| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| 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 mechanics question requires energy methods with elastic strings, friction work, and multi-stage analysis. Part (a) is standard energy conservation, but parts (b) and (c) require setting up equations for unknown distances and verifying equilibrium conditions—more sophisticated than typical M1/M2 problems but still follows established M3 patterns without requiring novel insight. |
| Spec | 3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F_r = \frac{1}{7} \times 0.4 \times 9.8 \; (= 0.56)\) (N) | B1 | Correct friction seen (may be in WD), \(g\) or 9.8 acceptable |
| \(\frac{1}{2} \times 0.4v^2 = \frac{1}{2} \times 0.4(1.8)^2 - 0.8 \times \text{"0.56"}\) | M1, A1, A1 | Work-Energy equation with 2 KE terms and WD by friction; one mark each for KE terms; ft their \(F_r\) |
| \(v^2 = 1.00 \Rightarrow v = 1.0\) or \(1.00\) (ms\(^{-1}\)) | A1 | \(v = 1.0\) or \(1.00\) (2 or 3 sf as \(g\) used) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 0.4(1.0)^2 = 0.56x + \frac{0.6x^2}{2(0.8)}\) | M1, A1, B1 | Work-Energy with KE, WD and EPE; EPE term of form \(\frac{\lambda x^2}{k \times \text{natural length}}\) with \(k=1\) or \(2\); correct EPE term |
| \(0.375x^2 + 0.56x - 0.2 = 0\) | DM1 | Reducing to 3-term quadratic in \(x\); depends on first M of (b) |
| \(x = 0.2977\ldots\) | A1 | |
| Total distance \(= 1.1\) (m) (or \(1.10\)) | A1 | 1.1 (m) or 1.10 (m), (2 or 3 sf as \(g\) used) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 0.4(1.8)^2 = 0.56y + \frac{0.6(y-0.8)^2}{2(0.8)}\) | M1, A1, B1 | First 4 marks as main scheme |
| \(0.375y^2 - 0.046y - 0.408 = 0\) | DM1 | |
| \(y = 1.0977\ldots\) | A1 | Award A1A1 if final answer correct from correct equation; A1A0 if correct but not rounded |
| \(y = 1.1\) or \(1.10\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times 0.4(1.8)^2 = 0.56(y+0.8) + \frac{0.6y^2}{2(0.8)}\) | M1, A1, B1 | |
| Rest as main scheme | DM1, A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = \frac{0.6 \times \text{"0.2977"}}{0.8} \; (= 0.223)\) | M1, A1ft | Use of Hooke's Law for their extension at C; correct tension, ft their extension |
| \(0.223 < 0.56\) — Tension less than \(F_{\max}\). Therefore particle will not move. | A1cso* | Correct conclusion from fully correct working including evaluation of tension |
## Question 6:
### Part 6a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F_r = \frac{1}{7} \times 0.4 \times 9.8 \; (= 0.56)$ (N) | B1 | Correct friction seen (may be in WD), $g$ or 9.8 acceptable |
| $\frac{1}{2} \times 0.4v^2 = \frac{1}{2} \times 0.4(1.8)^2 - 0.8 \times \text{"0.56"}$ | M1, A1, A1 | Work-Energy equation with 2 KE terms and WD by friction; one mark each for KE terms; ft their $F_r$ |
| $v^2 = 1.00 \Rightarrow v = 1.0$ or $1.00$ (ms$^{-1}$) | A1 | $v = 1.0$ or $1.00$ (2 or 3 sf as $g$ used) |
### Part 6b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 0.4(1.0)^2 = 0.56x + \frac{0.6x^2}{2(0.8)}$ | M1, A1, B1 | Work-Energy with KE, WD and EPE; EPE term of form $\frac{\lambda x^2}{k \times \text{natural length}}$ with $k=1$ or $2$; correct EPE term |
| $0.375x^2 + 0.56x - 0.2 = 0$ | DM1 | Reducing to 3-term quadratic in $x$; depends on first M of (b) |
| $x = 0.2977\ldots$ | A1 | |
| Total distance $= 1.1$ (m) (or $1.10$) | A1 | 1.1 (m) or 1.10 (m), (2 or 3 sf as $g$ used) |
**ALT 1** — Work from A to C with total distance as unknown:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 0.4(1.8)^2 = 0.56y + \frac{0.6(y-0.8)^2}{2(0.8)}$ | M1, A1, B1 | First 4 marks as main scheme |
| $0.375y^2 - 0.046y - 0.408 = 0$ | DM1 | |
| $y = 1.0977\ldots$ | A1 | Award A1A1 if final answer correct from correct equation; A1A0 if correct but not rounded |
| $y = 1.1$ or $1.10$ | A1 | |
**ALT 2** — Work from A to C with distance BC as unknown:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times 0.4(1.8)^2 = 0.56(y+0.8) + \frac{0.6y^2}{2(0.8)}$ | M1, A1, B1 | |
| Rest as main scheme | DM1, A1, A1 | |
### Part 6c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = \frac{0.6 \times \text{"0.2977"}}{0.8} \; (= 0.223)$ | M1, A1ft | Use of Hooke's Law for their extension at C; correct tension, ft their extension |
| $0.223 < 0.56$ — Tension less than $F_{\max}$. Therefore particle will not move. | A1cso* | Correct conclusion from fully correct working including evaluation of tension |
---\begin{enumerate}
\item A particle $P$ of mass 0.4 kg is attached to one end of a light elastic string, of natural length 0.8 m and modulus of elasticity 0.6 N . The other end of the string is fixed to a point $A$ on a rough horizontal table. The coefficient of friction between $P$ and the table is $\frac { 1 } { 7 }$
\end{enumerate}
The particle $P$ is projected from $A$, with speed $1.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, along the surface of the table.\\
After travelling 0.8 m from $A$, the particle passes through the point $B$ on the table.\\
(a) Find the speed of $P$ at the instant it passes through $B$.
The particle $P$ comes to rest at the point $C$ on the table, where $A B C$ is a straight line.\\
(b) Find the total distance travelled by $P$ as it moves directly from $A$ to $C$.\\
(c) Show that $P$ remains at rest at $C$.
\hfill \mbox{\textit{Edexcel M3 2021 Q6 [14]}}