| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Standard +0.8 This M3 question requires using the chain rule form a = v(dv/dx) to separate variables and integrate, then applying initial conditions. While the integration of v dv = (3/4)√(x+1) dx is straightforward, recognizing the correct approach and manipulating the result to the required form requires solid understanding of variable acceleration. Part (b) adds another integration step. This is moderately challenging for M3 level but follows standard variable force techniques. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(v\frac{dv}{dx} = \frac{3\sqrt{x+1}}{4}\) | M1 | Set up differential equation in \(v\) and \(x\) only. M0 if acceleration is \(\frac{dv}{dx}\) or \(\frac{dv}{dt}\). M0 if no differential equation |
| \(\frac{1}{2}v^2 = \frac{1}{2}(x+1)^{\frac{3}{2}}\ (+C)\) | M1A1 | Clear attempt to separate variables and integrate. At least one power must increase by 1. Correct integration, condone missing \(+C\) |
| \(x=15, v=8 \Rightarrow C=0\) so \(v=(x+1)^{\frac{3}{4}}\) | A1* | (4) Given answer from complete correct working. Must include use of boundary conditions and initial DE. A0 if \(+C\) only considered after square root |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dx}{dt} = (x+1)^{\frac{3}{4}}\) | M1 | Set up DE in \(x\) and \(t\) only, using answer from (a) |
| \(4(x+1)^{\frac{1}{4}} = t\ (+C)\) | M1A1 | Clear attempt to separate variables and integrate. At least one power must increase by 1. Correct integration, condone missing \(+C\) |
| \(x=15, t=0 \Rightarrow C=8\) so \(4v^{\frac{1}{4}} = t+8\) | M1 | Use of boundary conditions in integrated equation and use of (a) to form equation in \(v\) and \(t\). M0 if boundary conditions not used |
| \(t = 4v^{\frac{1}{3}} - 8\) | A1 | (5) Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dv}{dt} = \frac{3}{4}v^{\frac{1}{3}}\) | M1 | Set up DE in \(v\) and \(t\) only |
| \(3v^{\frac{1}{3}} = \frac{3}{4}t\ (+C)\) | M1A1 | Clear attempt to separate variables and integrate. At least one power must increase by 1. Correct integration |
| \(t=0, v=8 \Rightarrow C=6\) so \(3v^{\frac{1}{3}} = \frac{3}{4}t + 6\) | M1 | Use of boundary conditions to form equation in \(v\) and \(t\) |
| \(t = 4v^{\frac{1}{3}} - 8\) | A1 | (5) Correct answer |
## Question 3:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $v\frac{dv}{dx} = \frac{3\sqrt{x+1}}{4}$ | M1 | Set up differential equation in $v$ and $x$ only. M0 if acceleration is $\frac{dv}{dx}$ or $\frac{dv}{dt}$. M0 if no differential equation |
| $\frac{1}{2}v^2 = \frac{1}{2}(x+1)^{\frac{3}{2}}\ (+C)$ | M1A1 | Clear attempt to separate variables and integrate. At least one power must increase by 1. Correct integration, condone missing $+C$ |
| $x=15, v=8 \Rightarrow C=0$ so $v=(x+1)^{\frac{3}{4}}$ | A1* | (4) Given answer from complete correct working. Must include use of boundary conditions and initial DE. A0 if $+C$ only considered after square root |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = (x+1)^{\frac{3}{4}}$ | M1 | Set up DE in $x$ and $t$ only, using answer from (a) |
| $4(x+1)^{\frac{1}{4}} = t\ (+C)$ | M1A1 | Clear attempt to separate variables and integrate. At least one power must increase by 1. Correct integration, condone missing $+C$ |
| $x=15, t=0 \Rightarrow C=8$ so $4v^{\frac{1}{4}} = t+8$ | M1 | Use of boundary conditions in integrated equation and use of (a) to form equation in $v$ and $t$. M0 if boundary conditions not used |
| $t = 4v^{\frac{1}{3}} - 8$ | A1 | (5) Correct answer |
**OR:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dv}{dt} = \frac{3}{4}v^{\frac{1}{3}}$ | M1 | Set up DE in $v$ and $t$ only |
| $3v^{\frac{1}{3}} = \frac{3}{4}t\ (+C)$ | M1A1 | Clear attempt to separate variables and integrate. At least one power must increase by 1. Correct integration |
| $t=0, v=8 \Rightarrow C=6$ so $3v^{\frac{1}{3}} = \frac{3}{4}t + 6$ | M1 | Use of boundary conditions to form equation in $v$ and $t$ |
| $t = 4v^{\frac{1}{3}} - 8$ | A1 | (5) Correct answer |
---\begin{enumerate}
\item A particle $P$ is moving along the $x$-axis.
\end{enumerate}
At time $t$ seconds, where $t \geqslant 0$, the displacement of $P$ from the origin $O$ is $x$ metres and $P$ is moving with velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$ direction.
The acceleration of $P$ is $\frac { 3 \sqrt { x + 1 } } { 4 } \mathrm {~ms} ^ { - 2 }$ in the positive $x$ direction.\\
When $t = 0 , x = 15$ and $v = 8$\\
(a) Show that $v = ( x + 1 ) ^ { \frac { 3 } { 4 } }$\\
(b) Find $t$ in terms of $v$.
\hfill \mbox{\textit{Edexcel M3 2024 Q3 [9]}}