| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Force given resultant and one force |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question applying the sine and cosine rules to a force triangle. Students need to recognize the geometry (angles in the triangle are 150°, 50°, and therefore 180°-150°-50°=−20° which means the third angle is actually 30°+50°=80° on the other side), then apply sine rule twice. It's straightforward application of known techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(15\sin 30° = R\sin 50°\) giving \(R \approx 9.79 \text{ (N)}\) | M1 A1 DM1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \[\frac{R}{\sin 30°} = \frac{15}{\sin 50°} \text{ giving } R \approx 9.79 \text{ (N)}\] | M1 A1 DM1 A1 | (4) |
| (b) \((\rightarrow) X - 15\cos 30° = R\cos 50°\) giving \(X \approx 19.3 \text{ (N)}\) | M1 A2 f.t. DM1 A1 | (5) [9] |
| Answer | Marks | Guidance |
|---|---|---|
| \[\frac{X}{\sin 100°} = \frac{15}{\sin 50°} = \frac{R}{\sin 30°} \text{ giving } X \approx 19.3 \text{ (N)}\] | M1 A2 f.t. on \(R\) DM1 A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| giving \(X \approx 19.3 \text{ (N)}\) | M1 A2 f.t. on \(R\) DM1 A1 | (5) |
**(a)** $15\sin 30° = R\sin 50°$ giving $R \approx 9.79 \text{ (N)}$ | M1 A1 DM1 A1 | (4)
**Alternatives using sine rule in (a) or cosine rule in (b):**
$$\frac{R}{\sin 30°} = \frac{15}{\sin 50°} \text{ giving } R \approx 9.79 \text{ (N)}$$ | M1 A1 DM1 A1 | (4)
**(b)** $(\rightarrow) X - 15\cos 30° = R\cos 50°$ giving $X \approx 19.3 \text{ (N)}$ | M1 A2 f.t. DM1 A1 | (5) [9]
**Alternatives using sine rule in (a) or (b):**
$$\frac{X}{\sin 100°} = \frac{15}{\sin 50°} = \frac{R}{\sin 30°} \text{ giving } X \approx 19.3 \text{ (N)}$$ | M1 A2 f.t. on $R$ DM1 A1 | (5)
**OR: Cosine rule; any of:**
- $X^2 = R^2 + 15^2 - 2 \times 15 \times R\cos 100°$
- $R^2 = X^2 + 15^2 - 2 \times 15 \times X\cos 30°$
- $15^2 = R^2 + X^2 - 2xX \times R\cos 50°$
giving $X \approx 19.3 \text{ (N)}$ | M1 A2 f.t. on $R$ DM1 A1 | (5)
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9dbbbc01-fb66-460d-a42e-2c37ec8b451a-07_357_968_274_484}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Two forces $\mathbf { P }$ and $\mathbf { Q }$ act on a particle at a point $O$. The force $\mathbf { P }$ has magnitude 15 N and the force $\mathbf { Q }$ has magnitude $X$ newtons. The angle between $\mathbf { P }$ and $\mathbf { Q }$ is $150 ^ { \circ }$, as shown in Figure 1. The resultant of $\mathbf { P }$ and $\mathbf { Q }$ is $\mathbf { R }$.
Given that the angle between $\mathbf { R }$ and $\mathbf { Q }$ is $50 ^ { \circ }$, find
\begin{enumerate}[label=(\alph*)]
\item the magnitude of $\mathbf { R }$,
\item the value of $X$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2008 Q5 [9]}}