| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Elastic string with variable force |
| Difficulty | Challenging +1.8 This question combines elastic strings with variable resistance and requires energy methods with calculus (v dv/dx formulation). Part (i) involves deriving a differential equation from Newton's second law with three force components (weight, elastic tension, variable resistance). Part (ii) requires recognizing that maximum speed occurs when dv/dx = 0, solving a quadratic, then calculating two different energy forms. The variable resistance force (proportional to x²) adds complexity beyond standard elastic string problems, and the energy method approach is less routine than basic mechanics questions, though the individual steps are A-level standard. |
| Spec | 6.02h 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 |
|---|---|---|
| \(0.5v\frac{dv}{dx} = 0.5g - \frac{16x}{0.8} - 25x^2\) | M1 | Use Newton's Second Law vertically |
| \(v\frac{dv}{dx} = 10 - 40x - 50x^2\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int v\,dv = \int(10 - 40x - 50x^2)\,dx\) | M1 | Attempt to integrate |
| \(\frac{v^2}{2} = 10x - 20x^2 - \frac{50x^3}{3}(+c)\) | A1 | |
| \(0 = 10 - 40x - 50x^2\) | M1 | Put the acceleration equal to zero |
| \(x = 0.2\) (Ignore \(x = -1\) if seen) | A1 | |
| \(\frac{0.5v^2}{2} = \frac{8}{15} = 0.533\text{J}\) | B1 | Use \(KE = \frac{mv^2}{2}\) |
| \(16 \times \frac{0.2^2}{(2 \times 0.8)} = 0.4\text{J}\) | B1 | Use \(EE = \frac{\lambda x^2}{(2l)}\) |
## Question 4(i):
| $0.5v\frac{dv}{dx} = 0.5g - \frac{16x}{0.8} - 25x^2$ | M1 | Use Newton's Second Law vertically |
| $v\frac{dv}{dx} = 10 - 40x - 50x^2$ | A1 | AG |
## Question 4(ii):
| $\int v\,dv = \int(10 - 40x - 50x^2)\,dx$ | M1 | Attempt to integrate |
| $\frac{v^2}{2} = 10x - 20x^2 - \frac{50x^3}{3}(+c)$ | A1 | |
| $0 = 10 - 40x - 50x^2$ | M1 | Put the acceleration equal to zero |
| $x = 0.2$ (Ignore $x = -1$ if seen) | A1 | |
| $\frac{0.5v^2}{2} = \frac{8}{15} = 0.533\text{J}$ | B1 | Use $KE = \frac{mv^2}{2}$ |
| $16 \times \frac{0.2^2}{(2 \times 0.8)} = 0.4\text{J}$ | B1 | Use $EE = \frac{\lambda x^2}{(2l)}$ |
---4 A particle $P$ of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 16 N . The other end of the string is attached to a fixed point $O$. The particle $P$ is released from rest at the point 0.8 m vertically below $O$. When the extension of the string is $x \mathrm {~m}$, the downwards velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and a force of magnitude $25 x ^ { 2 } \mathrm {~N}$ opposes the motion of $P$.\\
(i) Show that, when $P$ is moving downwards, $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 10 - 40 x - 50 x ^ { 2 }$.\\
(ii) For the instant when $P$ has its greatest downwards speed, find the kinetic energy of $P$ and the elastic potential energy stored in the string.\\
\hfill \mbox{\textit{CAIE M2 2019 Q4 [8]}}