| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Horizontal elastic string on rough surface |
| Difficulty | Challenging +1.2 This is a multi-step energy problem involving elastic strings, friction, and kinematics. Part (a) requires setting up an energy equation with elastic PE, work done against friction, and KE, then solving a quadratic—standard M3 technique but with several components. Part (b) adds complexity by requiring consideration of the return journey with friction acting throughout. While methodical, it demands careful bookkeeping and understanding of energy principles beyond routine exercises, placing it moderately above average difficulty. |
| Spec | 3.03u Static equilibrium: on rough surfaces6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (a) resolve \(\uparrow\): \(R - mg = 0 \therefore R = 2g\) | M1 A1 | |
| friction \(= \mu R = \frac{10}{g} \times 2 \times 9.8 = 4\) | A1 | |
| work-energy: work done \(=\) loss of KE \(-\) gain of EPE | M1 | |
| \(\therefore Fx = \frac{1}{2}mu^2 - \frac{1}{2}x^2\) | M1 | |
| so \(4d = \frac{1}{2} \times 2 \times 5^2 - \frac{50(d-1)^2}{2-1}\) | A1 | |
| \(\therefore 4d = 25 - 25(d^2 - 2d + 1)\) | M1 | |
| giving \(25d^2 - 46d = 0, d(25d - 46) = 0\) | M1 | |
| \(\therefore d = 0\) (initially) or \(\frac{46}{25} = 1.84\) m | A1 | |
| (b) work-energy: work done \(=\) loss of EPE \(-\) gain of KE | M1 | |
| \(\therefore 4 \times \frac{46}{25} = \frac{50(d-1)^2}{2-1} - \frac{1}{2} \times 2 \times v^2\) | M1 A1 | |
| giving \(21^2 = (4 \times 46) + 25v^2\) | M1 | |
| so \(v^2 = \frac{252}{25} \therefore v = 3.2\) ms\(^{-1}\) (2sf) | A1 | (14) |
(a) resolve $\uparrow$: $R - mg = 0 \therefore R = 2g$ | M1 A1
friction $= \mu R = \frac{10}{g} \times 2 \times 9.8 = 4$ | A1
work-energy: work done $=$ loss of KE $-$ gain of EPE | M1
$\therefore Fx = \frac{1}{2}mu^2 - \frac{1}{2}x^2$ | M1
so $4d = \frac{1}{2} \times 2 \times 5^2 - \frac{50(d-1)^2}{2-1}$ | A1
$\therefore 4d = 25 - 25(d^2 - 2d + 1)$ | M1
giving $25d^2 - 46d = 0, d(25d - 46) = 0$ | M1
$\therefore d = 0$ (initially) or $\frac{46}{25} = 1.84$ m | A1
(b) work-energy: work done $=$ loss of EPE $-$ gain of KE | M1
$\therefore 4 \times \frac{46}{25} = \frac{50(d-1)^2}{2-1} - \frac{1}{2} \times 2 \times v^2$ | M1 A1
giving $21^2 = (4 \times 46) + 25v^2$ | M1
so $v^2 = \frac{252}{25} \therefore v = 3.2$ ms$^{-1}$ (2sf) | A1 | (14)
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**Total: (75)**7. A particle of mass 2 kg is attached to one end of a light elastic string of natural length 1 m and modulus of elasticity 50 N . The other end of the string is attached to a fixed point $O$ on a rough horizontal plane and the coefficient of friction between the particle and the plane is $\frac { 10 } { 49 }$.
The particle is projected from $O$ along the plane with an initial speed of $5 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the greatest distance from $O$ which the particle reaches is 1.84 m .
\item Find, correct to 2 significant figures, the speed at which the particle returns to $O$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [14]}}