| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Maximum/minimum speed in elastic motion |
| Difficulty | Standard +0.8 This is a multi-part vertical elastic string problem requiring energy conservation with careful accounting of gravitational PE, elastic PE, and KE. Part (i) involves deriving a specific relationship (algebraically demanding), part (ii) requires differentiation or completing the square to find maximum speed, and part (iii) needs force analysis at the lowest point. The combination of elastic energy, gravity, and calculus techniques makes this moderately challenging, above typical A-level questions but accessible with solid M2 knowledge. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EE gain \(= 20x^2/(2 \times 1.25)\) | B1 | |
| PE loss \(= 0.5g(1.25 + x)\) | B1 | |
| \(\left[\frac{1}{2}(0.5)v^2 = 6.25 + 5x - 8x^2\right]\) | M1 | For using KE gain = PE loss - EE gain |
| \(v^2 = -32x^2 + 20x + 25\) | A1 | AG. Subtotal: 4 |
| ALTERNATIVE: \(\left[0.5v(dv/dx) = -20x/1.25 + 0.5g\right]\) | M1 | For using Newton's second law with \(A = v(dv/dx)\) and \(T = \lambda x/L\) |
| \(\left[\frac{1}{2}(0.5)v^2 = -10x^2/1.25 + 5x + c\right]\) | M1 | For integrating and using \(\frac{1}{2}(0.5)v(0)^2 = 0.5g \times 1.25\) |
| \(c = 6.25\) | A1 | |
| \(v^2 = -32x^2 + 20x + 25\) | A1 | AG. (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v^2 = -32(x - 5/16)^2 + 28.125]\) | M1 | For obtaining \(v^2\) in the form \(a(x-b)^2 + c\) |
| \([v_{\max} = \sqrt{28.125}]\) | M1 | For substituting \(v_{\max} = \sqrt{c}\) |
| Maximum speed is \(5.30\text{ ms}^{-1}\) | A1 | Subtotal: 3 |
| ALTERNATIVE 1: \([-64x + 20 = 0 \Rightarrow x = 5/16]\) | M1 | For solving \(d(v^2)/dx\) for \(x\) |
| \([v_{\max}^2 = -32(5/16)^2 + 20(5/16) + 25]\) | M1 | For substituting \(x\) found into \(v^2(x)\) |
| Maximum speed is \(5.30\text{ ms}^{-1}\) | A1 | (3) |
| ALTERNATIVE 2: \([T = mg = 5,\ T = \lambda x/L = 20x/1.25 \Rightarrow x = 5/16]\) | M1 | Using \(a = 0\) at maximum speed |
| M1 | For substituting \(x = 5/16\) in \(v^2\) | |
| Maximum speed is \(5.30\text{ ms}^{-1}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([-32x^2 + 20x + 25 = 0 \Rightarrow x = 1.25]\) | M1 | For attempting to solve \(v = 0\) |
| \(\left[a = v(dv/dx) = \frac{1}{2}d(v^2)/dx = -32x + 10\right]\) | M1 | For using \(a = \frac{1}{2}d(v^2)/dx\) |
| \(a = -32 \times 1.25 + 10\) | A1ft | |
| Acceleration is \(30\text{ ms}^{-2}\) (upwards) | A1 | Subtotal: 4 |
| ALTERNATIVE 1: \([-32x^2 + 20x + 25 = 0 \Rightarrow x = 1.25]\) | M1 | For attempting to solve \(v = 0\) |
| \([0.5g - 20x/1.25 = 0.5a]\) | M1 | For using Newton's second law |
| \(a = g - 2(20 \times 1.25/1.25)\) | A1ft | |
| Acceleration is \(30\text{ ms}^{-2}\) (upwards) | A1 | (4) |
| ALTERNATIVE 2: \([-32x^2 + 20x + 25 = 0 \Rightarrow x = 1.25]\) | M1 | For attempting to solve \(v = 0\) |
| M1 | Using Newton's 2nd Law | |
| \(20 - 5 = 0.5a\) or \(5 - 20 = 0.5a\) | A1ft | |
| Acceleration is \(30\text{ ms}^{-2}\) | A1 | If \(a = -30\) then direction should be explained. (4). Total: 11 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| EE gain $= 20x^2/(2 \times 1.25)$ | B1 | |
| PE loss $= 0.5g(1.25 + x)$ | B1 | |
| $\left[\frac{1}{2}(0.5)v^2 = 6.25 + 5x - 8x^2\right]$ | M1 | For using KE gain = PE loss - EE gain |
| $v^2 = -32x^2 + 20x + 25$ | A1 | AG. **Subtotal: 4** |
| **ALTERNATIVE:** $\left[0.5v(dv/dx) = -20x/1.25 + 0.5g\right]$ | M1 | For using Newton's second law with $A = v(dv/dx)$ and $T = \lambda x/L$ |
| $\left[\frac{1}{2}(0.5)v^2 = -10x^2/1.25 + 5x + c\right]$ | M1 | For integrating and using $\frac{1}{2}(0.5)v(0)^2 = 0.5g \times 1.25$ |
| $c = 6.25$ | A1 | |
| $v^2 = -32x^2 + 20x + 25$ | A1 | AG. **(4)** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v^2 = -32(x - 5/16)^2 + 28.125]$ | M1 | For obtaining $v^2$ in the form $a(x-b)^2 + c$ |
| $[v_{\max} = \sqrt{28.125}]$ | M1 | For substituting $v_{\max} = \sqrt{c}$ |
| Maximum speed is $5.30\text{ ms}^{-1}$ | A1 | **Subtotal: 3** |
| **ALTERNATIVE 1:** $[-64x + 20 = 0 \Rightarrow x = 5/16]$ | M1 | For solving $d(v^2)/dx$ for $x$ |
| $[v_{\max}^2 = -32(5/16)^2 + 20(5/16) + 25]$ | M1 | For substituting $x$ found into $v^2(x)$ |
| Maximum speed is $5.30\text{ ms}^{-1}$ | A1 | **(3)** |
| **ALTERNATIVE 2:** $[T = mg = 5,\ T = \lambda x/L = 20x/1.25 \Rightarrow x = 5/16]$ | M1 | Using $a = 0$ at maximum speed |
| | M1 | For substituting $x = 5/16$ in $v^2$ |
| Maximum speed is $5.30\text{ ms}^{-1}$ | A1 | **(3)** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[-32x^2 + 20x + 25 = 0 \Rightarrow x = 1.25]$ | M1 | For attempting to solve $v = 0$ |
| $\left[a = v(dv/dx) = \frac{1}{2}d(v^2)/dx = -32x + 10\right]$ | M1 | For using $a = \frac{1}{2}d(v^2)/dx$ |
| $a = -32 \times 1.25 + 10$ | A1ft | |
| Acceleration is $30\text{ ms}^{-2}$ (upwards) | A1 | **Subtotal: 4** |
| **ALTERNATIVE 1:** $[-32x^2 + 20x + 25 = 0 \Rightarrow x = 1.25]$ | M1 | For attempting to solve $v = 0$ |
| $[0.5g - 20x/1.25 = 0.5a]$ | M1 | For using Newton's second law |
| $a = g - 2(20 \times 1.25/1.25)$ | A1ft | |
| Acceleration is $30\text{ ms}^{-2}$ (upwards) | A1 | **(4)** |
| **ALTERNATIVE 2:** $[-32x^2 + 20x + 25 = 0 \Rightarrow x = 1.25]$ | M1 | For attempting to solve $v = 0$ |
| | M1 | Using Newton's 2nd Law |
| $20 - 5 = 0.5a$ or $5 - 20 = 0.5a$ | A1ft | |
| Acceleration is $30\text{ ms}^{-2}$ | A1 | If $a = -30$ then direction should be explained. **(4). Total: 11** |
---6 One end of a light elastic string of natural length 1.25 m and modulus of elasticity 20 N is attached to a fixed point $O$. A particle $P$ of mass 0.5 kg is attached to the other end of the string. $P$ is held at rest at $O$ and then released. When the extension of the string is $x \mathrm {~m}$ the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that $v ^ { 2 } = - 32 x ^ { 2 } + 20 x + 25$.\\
(ii) Find the maximum speed of $P$.\\
(iii) Find the acceleration of $P$ when it is at its lowest point.
\hfill \mbox{\textit{CAIE M2 2008 Q6 [11]}}