| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Challenging +1.2 This M3 question requires using v dv/dx = a to find acceleration, then solving for x, followed by separating variables to integrate and find v(t). While it involves multiple mechanics techniques and careful algebraic manipulation of fractional powers, the methods are standard for M3 and the steps are clearly signposted by the two-part structure. More challenging than average due to the algebraic complexity and chain of reasoning, but follows predictable M3 patterns. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a=v\frac{dv}{dx}\) | M1 | Use of \(a=v\frac{dv}{dx}\) or \(a=\frac{d}{dx}\left(\frac{1}{2}v^2\right)\). Evidence of differentiation, power decreasing by 1. Should see product of terms. |
| \(=\frac{3}{2}(2x+1)^{\frac{1}{2}}\times 2\times(2x+1)^{\frac{3}{2}}=3(2x+1)^2\) | A1 | Correct differentiation |
| \(3(2x+1)^2=243\) | M1 | Independent. Use result from differentiation and put \(a=243\) then solve for \(x\) |
| \(x=4\) | A1 | Cao. If \(-5\) is seen it must be rejected or \(4\) must be clearly identified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2x+1)^{\frac{3}{2}}=\frac{dx}{dt}\) OR \(a=3v^{\frac{4}{3}}=\frac{dv}{dt}\) | M1 A1 | Use of \(v=\frac{dx}{dt}\) to obtain DE in \(x\) and \(t\), OR use of \(a=\frac{dv}{dt}\) to obtain DE in \(v\) and \(t\) |
| \(\int dt=\int(2x+1)^{-\frac{3}{2}}dx\) OR \(\int 3\,dt=\int v^{-\frac{4}{3}}dv\) | M1 | Separate and integrate (evidence of integration, power increasing by 1) |
| \(t=-(2x+1)^{-\frac{1}{2}}(+C)\) OR \(3t+(C)=-3v^{-\frac{1}{3}}\) | A1 | Correct integration, condone missing \(C\) |
| \(t=0,\,x=0\Rightarrow C=1\) OR \(t=0,\,x=0\Rightarrow v=1\Rightarrow C=-3\) | M1 | Use \(t=0\), \(x=0\) to obtain value of \(C\) and obtain equation in \(v\) and \(t\) only |
| \(v=\frac{1}{(1-t)^3}\) | A1 | Cao. Accept \(v=(1-t)^{-3}\) or \(\frac{-1}{(t-1)^3}\) or \(-(t-1)^{-3}\) |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a=v\frac{dv}{dx}$ | M1 | Use of $a=v\frac{dv}{dx}$ or $a=\frac{d}{dx}\left(\frac{1}{2}v^2\right)$. Evidence of differentiation, power decreasing by 1. Should see product of terms. |
| $=\frac{3}{2}(2x+1)^{\frac{1}{2}}\times 2\times(2x+1)^{\frac{3}{2}}=3(2x+1)^2$ | A1 | Correct differentiation |
| $3(2x+1)^2=243$ | M1 | Independent. Use result from differentiation and put $a=243$ then solve for $x$ |
| $x=4$ | A1 | Cao. If $-5$ is seen it must be rejected or $4$ must be clearly identified |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2x+1)^{\frac{3}{2}}=\frac{dx}{dt}$ **OR** $a=3v^{\frac{4}{3}}=\frac{dv}{dt}$ | M1 A1 | Use of $v=\frac{dx}{dt}$ to obtain DE in $x$ and $t$, OR use of $a=\frac{dv}{dt}$ to obtain DE in $v$ and $t$ |
| $\int dt=\int(2x+1)^{-\frac{3}{2}}dx$ **OR** $\int 3\,dt=\int v^{-\frac{4}{3}}dv$ | M1 | Separate and integrate (evidence of integration, power increasing by 1) |
| $t=-(2x+1)^{-\frac{1}{2}}(+C)$ **OR** $3t+(C)=-3v^{-\frac{1}{3}}$ | A1 | Correct integration, condone missing $C$ |
| $t=0,\,x=0\Rightarrow C=1$ **OR** $t=0,\,x=0\Rightarrow v=1\Rightarrow C=-3$ | M1 | Use $t=0$, $x=0$ to obtain value of $C$ and obtain equation in $v$ and $t$ only |
| $v=\frac{1}{(1-t)^3}$ | A1 | Cao. Accept $v=(1-t)^{-3}$ or $\frac{-1}{(t-1)^3}$ or $-(t-1)^{-3}$ |
---\begin{enumerate}
\item In this question you must show all stages in your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
A particle $P$ is moving along the $x$-axis.\\
At time $t$ seconds, where $0 \leqslant t \leqslant \frac { 2 } { 3 } , P$ is $x$ metres from the origin 0 and is moving with velocity $\mathrm { v } \mathrm { m } \mathrm { s } ^ { - 1 }$ in the positive x direction where
$$v = ( 2 x + 1 ) ^ { \frac { 3 } { 2 } }$$
When $\mathrm { t } = 0 , \mathrm { P }$ passes through 0 .\\
(a) Find the value of x when the acceleration of P is $243 \mathrm {~m} \mathrm {~s} ^ { - 2 }$\\
(b) Find v in terms of t .
\hfill \mbox{\textit{Edexcel M3 2023 Q4 [10]}}