| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string with compression (spring) |
| Difficulty | Standard +0.3 This is a standard M3 elastic spring problem requiring (a) application of Hooke's law and Newton's second law with a compressed spring, and (b) energy conservation to find the distance traveled. Both parts follow routine procedures taught in M3 with no novel insight required, making it slightly easier than average. |
| 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 principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Weight + thrust = mass × acceleration | M1 | |
| \(0.5\times g + \frac{20\times 1}{2} = 0.5a\) | B1(thrust), A1ft | |
| \(a = g + 20 = 29.8 \approx 30\ (\text{m s}^{-2})\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Change in GPE \(= mg(x+1)\) | B1 | |
| EPE at B \(= \frac{20\times 1^2}{2\times 2}\) or EPE at C \(= \frac{20\times x^2}{2\times 2}\) | B1 | |
| Conservation of energy: \(\frac{20\times 1^2}{2\times 2} + mgh = \frac{20\times x^2}{2\times 2}\), \(h = x+1\) | M1A1 | |
| \(5 + 0.5g(x+1) = 5x^2\) | ||
| \(5x^2 - 0.5gx - (5 + 0.5g) = 0\) | ||
| \(x = \frac{0.5g + \sqrt{(0.5g)^2 + 20(5+0.5g)}}{10} = 1.98\) | M1dep | |
| Distance \(BC = 1 + 1.98 = 2.98\ \text{(m)}\) | A1 | (6) [10] |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Weight + thrust = mass × acceleration | M1 | |
| $0.5\times g + \frac{20\times 1}{2} = 0.5a$ | B1(thrust), A1ft | |
| $a = g + 20 = 29.8 \approx 30\ (\text{m s}^{-2})$ | A1 | **(4)** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Change in GPE $= mg(x+1)$ | B1 | |
| EPE at B $= \frac{20\times 1^2}{2\times 2}$ or EPE at C $= \frac{20\times x^2}{2\times 2}$ | B1 | |
| Conservation of energy: $\frac{20\times 1^2}{2\times 2} + mgh = \frac{20\times x^2}{2\times 2}$, $h = x+1$ | M1A1 | |
| $5 + 0.5g(x+1) = 5x^2$ | | |
| $5x^2 - 0.5gx - (5 + 0.5g) = 0$ | | |
| $x = \frac{0.5g + \sqrt{(0.5g)^2 + 20(5+0.5g)}}{10} = 1.98$ | M1dep | |
| Distance $BC = 1 + 1.98 = 2.98\ \text{(m)}$ | A1 | **(6) [10]** |
---3. A particle $P$ of mass 0.5 kg is attached to one end of a light elastic spring, of natural length 2 m and modulus of elasticity 20 N . The other end of the spring is attached to a fixed point $A$. The particle $P$ is held at rest at the point $B$, which is 1 m vertically below $A$, and then released.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ immediately after it is released from rest.
The particle comes to instantaneous rest for the first time at the point $C$.
\item Find the distance $B C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q3 [10]}}