Standard +0.3 This question requires students to understand that a bearing of 045° means equal i and j components, set up the equation 16-t²=31-8t, solve the resulting quadratic, and calculate speed using Pythagoras. It's slightly above average due to the vector bearing interpretation and quadratic solving, but follows a standard mechanics template with no novel insight required.
2 The directions of the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are east and north.
The velocity of a particle, \(\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }\), at time \(t \mathrm {~s}\) is given by
$$\mathbf { v } = \left( 16 - t ^ { 2 } \right) \mathbf { i } + ( 31 - 8 t ) \mathbf { j }$$
Find the time at which the particle is travelling on a bearing of \(045 ^ { \circ }\) and the speed of the particle at this time. [0pt]
[6]
Relevant attempt at Newton's 2nd Law. Total force must be expressed as a vector. Allow weight missing, wrong component or wrong sign. Condone \(mg\) in place of \(m\) for this mark only.
Magnitude of acceleration is \(7.5\) m s\(^{-2}\)
B1
CAO apart from using \(g=9.8 \Rightarrow a=7.7\)
Direction is vertically upwards
B1
## Question 2:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{p}: \sqrt{(-1)^2+(-1)^2+5^2}=\sqrt{27}$ | M1 | Use of Pythagoras |
| $\mathbf{q}: \sqrt{(-1)^2+(-4)^2+2^2}=\sqrt{21}$ | | Note magnitudes are 5.196, 4.583 and 5.385 respectively |
| $\mathbf{r}: \sqrt{2^2+5^2+0^2}=\sqrt{29}$ | | |
| Greatest magnitude: $\mathbf{r}$ | A1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Weight} = \begin{pmatrix} 0 \\ 0 \\ -4 \end{pmatrix}$ | B1 | Condone $g=9.8$ giving weight $\begin{pmatrix}0\\0\\-3.92\end{pmatrix}$ N. Accept $4\downarrow$ |
| $\mathbf{p}+\mathbf{q}+\mathbf{r}+\text{weight}=\begin{pmatrix}0\\0\\3\end{pmatrix}$ | | $g=9.8$ gives $\begin{pmatrix}0\\0\\3.08\end{pmatrix}$ |
| $0.4\mathbf{a}=\begin{pmatrix}0\\0\\3\end{pmatrix}$ | B1 | Relevant attempt at Newton's 2nd Law. Total force must be expressed as a vector. Allow weight missing, wrong component or wrong sign. Condone $mg$ in place of $m$ for this mark only. |
| Magnitude of acceleration is $7.5$ m s$^{-2}$ | B1 | CAO apart from using $g=9.8 \Rightarrow a=7.7$ |
| Direction is vertically upwards | B1 | |
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2 The directions of the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are east and north.\\
The velocity of a particle, $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, at time $t \mathrm {~s}$ is given by
$$\mathbf { v } = \left( 16 - t ^ { 2 } \right) \mathbf { i } + ( 31 - 8 t ) \mathbf { j }$$
Find the time at which the particle is travelling on a bearing of $045 ^ { \circ }$ and the speed of the particle at this time.\\[0pt]
[6]
\hfill \mbox{\textit{OCR MEI M1 Q2 [6]}}