| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.8 This M3 question requires applying F=ma with v as a function of x (not t), necessitating the chain rule a = v(dv/dx). Part (a) involves careful differentiation and substitution. Part (b) requires separating variables and integrating a rational function. While systematic, it demands strong calculus skills and understanding of variable force mechanics beyond standard M1/M2 content. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = \frac{12}{x+3}\) | — | Given |
| \(\frac{dv}{dx} = -\frac{12}{(x+3)^2}\) | M1 | Attempt differentiation of \(v = \frac{12}{x+3}\) or \(\frac{1}{2}v^2 = \frac{72}{(x+3)^2}\) wrt \(x\); \((x+3)^{-2}\) or \((x+3)^{-3}\) (oe). Both sides must be differentiated wrt \(x\) |
| \(F = 0.5v\frac{dv}{dx} = 0.5 \times \frac{12}{x+3} \times -\frac{12}{(x+3)^2}\) | DM1 | Use NL2 with accel \(v\frac{dv}{dx}\) as obtained. Must include mass. Dependent on first M |
| A1 | Correct expression for \(F\) with correct mass and correct acceleration | |
| \(x=3\): \( | F | = 0.5 \times \frac{12}{6} \times \frac{12}{6^2} = \frac{1}{3}\) N |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int(x+3)dx = \int 12\, dt\) | M1 | Use \(v = \frac{dx}{dt}\) and attempt the integration |
| \(\frac{1}{2}x^2 + 3x = 12t\ (+c)\) | A1 | Correct integration; constant of integration not needed |
| \(x=4,\ t=2\): \(8+12=24+c\), \(c=-4\) | DM1 | Use given values to obtain value for \(c\). Dependent on first M |
| \(x=10\): \(50+30=12t-4\) | DM1 | Use \(x=10\) to obtain linear equation for \(t\). Dependent on first but not second M |
| \(t=7\) | A1cao (5) | cao \(t=7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_2^T 12\,dt = \int_4^{10}(x+3)\,dx\) | M1A1 | As main scheme - limits not needed |
| \(12(T-2) = \left[\frac{x^2}{2}+3x\right]_4^{10}\) | DM1 | Correct limits shown |
| \(12(T-2)=80-20\), \(T=7\) | DM1, A1 | Substitute limits; \(T=7\) |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = \frac{12}{x+3}$ | — | Given |
| $\frac{dv}{dx} = -\frac{12}{(x+3)^2}$ | M1 | Attempt differentiation of $v = \frac{12}{x+3}$ or $\frac{1}{2}v^2 = \frac{72}{(x+3)^2}$ wrt $x$; $(x+3)^{-2}$ or $(x+3)^{-3}$ (oe). Both sides must be differentiated wrt $x$ |
| $F = 0.5v\frac{dv}{dx} = 0.5 \times \frac{12}{x+3} \times -\frac{12}{(x+3)^2}$ | DM1 | Use NL2 with accel $v\frac{dv}{dx}$ as obtained. Must include mass. Dependent on first M |
| | A1 | Correct expression for $F$ with correct mass and correct acceleration |
| $x=3$: $|F| = 0.5 \times \frac{12}{6} \times \frac{12}{6^2} = \frac{1}{3}$ N | A1 (4) | Use $x=3$ to obtain correct magnitude $\frac{1}{3}$, 0.33 or better. Must be positive |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int(x+3)dx = \int 12\, dt$ | M1 | Use $v = \frac{dx}{dt}$ and attempt the integration |
| $\frac{1}{2}x^2 + 3x = 12t\ (+c)$ | A1 | Correct integration; constant of integration not needed |
| $x=4,\ t=2$: $8+12=24+c$, $c=-4$ | DM1 | Use given values to obtain value for $c$. Dependent on first M |
| $x=10$: $50+30=12t-4$ | DM1 | Use $x=10$ to obtain linear equation for $t$. Dependent on first but **not** second M |
| $t=7$ | A1cao (5) | cao $t=7$ |
**ALT (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_2^T 12\,dt = \int_4^{10}(x+3)\,dx$ | M1A1 | As main scheme - limits not needed |
| $12(T-2) = \left[\frac{x^2}{2}+3x\right]_4^{10}$ | DM1 | Correct limits shown |
| $12(T-2)=80-20$, $T=7$ | DM1, A1 | Substitute limits; $T=7$ |
---\begin{enumerate}
\item A particle $P$ of mass 0.5 kg is moving along the positive $x$-axis under the action of a resultant force. The force acts along the $x$-axis. At time $t$ seconds, $P$ is $x$ metres from the origin $O$ and is moving away from $O$ in the positive $x$ direction with speed $\frac { 12 } { x + 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(a) Find the magnitude of the force acting on $P$ when $x = 3$
\end{enumerate}
Given that $x = 4$ when $t = 2$\\
(b) find the value of $t$ when $x = 10$\\
\hfill \mbox{\textit{Edexcel M3 2016 Q1 [9]}}