| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on rough inclined plane |
| Difficulty | Standard +0.8 This M3 question requires energy methods with elastic potential energy, work done against friction, and gravitational PE on an incline. Students must handle the spring being compressed initially then extended, calculate work done by variable friction (changes direction at natural length), and combine multiple energy terms. More complex than standard M3 questions but follows established techniques without requiring novel insight. |
| Spec | 6.02g 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\Delta\text{GPE} = 2g \times 0.7\sin 30 (= 6.86)\) | B1 | Correct unsimplified change in GPE |
| For a correct EPE term \(\frac{12(0.5)^2}{2\times0.8}\) or \(\frac{12(0.2)^2}{2\times0.8}\) | B1 | Correct EPE term |
| \(WD = 2g\cos30 \times 0.3 \times 0.7 (= 3.56)\) | M1A1 | Complete method for Work Done; correct unsimplified expression |
| \(\frac{12(0.5)^2}{2\times0.8} + 6.86 = \frac{12(0.2)^2}{2\times0.8} + 0.7\times5.09 + \frac{1}{2}\times2v^2\) | M1A1 | Energy equation with 2 EPE terms, \(\Delta\)GPE, WD and KE; dimensionally correct; fully correct unsimplified equation |
| \(v^2 = 4.87 \Rightarrow v = 2.21\text{ (ms}^{-1})\) | A1 | 2.21 or 2.2, cao |
## Question 3:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\Delta\text{GPE} = 2g \times 0.7\sin 30 (= 6.86)$ | B1 | Correct unsimplified change in GPE |
| For a correct EPE term $\frac{12(0.5)^2}{2\times0.8}$ or $\frac{12(0.2)^2}{2\times0.8}$ | B1 | Correct EPE term |
| $WD = 2g\cos30 \times 0.3 \times 0.7 (= 3.56)$ | M1A1 | Complete method for Work Done; correct unsimplified expression |
| $\frac{12(0.5)^2}{2\times0.8} + 6.86 = \frac{12(0.2)^2}{2\times0.8} + 0.7\times5.09 + \frac{1}{2}\times2v^2$ | M1A1 | Energy equation with 2 EPE terms, $\Delta$GPE, WD and KE; dimensionally correct; fully correct unsimplified equation |
| $v^2 = 4.87 \Rightarrow v = 2.21\text{ (ms}^{-1})$ | A1 | 2.21 or 2.2, cao |
**Note:** Answers using constant acceleration score no marks. **[7]**
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\caption{Figure 3}
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A particle $P$ of mass 2 kg is attached to one end of a light elastic spring, of natural length 0.8 m and modulus of elasticity 12 N . The other end of the spring is attached to a fixed point $A$ on a rough plane. The plane is inclined at $30 ^ { \circ }$ to the horizontal. Initially $P$ is held at rest on the plane at the point $B$, where $B$ is below $A$, with $A B = 0.3 \mathrm {~m}$ and $A B$ lies along a line of greatest slope of the plane. The point $C$ lies on the plane with $A C = 1 \mathrm {~m}$, as shown in Figure 3.
The coefficient of friction between $P$ and the plane is 0.3
After being released $P$ passes through the point $C$.
Find the speed of $P$ at the instant it passes through $C$.
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\hfill \mbox{\textit{Edexcel M3 2020 Q3 [7]}}