| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.3 This is a standard M1 triangle of forces problem requiring application of the cosine rule to find an unknown force magnitude, followed by a straightforward vector subtraction. While it involves two parts and algebraic manipulation of the cosine rule result, the method is routine and well-practiced in M1, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.10c Magnitude and direction: of vectors3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((X - 20\cos60°)^2 + (20\cos30°)^2 = (3X)^2\) | M1 A1 | M1 for complete method giving equation in \(X\) only; two components, squaring and adding, equating to \((3X)^2\); x-component must be a difference; A1 correct unsimplified equation in \(X\) only |
| \(8X^2 + 20X - 400 = 0\) | A1 | Correct fully numerical 3-term quadratic \(= 0\) |
| \(X = \dfrac{-5 \pm \sqrt{25+800}}{4} = 5.93\) (3 SF) | M1A1 | M1 independent, for solving 3-term quadratic; A1 for 5.93 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((3X)^2 = 20^2 + X^2 - 2 \cdot 20X\cos60°\) | M1A1 | M1 for cosine rule with \(\cos60°\) (M0 if \(\cos120°\) used) |
| \(8X^2 + 20X - 400 = 0\) | A1 | Correct 3-term quadratic |
| \(X = \dfrac{-5 \pm \sqrt{25+800}}{4} = 5.93\) (3SF) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \( | \mathbf{P}-\mathbf{Q} | ^2 = 20^2 + X^2 - 2X \times 20 \times \cos120°\) |
| \( | \mathbf{P}-\mathbf{Q} | = 23.5\) N (3SF) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \( | \mathbf{P}-\mathbf{Q} | ^2 = (X+20\cos60°)^2 + (20\cos30°)^2\) |
| \( | \mathbf{P}-\mathbf{Q} | = 23.5\) N (3SF) |
# Question 6:
## Part 6a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(X - 20\cos60°)^2 + (20\cos30°)^2 = (3X)^2$ | M1 A1 | M1 for complete method giving equation in $X$ only; two components, squaring and adding, equating to $(3X)^2$; x-component must be a difference; A1 correct unsimplified equation in $X$ only |
| $8X^2 + 20X - 400 = 0$ | A1 | Correct fully numerical 3-term quadratic $= 0$ |
| $X = \dfrac{-5 \pm \sqrt{25+800}}{4} = 5.93$ (3 SF) | M1A1 | M1 independent, for solving 3-term quadratic; A1 for 5.93 |
**Alternative (cosine rule):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(3X)^2 = 20^2 + X^2 - 2 \cdot 20X\cos60°$ | M1A1 | M1 for cosine rule with $\cos60°$ (M0 if $\cos120°$ used) |
| $8X^2 + 20X - 400 = 0$ | A1 | Correct 3-term quadratic |
| $X = \dfrac{-5 \pm \sqrt{25+800}}{4} = 5.93$ (3SF) | M1A1 | |
## Part 6b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $|\mathbf{P}-\mathbf{Q}|^2 = 20^2 + X^2 - 2X \times 20 \times \cos120°$ | M1A1 | M1 for cosine rule with $\cos120°$ (M0 if $\cos60°$); A1 correct expression for $|\mathbf{P}-\mathbf{Q}|$ in terms of $X$ |
| $|\mathbf{P}-\mathbf{Q}| = 23.5$ N (3SF) | DM1 A1 | DM1 dependent on first M1, substituting their $X$ and solving for $|\mathbf{P}-\mathbf{Q}|$ |
**Alternative (components):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $|\mathbf{P}-\mathbf{Q}|^2 = (X+20\cos60°)^2 + (20\cos30°)^2$ | M1A1 | M1 complete method, two components, squaring and adding; x-component must be a sum |
| $|\mathbf{P}-\mathbf{Q}| = 23.5$ N (3SF) | DM1 A1 | DM1 dependent on first M1 |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ed659098-c1cf-4ee1-a12a-bf8b6c42db95-11_472_908_285_520}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Two forces $\mathbf { P }$ and $\mathbf { Q }$ act on a particle at $O$. The angle between the lines of action of $\mathbf { P }$ and $\mathbf { Q }$ is $120 ^ { \circ }$ as shown in Figure 4. The force $\mathbf { P }$ has magnitude 20 N and the force $\mathbf { Q }$ has magnitude $X$ newtons. The resultant of $\mathbf { P }$ and $\mathbf { Q }$ is the force $\mathbf { R }$.
Given that the magnitude of $\mathbf { R }$ is $3 X$ newtons, find, giving your answers to 3 significant figures
\begin{enumerate}[label=(\alph*)]
\item the value of $X$,
\item the magnitude of $( \mathbf { P } - \mathbf { Q } )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q6 [9]}}