| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question requiring standard application of kinematic equations (finding acceleration from velocity change) and Newton's second law (F=ma), followed by basic vector magnitude and angle calculations. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the vector component work. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(a = \frac{\Delta v}{t} = \frac{1}{2}([4i - 7j] - (-2i + j)) = 3i - 4j\) | M1 A1 |
| (b) | \(F = ma = 5(3i - 4j)\) | M1 |
| mag. of \(F = 5\sqrt{3^2 + 4^2} = 25\) N | M1 A1 | |
| req'd angle \(= \tan^{-1}\frac{4}{3} = 37°\) to nearest degree | M1 A1 | (7) |
(a) | $a = \frac{\Delta v}{t} = \frac{1}{2}([4i - 7j] - (-2i + j)) = 3i - 4j$ | M1 A1 |
(b) | $F = ma = 5(3i - 4j)$ | M1 |
| mag. of $F = 5\sqrt{3^2 + 4^2} = 25$ N | M1 A1 |
| req'd angle $= \tan^{-1}\frac{4}{3} = 37°$ to nearest degree | M1 A1 | (7) |A constant force, $\mathbf{F}$, acts on a particle, $P$, of mass 5 kg causing its velocity to change from $(-2\mathbf{i} + \mathbf{j})$ m s$^{-1}$ to $(4\mathbf{i} - 7\mathbf{j})$ m s$^{-1}$ in 2 seconds.
\begin{enumerate}[label=(\alph*)]
\item Find, in the form $a\mathbf{i} + b\mathbf{j}$, the acceleration of $P$. [2 marks]
\item Show that the magnitude of $\mathbf{F}$ is 25 N and find, to the nearest degree, the acute angle between the line of action of $\mathbf{F}$ and the vector $\mathbf{j}$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q1 [7]}}