correct inequality/equation for maximum friction. |
| \(F = T = \frac{3mg}{l}\times\frac{1}{3} \leq 4\mu mg\) | M1, A1ft | Resolve horizontally using Hooke's Law; correct formula; attempt at extension. Correct inequality following their friction. |
| \(\mu \geq \frac{1}{4}\) | A1cso (4) | Obtain given inequality with no errors. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| EPE \(= \frac{3mgl^2}{2l}\) | B1 | Correct EPE when extension is \(l\). |
| Work done against friction \(= \frac{2}{5}\times 4mg\times l\) | B1 | Correct work done against friction. |
| \(\frac{1}{2}\times 4mv^2 = \frac{3mgl^2}{2l} - \frac{2}{5}\times 4mg\times l\) | M1A1ft | Attempt energy equation with KE, EPE and WD terms; EPE of form \(k\frac{\lambda x^2}{l}\). Correct equation following their EPE and WD terms. |
| \(v^2 \leq 0 \Rightarrow\) string does not become slack | A1cso (5) [9] | Solve/state \(v^2\leq 0\) from fully correct equation; correct conclusion; no errors in working. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| EPE lost \(= \frac{3mgl^2}{2l} - \frac{3mgx^2}{2l}\) | B1 | Either EPE term correct. |
| Work done against friction \(= \frac{2}{5}\times 4mg\times(l-x)\) | B1 | Correct WD against friction. |
| \(\frac{3mgl^2}{2l} - \frac{3mgx^2}{2l} = \frac{8}{5}mg(l-x)\) | M1A1ft | Energy equation with difference of EPE terms and WD; EPE terms of form \(k\frac{\lambda x^2}{l}\). |
| \(x = \frac{1}{15}l\); positive extension at \(x=\frac{1}{15}l\); at rest before string becomes slack | A1cso | Solve for \(x\); state conclusion; no errors. |
# Question 5:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $F \leq \mu\cdot 4mg$ | B1 | Correct inequality/equation for maximum friction. |
| $F = T = \frac{3mg}{l}\times\frac{1}{3} \leq 4\mu mg$ | M1, A1ft | Resolve horizontally using Hooke's Law; correct formula; attempt at extension. Correct inequality following their friction. |
| $\mu \geq \frac{1}{4}$ | A1cso (4) | Obtain given inequality with no errors. |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| EPE $= \frac{3mgl^2}{2l}$ | B1 | Correct EPE when extension is $l$. |
| Work done against friction $= \frac{2}{5}\times 4mg\times l$ | B1 | Correct work done against friction. |
| $\frac{1}{2}\times 4mv^2 = \frac{3mgl^2}{2l} - \frac{2}{5}\times 4mg\times l$ | M1A1ft | Attempt energy equation with KE, EPE and WD terms; EPE of form $k\frac{\lambda x^2}{l}$. Correct equation following their EPE and WD terms. |
| $v^2 \leq 0 \Rightarrow$ string does not become slack | A1cso (5) [9] | Solve/state $v^2\leq 0$ from fully correct equation; correct conclusion; no errors in working. |
**Alternative (extension is $x$ when particle comes to rest):**
| Working/Answer | Marks | Guidance |
|---|---|---|
| EPE lost $= \frac{3mgl^2}{2l} - \frac{3mgx^2}{2l}$ | B1 | Either EPE term correct. |
| Work done against friction $= \frac{2}{5}\times 4mg\times(l-x)$ | B1 | Correct WD against friction. |
| $\frac{3mgl^2}{2l} - \frac{3mgx^2}{2l} = \frac{8}{5}mg(l-x)$ | M1A1ft | Energy equation with difference of EPE terms and WD; EPE terms of form $k\frac{\lambda x^2}{l}$. |
| $x = \frac{1}{15}l$; positive extension at $x=\frac{1}{15}l$; at rest before string becomes slack | A1cso | Solve for $x$; state conclusion; no errors. |\begin{enumerate}
\item A particle $P$ of mass $4 m$ is attached to one end of a light elastic string of natural length $l$ and modulus of elasticity 3 mg . The other end of the string is attached to a fixed point $O$ on a rough horizontal table. The particle lies at rest at the point $A$ on the table, where $O A = \frac { 4 } { 3 } l$. The coefficient of friction between $P$ and the table is $\mu$.\\
(a) Show that $\mu \geqslant \frac { 1 } { 4 }$
\end{enumerate}
The particle is now moved along the table to the point $B$, where $O B = 2 l$, and released from rest. Given that $\mu = \frac { 2 } { 5 }$\\
(b) show that $P$ comes to rest before the string becomes slack.\\
(5)\\
\hfill \mbox{\textit{Edexcel M3 2017 Q5 [9]}}