| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Standard +0.8 This M3 question requires applying F=ma with position-dependent force, using the chain rule (a = v dv/dx), separating variables, and integrating. It also requires recognizing that maximum speed occurs when a=0 to find the constant. This is a standard M3 technique but requires multiple connected steps and careful algebraic manipulation, making it moderately harder than average A-level questions. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3v\frac{dv}{dx} = \frac{9}{2}(26-x)\) | M1 | NL2 with accel in correct form - can be implied by subsequent working |
| \(\frac{d(\frac{1}{2}v^2)}{dx} = \frac{3}{2}(26-x)\) | ||
| \(\frac{1}{2}v^2 = \frac{3}{2}\left(26x - \frac{1}{2}x^2\right) + c\) | M1A1 | M1 integrate equation wrt \(x\); A1 correct result after integrating - constant not needed |
| Max speed when accel is zero i.e. when \(x = 26\) | B1 | Deduce max speed occurs when \(x = 26\) |
| \(\frac{1}{2} \times 32^2 = \frac{3}{2} \times \frac{1}{2} \times 26^2 + c \Rightarrow c = 5\) | A1 | Correct value for the constant |
| \(v^2 = 3\left(26x - \frac{1}{2}x^2\right) + 10\) | A1 (6) | Correct expression for \(v^2\) - can be in any form |
# Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3v\frac{dv}{dx} = \frac{9}{2}(26-x)$ | M1 | NL2 with accel in correct form - can be implied by subsequent working |
| $\frac{d(\frac{1}{2}v^2)}{dx} = \frac{3}{2}(26-x)$ | | |
| $\frac{1}{2}v^2 = \frac{3}{2}\left(26x - \frac{1}{2}x^2\right) + c$ | M1A1 | M1 integrate equation wrt $x$; A1 correct result after integrating - constant not needed |
| Max speed when accel is zero i.e. when $x = 26$ | B1 | Deduce max speed occurs when $x = 26$ |
| $\frac{1}{2} \times 32^2 = \frac{3}{2} \times \frac{1}{2} \times 26^2 + c \Rightarrow c = 5$ | A1 | Correct value for the constant |
| $v^2 = 3\left(26x - \frac{1}{2}x^2\right) + 10$ | A1 (6) | Correct expression for $v^2$ - can be in any form |
**ALT for last 3 marks:** M1 (B1 on e-pen) Complete square and equate constant part to $32^2$ or use max of quadratic $= \frac{4ac-b^2}{4a}$; A1 correct $c$; A1 correct expression for $v^2$
---\begin{enumerate}
\item A particle $P$ of mass 3 kg is moving along the horizontal $x$-axis. At time $t = 0 , P$ passes through the origin $O$ moving in the positive $x$ direction. At time $t$ seconds, $O P = x$ metres and the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t$ seconds, the resultant force acting on $P$ is $\frac { 9 } { 2 } ( 26 - x ) \mathrm { N }$, measured in the positive $x$ direction. For $t > 0$ the maximum speed of $P$ is $32 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
Find $v ^ { 2 }$ in terms of $x$.\\
\hfill \mbox{\textit{Edexcel M3 2015 Q1 [6]}}