| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Particle motion - velocity/time (dv/dt = f(v,t)) |
| Difficulty | Standard +0.3 This is a standard mechanics differential equation problem requiring Newton's second law setup, separation of variables, and integration. The question guides students through each step explicitly (showing a given form, then solving), making it slightly easier than average. The mathematics is routine for M2 level—resolving forces, separating variables, and integrating to find distance—but requires careful bookkeeping across multiple parts. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For using Newton's second law | M1 | |
| \(120 - 8v - 80 \times 10 \times 0.1 = 80a\) | A1 | |
| \(\frac{1}{5-v}\frac{dv}{dt} = \frac{1}{10}\) from correct working | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For separating the variables and attempting to integrate | M1 | |
| \(-\ln(5-v) = \frac{1}{10}t + (C)\) | A1 | |
| For using \(v(0) = 0\) to find \(C\) (or equivalent by using limits) \((C = -\ln 5)\) | M1 | |
| For converting the equation from logarithmic to exponential form (allow even if \(+C\) omitted) \(\left(5 \div (5-v) = e^{t/10}\right)\) | M1 | |
| \(v = 5(1 - e^{-t/10})\) from correct working | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For using \(v = \frac{dx}{dt}\) and attempting to integrate | M1 | |
| \(x = 5(t + 10e^{-t/10}) + (C)\) | A1ft | |
| For using \(x(0) = 0\) to find \((C)\) \((= -50)\), then substituting \(t = 20\) (or equivalent using limits) | M1 | |
| Length is 56.8 m | A1 | |
| OR alternative: For using Newton's second law with \(a = v\frac{dv}{dx}\), separating variables and attempting to integrate | M1 | |
| \(-v - 5\ln(5-v) = \frac{x}{10} + C\) | A1 | |
| For using \(v = 0\) when \(x = 0\) to find \(C\ (= -5\ln 5)\), then substituting \(t = 20\) into \(v(t)\): \(v(20) = 5(1-e^{-2}) = 4.3233\), and substituting \(v(20)\) into above: \((x = -50(1-e^{-2}) + 50 \times 2 = 50 + 50e^{-2})\) | M1 | |
| Length is 56.8 m | A1 |
# Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| For using Newton's second law | M1 | |
| $120 - 8v - 80 \times 10 \times 0.1 = 80a$ | A1 | |
| $\frac{1}{5-v}\frac{dv}{dt} = \frac{1}{10}$ from correct working | A1 | |
# Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| For separating the variables and attempting to integrate | M1 | |
| $-\ln(5-v) = \frac{1}{10}t + (C)$ | A1 | |
| For using $v(0) = 0$ to find $C$ (or equivalent by using limits) $(C = -\ln 5)$ | M1 | |
| For converting the equation from logarithmic to exponential form (allow even if $+C$ omitted) $\left(5 \div (5-v) = e^{t/10}\right)$ | M1 | |
| $v = 5(1 - e^{-t/10})$ from correct working | A1 | |
# Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| For using $v = \frac{dx}{dt}$ and attempting to integrate | M1 | |
| $x = 5(t + 10e^{-t/10}) + (C)$ | A1ft | |
| For using $x(0) = 0$ to find $(C)$ $(= -50)$, then substituting $t = 20$ (or equivalent using limits) | M1 | |
| Length is 56.8 m | A1 | |
| **OR alternative:** For using Newton's second law with $a = v\frac{dv}{dx}$, separating variables and attempting to integrate | M1 | |
| $-v - 5\ln(5-v) = \frac{x}{10} + C$ | A1 | |
| For using $v = 0$ when $x = 0$ to find $C\ (= -5\ln 5)$, then substituting $t = 20$ into $v(t)$: $v(20) = 5(1-e^{-2}) = 4.3233$, and substituting $v(20)$ into above: $(x = -50(1-e^{-2}) + 50 \times 2 = 50 + 50e^{-2})$ | M1 | |
| Length is 56.8 m | A1 | |6 A cyclist and his machine have a total mass of 80 kg . The cyclist starts from rest and rides from the bottom to the top of a straight slope inclined at an angle $\theta$ to the horizontal, where $\sin \theta = 0.1$. The cyclist exerts a constant force of magnitude 120 N . There is a resisting force of magnitude $8 v \mathrm {~N}$ acting on the cyclist, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the cyclist's speed at time $t \mathrm {~s}$ after the start.\\
(i) Show that $\left( \frac { 1 } { 5 - v } \right) \frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 1 } { 10 }$.\\
(ii) Solve this differential equation and hence show that $v = 5 \left( 1 - \mathrm { e } ^ { - \frac { 1 } { 10 } t } \right)$.\\
(iii) Given that the cyclist takes 20 s to reach the top of the slope, find the length of the slope.
\hfill \mbox{\textit{CAIE M2 2003 Q6 [12]}}