| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Standard +0.3 This is a straightforward Further Maths mechanics question requiring differentiation of an exponential function to find acceleration, solving for k using given conditions, then integration to find displacement. While it involves exponentials and logarithms, the techniques are standard and the multi-step process is clearly signposted with no novel insight required. It's slightly easier than average even for FM2 due to its routine nature. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(a = \frac{\mathrm{d}v}{\mathrm{d}t}\) | M1 | Use the model and differentiate \(v\) to obtain \(a\). Obtain form \(pe^{2t}+qe^{t}-k\) |
| \(a = 2e^{2t}+6e^{t}-k\) | A1 | Correct only |
| Substitute \(t = \ln 2\) into their acceleration and solve for \(k\) | M1 | \((2\times4+6\times2-k=0)\) Their acceleration must come from an attempt to differentiate |
| \(k = 20\) | A1 | Correct only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(v = \frac{\mathrm{d}x}{\mathrm{d}t}\) | M1 | Integrate \(v\) to obtain \(x\). Obtain form \(ae^{2t}+be^{t}+ct^2\) |
| \(x = \frac{1}{2}e^{2t}+6e^{t}-\frac{k}{2}t^{2}(+C)\) | A1ft | Allow in \(k\) or their \(k\) |
| Correct use of boundary conditions in their \(x\) | M1 | Use the model to evaluate constant of integration or use boundary conditions as limits in a definite integral. Their \(x\) must come from an attempt to integrate |
| \(x = 2+12-10(\ln 2)^{2}-\left(\frac{1}{2}+6\right)\left(=7.5-10(\ln 2)^{2}\right)= 2.7\) to 2 s.f. | A1 | 2sf only |
## Question 2:
### Part 2a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $a = \frac{\mathrm{d}v}{\mathrm{d}t}$ | M1 | Use the model and differentiate $v$ to obtain $a$. Obtain form $pe^{2t}+qe^{t}-k$ |
| $a = 2e^{2t}+6e^{t}-k$ | A1 | Correct only |
| Substitute $t = \ln 2$ into their acceleration and solve for $k$ | M1 | $(2\times4+6\times2-k=0)$ Their acceleration must come from an attempt to differentiate |
| $k = 20$ | A1 | Correct only |
### Part 2b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $v = \frac{\mathrm{d}x}{\mathrm{d}t}$ | M1 | Integrate $v$ to obtain $x$. Obtain form $ae^{2t}+be^{t}+ct^2$ |
| $x = \frac{1}{2}e^{2t}+6e^{t}-\frac{k}{2}t^{2}(+C)$ | A1ft | Allow in $k$ or their $k$ |
| Correct use of boundary conditions in their $x$ | M1 | Use the model to evaluate constant of integration or use boundary conditions as limits in a definite integral. Their $x$ must come from an attempt to integrate |
| $x = 2+12-10(\ln 2)^{2}-\left(\frac{1}{2}+6\right)\left(=7.5-10(\ln 2)^{2}\right)= 2.7$ to 2 s.f. | A1 | 2sf only |
---\begin{enumerate}
\item A particle $P$ is moving along the $x$-axis.
\end{enumerate}
At time $t$ seconds, $t \geqslant 0 , P$ has acceleration $a \mathrm {~ms} ^ { - 2 }$ and velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the direction of $x$ increasing, where
$$v = \mathrm { e } ^ { 2 t } + 6 \mathrm { e } ^ { t } - k t$$
and $k$ is a positive constant.\\
When $t = \ln 2$, $a = 0$\\
(a) Find the value of $k$.
When $t = 0$, the particle passes through the fixed point $A$.\\
When $t = \ln 2$, the particle is $d$ metres from $A$.\\
(b) Showing all stages of your working, find the value of $d$ correct to 2 significant figures.\\[0pt]
[Solutions relying entirely on calculator technology are not acceptable.]
\hfill \mbox{\textit{Edexcel FM2 AS 2023 Q2 [8]}}