| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Elastic string with variable force |
| Difficulty | Challenging +1.8 This is a challenging M2 mechanics problem requiring multiple techniques: deriving a differential equation using F=ma with variable resistance, solving it using v dv/dx = a, applying energy methods with elastic strings, and verifying motion constraints. The variable resistance force (0.1x²) and multi-stage motion analysis elevate this above standard elastic string questions, but the individual steps follow established M2 methods without requiring novel insight. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5v\frac{dv}{dx} = -0.5g - 0.1x^2\) | M1 | Use Newton's Second Law vertically |
| \(v\frac{dv}{dx} = -10 - 0.2x^2\) | A1 (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int v\, dv = \int(-10 - 0.2x^2)\, dx\) | M1 | Attempt to integrate the expression in part (i) |
| \(\frac{v^2}{2} - 10 \times \frac{-0.2x^3}{3} + c\) | A1 | |
| \(\left[\frac{v^2}{2} = -10 - \frac{0.2}{3} + 18\right]\) | M1 | Either use limits or find \(c\) and put \(x = 1\) |
| \(v = 3.98\ (329\ldots)\ \text{ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.5v\frac{dv}{dx} = -0.5g - 0.1x^2 - \frac{16(x-1)}{1}\) | M1 | Use Newton's Second Law vertically when string becomes taut |
| \(v\frac{dv}{dx} = -10 - 0.2x^2 - 32x + 32 = 22 - 32x - 0.2x^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int v\, dv = \int(22 - 32x - 0.2x^2)\, dx\) | M1 | Attempt to integrate after the string becomes taut |
| \(\frac{v^2}{2} = 22x - \frac{32x^2}{2} - \frac{0.2x^3}{3} + k\) | A1 | |
| \(x = 1,\ v = 3.98\ (329\ldots)\) hence \(k = 2\). Now put \(x = 1.5\): \(22\times 1.5 - 32\times\frac{1.5^2}{2} - 0.2\times\frac{1.5^3}{3} + 2 = -1.225\) | M1 | Either use limits or find \(k\) and put \(x = 1.5\) |
| As \(\frac{v^2}{2}\) cannot be negative, \(P\) comes to rest before the extension of the string is \(0.5\). | A1 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5v\frac{dv}{dx} = -0.5g - 0.1x^2$ | M1 | Use Newton's Second Law vertically |
| $v\frac{dv}{dx} = -10 - 0.2x^2$ | A1 (AG) | |
---
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int v\, dv = \int(-10 - 0.2x^2)\, dx$ | M1 | Attempt to integrate the expression in part (i) |
| $\frac{v^2}{2} - 10 \times \frac{-0.2x^3}{3} + c$ | A1 | |
| $\left[\frac{v^2}{2} = -10 - \frac{0.2}{3} + 18\right]$ | M1 | Either use limits or find $c$ and put $x = 1$ |
| $v = 3.98\ (329\ldots)\ \text{ms}^{-1}$ | A1 | |
---
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5v\frac{dv}{dx} = -0.5g - 0.1x^2 - \frac{16(x-1)}{1}$ | M1 | Use Newton's Second Law vertically when string becomes taut |
| $v\frac{dv}{dx} = -10 - 0.2x^2 - 32x + 32 = 22 - 32x - 0.2x^2$ | A1 | |
---
## Question 7(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int v\, dv = \int(22 - 32x - 0.2x^2)\, dx$ | M1 | Attempt to integrate after the string becomes taut |
| $\frac{v^2}{2} = 22x - \frac{32x^2}{2} - \frac{0.2x^3}{3} + k$ | A1 | |
| $x = 1,\ v = 3.98\ (329\ldots)$ hence $k = 2$. Now put $x = 1.5$: $22\times 1.5 - 32\times\frac{1.5^2}{2} - 0.2\times\frac{1.5^3}{3} + 2 = -1.225$ | M1 | Either use limits or find $k$ and put $x = 1.5$ |
| As $\frac{v^2}{2}$ cannot be negative, $P$ comes to rest before the extension of the string is $0.5$. | A1 | |7 A particle $P$ of mass 0.5 kg is attached to a fixed point $O$ by a light elastic string of natural length 1 m and modulus of elasticity 16 N . The particle $P$ is projected vertically upwards from $O$ with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. A resisting force of magnitude $0.1 x ^ { 2 } \mathrm {~N}$ acts on $P$ when $P$ has displacement $x \mathrm {~m}$ above $O$. After projection the upwards velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that, before the string becomes taut, $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 10 - 0.2 x ^ { 2 }$.\\
(ii) Find the velocity of $P$ at the instant the string becomes taut.\\
(iii) Find an expression for the acceleration of $P$ while it is moving upwards after the string becomes taut.\\
(iv) Verify that $P$ comes to instantaneous rest before the extension of the string is 0.5 m .\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M2 2019 Q7 [12]}}