| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2020 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question on variable acceleration requiring students to form and solve a differential equation using given initial conditions. While it involves calculus and requires careful algebraic manipulation, the structure is formulaic: find constants from conditions, separate variables, and integrate. The 'show that' part guides students significantly, making this above-average difficulty but not requiring novel insight. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Use of initial condition to find \(p\) | M1 | Use initial conditions |
| \(t=0, v=0\), acceleration \(= 3 \Rightarrow p = 3\) | A1 | cao |
| Use \(v=4\), acceleration \(= \frac{1}{2}\) | M1 | Use second condition |
| \(q = -\frac{5}{8}\) | A1 | cao |
| Use acceleration \(= \frac{dv}{dt}\) and rearrange | M1 | Use appropriate derivative and rearrange |
| \(8\frac{dv}{dt} = (24 - 5v)\) * | A1* | Correct given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| Separate the variables and integrate | M1 | Separate variables and integrate |
| \(8\int \frac{dv}{(24-5v)} = \int dt\) | A1 | Correct integration (\(C\) not required) |
| \(-\frac{8}{5}\ln(24-5v) = t + C\) | A1 | |
| Use \(t=0, v=0\) to give \(C = -\frac{8}{5}\ln 24\) | M1 | Use of limits or initial conditions to find \(C\) |
| Substitute \(v=4\) and find and simplify \(T\) | M1 | Use \(v=4\) to find \(T\) |
| \(T = \frac{8}{5}\ln 6\) | A1 | Correct answer (single log) |
## Question 3:
### Part (a)
| Working | Mark | Notes |
|---------|------|-------|
| Use of initial condition to find $p$ | M1 | Use initial conditions |
| $t=0, v=0$, acceleration $= 3 \Rightarrow p = 3$ | A1 | cao |
| Use $v=4$, acceleration $= \frac{1}{2}$ | M1 | Use second condition |
| $q = -\frac{5}{8}$ | A1 | cao |
| Use acceleration $= \frac{dv}{dt}$ and rearrange | M1 | Use appropriate derivative and rearrange |
| $8\frac{dv}{dt} = (24 - 5v)$ * | A1* | Correct given answer |
### Part (b)
| Working | Mark | Notes |
|---------|------|-------|
| Separate the variables and integrate | M1 | Separate variables and integrate |
| $8\int \frac{dv}{(24-5v)} = \int dt$ | A1 | Correct integration ($C$ not required) |
| $-\frac{8}{5}\ln(24-5v) = t + C$ | A1 | |
| Use $t=0, v=0$ to give $C = -\frac{8}{5}\ln 24$ | M1 | Use of limits or initial conditions to find $C$ |
| Substitute $v=4$ and find and simplify $T$ | M1 | Use $v=4$ to find $T$ |
| $T = \frac{8}{5}\ln 6$ | A1 | Correct answer (single log) |
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This appears to be a blank back page of an exam paper or mark scheme document.\begin{enumerate}
\item At time $t = 0$, a toy electric car is at rest at a fixed point $O$. The car then moves in a horizontal straight line so that at time $t$ seconds $( t > 0 )$ after leaving $O$, the velocity of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the acceleration of the car is modelled as $( p + q v ) \mathrm { ms } ^ { - 2 }$, where $p$ and $q$ are constants.
\end{enumerate}
When $t = 0$, the acceleration of the car is $3 \mathrm {~ms} ^ { - 2 }$\\
When $t = T$, the acceleration of the car is $\frac { 1 } { 2 } \mathrm {~ms} ^ { - 2 }$ and $v = 4$\\
(a) Show that
$$8 \frac { \mathrm {~d} v } { \mathrm {~d} t } = ( 24 - 5 v )$$
(b) Find the exact value of $T$, simplifying your answer.
\hfill \mbox{\textit{Edexcel FM2 AS 2020 Q3 [12]}}