| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Forces as vectors |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question requiring basic vector operations: calculating three magnitudes using Pythagoras, identifying the largest, then applying F=ma with weight included. All steps are routine with no problem-solving insight needed, making it slightly easier than average A-level standard. |
| Spec | 1.10c Magnitude and direction: of vectors3.03d Newton's second law: 2D vectors3.03f Weight: W=mg |
| Answer | Marks |
|---|---|
| B1 | \(v=0\) when it arrives. \(150000(t - \frac{t^2}{4}) = 0\) \(\Rightarrow t = 4\) (on arrival) |
| Answer | Marks |
|---|---|
| M1 | Distance travelled \(s = \int v \, dt\). \(s = 150000(\frac{t^2}{2} - \frac{t^3}{12}) + c\) |
| A1 | When \(t = 4\), \(s = 400000\) |
| M1 | The journey is \(400000\) km (dependent on previous M mark) |
| A1 | If \(400000\) seen award the previous mark |
| Answer | Marks |
|---|---|
| B1 | For maximum speed \(a = \frac{dv}{dt} = 0\). \(\frac{dv}{dt} = 150000(1 - \frac{t}{2})\) \(\Rightarrow t = 2\) |
| B1 | \(v = 150000(2 - \frac{1}{4} \times 2^2) = 150000\). Maximum speed is \(150000\) kmh\(^{-1}\) |
# Question 1
## (i)
B1 | $v=0$ when it arrives. $150000(t - \frac{t^2}{4}) = 0$ $\Rightarrow t = 4$ (on arrival)
Note: Award this mark for substituting $t = 4$ to obtain $v = 0$. Condone omission of $t = 0$.
## (ii)
M1 | Distance travelled $s = \int v \, dt$. $s = 150000(\frac{t^2}{2} - \frac{t^3}{12}) + c$
A1 | When $t = 4$, $s = 400000$
M1 | The journey is $400000$ km (dependent on previous M mark)
A1 | If $400000$ seen award the previous mark
Note: Do not accept multiplication by $t$. Substituting their $t = 4$.
## (iii)
B1 | For maximum speed $a = \frac{dv}{dt} = 0$. $\frac{dv}{dt} = 150000(1 - \frac{t}{2})$ $\Rightarrow t = 2$
B1 | $v = 150000(2 - \frac{1}{4} \times 2^2) = 150000$. Maximum speed is $150000$ kmh$^{-1}$
Note: $t = 2$ seen. Accept a trial and error method. CAO1 In this question take $\boldsymbol { g } = \mathbf { 1 0 }$.\\
The directions of the unit vectors $\left( \begin{array} { l } 1 \\ 0 \\ 0 \end{array} \right) , \left( \begin{array} { l } 0 \\ 1 \\ 0 \end{array} \right)$ and $\left( \begin{array} { l } 0 \\ 0 \\ 1 \end{array} \right)$ are east, north and vertically upwards.\\
Forces $\mathbf { p } , \mathbf { q }$ and $\mathbf { r }$ are given by $\mathbf { p } = \left( \begin{array} { r } - 1 \\ - 1 \\ 5 \end{array} \right) \mathrm { N } , \mathbf { q } = \left( \begin{array} { r } - 1 \\ - 4 \\ 2 \end{array} \right) \mathrm { N }$ and $\mathbf { r } = \left( \begin{array} { l } 2 \\ 5 \\ 0 \end{array} \right) \mathrm { N }$.\\
(i) Find which of $\mathbf { p } , \mathbf { q }$ and $\mathbf { r }$ has the greatest magnitude.\\
(ii) A particle has mass 0.4 kg . The forces acting on it are $\mathbf { p } , \mathbf { q } , \mathbf { r }$ and its weight.
Find the magnitude of the particle's acceleration and describe the direction of this acceleration.
\hfill \mbox{\textit{OCR MEI M1 Q1 [6]}}