| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Challenging +1.2 This M3 question requires using v(dv/dx) = a to integrate acceleration with respect to position, then separating variables to find x(t). While it involves multiple integration steps and algebraic manipulation, the technique is standard for M3 and the integrations are straightforward (power rule). The 'show that' part in (b) provides the target answer, reducing problem-solving demand. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(a = v\frac{dv}{dx} = -\frac{2}{(2x+1)^3}\), separate and integrate | M1 | Allow omission of \(-\) sign, powers increasing by 1 |
| \(\frac{1}{2}v^2 = \frac{1}{2(2x+1)^2} + C\) | A1 | Correct equation, allow omission of \(C\) |
| \(x=0, v=1 \Rightarrow C = 0\) | M1 | Use of initial conditions to find \(C\) |
| \(v = \frac{1}{(2x+1)}\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dx}{dt} = \frac{1}{(2x+1)}\), separate and integrate | M1 | Powers increasing by 1 |
| \(x^2 + x + D = t\) | A1 | Correct equation, allow omission of \(D\) |
| Complete the square: \((x+\frac{1}{2})^2 - \frac{1}{4} = t\) or use quadratic formula: \(x = \frac{-1 \pm \sqrt{1+4t}}{2}\) | M1 | Complete the square or use quadratic formula |
| \(x = \frac{1}{2}(\sqrt{(4t+1)}-1)\)* | A1* | Given answer correctly obtained, with at least one line of working and justification of positive root e.g. \(x > 0\) |
# Question 3:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $a = v\frac{dv}{dx} = -\frac{2}{(2x+1)^3}$, separate and integrate | M1 | Allow omission of $-$ sign, powers increasing by 1 |
| $\frac{1}{2}v^2 = \frac{1}{2(2x+1)^2} + C$ | A1 | Correct equation, allow omission of $C$ |
| $x=0, v=1 \Rightarrow C = 0$ | M1 | Use of initial conditions to find $C$ |
| $v = \frac{1}{(2x+1)}$ | A1 | cso |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \frac{1}{(2x+1)}$, separate and integrate | M1 | Powers increasing by 1 |
| $x^2 + x + D = t$ | A1 | Correct equation, allow omission of $D$ |
| Complete the square: $(x+\frac{1}{2})^2 - \frac{1}{4} = t$ or use quadratic formula: $x = \frac{-1 \pm \sqrt{1+4t}}{2}$ | M1 | Complete the square or use quadratic formula |
| $x = \frac{1}{2}(\sqrt{(4t+1)}-1)$* | A1* | Given answer correctly obtained, with at least one line of working and justification of positive root e.g. $x > 0$ |
---\begin{enumerate}
\item A particle $P$ is moving along the $x$-axis. At time $t$ seconds, where $t \geqslant 0 , P$ is $x$ metres from the origin $O$ and is moving with speed $v \mathrm {~ms} ^ { - 1 }$
\end{enumerate}
The acceleration of $P$ has magnitude $\frac { 2 } { ( 2 x + 1 ) ^ { 3 } } \mathrm {~ms} ^ { - 2 }$ and is directed towards $O$\\
When $t = 0 , P$ passes through $O$ in the positive $x$ direction with speed $1 \mathrm {~ms} ^ { - 1 }$\\
(a) Find $v$ in terms of $x$\\
(b) Show that $x = \frac { 1 } { 2 } ( \sqrt { ( 4 t + 1 ) } - 1 )$
\hfill \mbox{\textit{Edexcel M3 2022 Q3 [8]}}