| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Velocity from acceleration and initial conditions |
| Difficulty | Moderate -0.3 This is a standard A-level mechanics question involving differentiation of position vectors to find velocity and acceleration. Part (a) uses bearing geometry (routine), part (b) requires differentiation and magnitude calculation (standard), and part (c) involves setting acceleration parallel to a given vector (slightly more challenging but still a common technique). The question is slightly easier than average due to straightforward differentiation and clear step-by-step structure. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2c\mathbf{i} - 6\mathbf{j}\) or as column vector, seen or implied | B1 | 1.1b |
| ALT 1: Diagram showing right-angled triangle with sides \(2c\) and \(6\) at \(45°\) AND either \(\tan 45° = \frac{2c}{6} \Rightarrow 2c=6\) or states isosceles triangle so \(2c=6\) | M1 | 3.1a - N.B. Must see justification for equation. N.B. M0 if using wrong bearing |
| ALT 2: \(\tan 135° = \frac{2c}{-6} \Rightarrow 2c=6\) | M1 | 3.1a - N.B. M0 if using wrong bearing |
| \(c = 3\) * | A1* | 2.2a |
| SC1 M1A0: \(2c\mathbf{i}-6\mathbf{j} = k(\mathbf{i}-\mathbf{j}) \Rightarrow 2c=6\) or \(\mathbf{i}\)-cpt \(= -\mathbf{j}\)-cpt \(\Rightarrow 2c=6\) (must see justification) | ||
| SC2 M1A0: \(\tan 45° = \frac{2c}{6}\) or \(\frac{6}{2c} \Rightarrow 2c=6\) (must see justification) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(ct^{\frac{1}{2}} = 2c\) and \((-)\ \frac{3t^2}{8} = (-)6\) when \(t=4\), seen or implied | B1 | |
| ALT 3: Diagram with \(ct^{\frac{1}{2}}\) and \(\frac{3t^2}{8}\) at \(45°\) AND either \(\tan 45° = \frac{\frac{3t^2}{8}}{ct^{\frac{1}{2}}} \Rightarrow 2c=6\) when \(t=4\) or states isosceles triangle so \(ct^{\frac{1}{2}} = \frac{3t^2}{8} \Rightarrow 2c=6\) when \(t=4\) | M1 | N.B. M0 if using wrong bearing |
| \(c=3\) | A1* | |
| SC3 M1A0: \((ct^{\frac{1}{2}}\mathbf{i} - \frac{3t^2}{8}\mathbf{j}) = k(\mathbf{i}-\mathbf{j}) \Rightarrow 2c=6\) when \(t=4\) or \(\mathbf{i}\)-cpt \(= -\mathbf{j}\)-cpt \(\Rightarrow ct^{\frac{1}{2}} = \frac{3t^2}{8} \Rightarrow 2c=6\) when \(t=4\) (must see justification) | ||
| SC4 M1A0: \(\tan 45° = \frac{\frac{3t^2}{8}}{ct^{\frac{1}{2}}} \Rightarrow 2c=6\) when \(t=4\) (must see justification) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2c\mathbf{i} - 6\mathbf{j}\) seen or implied | B1 | B0 for \(r = 2c - 6\) if no evidence of components |
| Use bearing to obtain correct diagram showing right-angled triangle with at least one 45° angle marked; ALT 2: Use \(\tan 135° = \frac{2c}{-6} \Rightarrow 2c = 6\) | M1 | Must see justification for equation |
| bearing \(= 45° + 90° = 135°\) | A1* | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate \(\mathbf{r}\) wrt \(t\) to obtain \(\mathbf{v}\) | M1 | Both powers of \(t\) decreasing by 1; M0 if \(\mathbf{i}\) or \(\mathbf{j}\) missing |
| \(\mathbf{v} = \frac{3}{2}t^{-\frac{1}{2}}\mathbf{i} - \frac{3}{4}t\mathbf{j}\) | A1 | Correct unsimplified derivative or two correct components |
| Put \(t=4\) into both components and use Pythagoras: \(\sqrt{\left(\frac{3}{4}\right)^2 + (-3)^2}\) | M1 | Must use attempted derivative of \(\mathbf{r}\); allow missing \(-\) sign |
| \(\sqrt{\frac{153}{16}}\) or \(\frac{\sqrt{153}}{4}\) or \(\frac{3\sqrt{17}}{4}\) or \(3\sqrt{\frac{17}{16}} = 3.0923\ldots\) (m s\(^{-1}\)) | A1 | Accept 3.1 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate \(\mathbf{v}\) wrt \(t\) to obtain \(\mathbf{a}\) | M1 | Both powers of \(t\) decreasing by 1; M0 if \(\mathbf{i}\) or \(\mathbf{j}\) missing |
| \(\mathbf{a} = -\frac{3}{4}t^{-\frac{3}{2}}\mathbf{i} - \frac{3}{4}\mathbf{j}\) | A1 | Correct unsimplified derivative |
| \(\dfrac{-\frac{3}{4}T^{-\frac{1}{2}}}{-\frac{3}{4}} = \frac{-1}{-27}\) | M1 | Use of appropriate ratio using attempted derivative of \(\mathbf{v}\); condone sign error and reciprocal |
| \((T =)\ 9\) | A1 | cao (allow \(t\) instead of \(T\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using horizontal motion \(s = ut\), with 35 resolved: \(x = 35\cos\alpha \times t\) | M1, A1 | Correct terms, condone sin/cos confusion and sign errors |
| Using vertical motion \(s = ut + \frac{1}{2}at^2\), with 35 resolved: \(y = 35\sin\alpha \times t - \frac{1}{2}gt^2\) | M1, A1 | Available if different letter used for \(y\) provided not same as \(x\) |
| Eliminate \(t\): \(y = 35\sin\alpha \times \frac{x}{35\cos\alpha} - \frac{1}{2}g\left(\frac{x}{35\cos\alpha}\right)^2\) | DM1 | Dependent on both previous M marks |
| \(y = \frac{3}{4}x - \frac{1}{160}x^2\) | A1* | Given answer; must show at least one further line with trig ratios and \(g = 9.8\) explicitly seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| ALT 1: \(0 = \frac{3}{4}x - \frac{1}{160}x^2\) and solve for \(x\) | M1 | ALT 2: \(\frac{dy}{dx} = \frac{3}{4} - \frac{x}{80} = 0 \Rightarrow x = 60\) and \(OA = 2 \times 60\); ALT 3: complete suvat method |
| \((OA =)\ 120\) (m) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| ALT 1: \(H = \frac{3}{4}\times 60 - \frac{1}{160}\times 60^2\) | M1 | ALT 2: complete square; ALT 3: \(\frac{dy}{dx}=0 \Rightarrow x=60\); ALT 4: suvat \(H = \frac{(35\sin\alpha)^2}{2g}\) |
| \((H =)\ 22.5\) | A1 | Accept 23 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H\) is greater (or \(K\) is smaller), as air resistance would slow the particle down | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. the inaccuracy of using \(9.8\ \text{m s}^{-2}\) for \(g\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Justification for \(H\) greater (or \(K\) smaller): air resistance will provide an extra force acting against the stone / opposing the motion | B1 | Also accept: air resistance would take away energy from the stone; work would be done against air resistance; air resistance will mean the acceleration will be less (or deceleration greater); air resistance would reduce the velocity/speed. Allow 'it' for air resistance. B0 if no justification given. B0 for air resistance would reduce the initial velocity. B0 for air resistance will limit the vertical height. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Any single correct answer e.g. the inaccuracy of using \(9.8\ \text{ms}^{-2}\) for \(g\); wind or weather effects; the spin of the stone; the size (or shape or surface area) of the stone; the stone is still modelled as a particle | B1 | Allow if stone referred to as e.g. a ball. B0 if any incorrect extras. Unacceptable: mass or weight of stone; air resistance; ground may not be horizontal. B0 for consequences of air resistance e.g. path won't be a parabola / path won't be symmetrical. |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2c\mathbf{i} - 6\mathbf{j}$ or as column vector, seen or implied | B1 | 1.1b |
| **ALT 1:** Diagram showing right-angled triangle with sides $2c$ and $6$ at $45°$ **AND either** $\tan 45° = \frac{2c}{6} \Rightarrow 2c=6$ **or states** isosceles triangle so $2c=6$ | M1 | 3.1a - N.B. Must see justification for equation. N.B. M0 if using wrong bearing |
| **ALT 2:** $\tan 135° = \frac{2c}{-6} \Rightarrow 2c=6$ | M1 | 3.1a - N.B. M0 if using wrong bearing |
| $c = 3$ * | A1* | 2.2a |
| **SC1 M1A0:** $2c\mathbf{i}-6\mathbf{j} = k(\mathbf{i}-\mathbf{j}) \Rightarrow 2c=6$ **or** $\mathbf{i}$-cpt $= -\mathbf{j}$-cpt $\Rightarrow 2c=6$ (must see justification) | | |
| **SC2 M1A0:** $\tan 45° = \frac{2c}{6}$ or $\frac{6}{2c} \Rightarrow 2c=6$ (must see justification) | | |
**When $t=4$ substituted at end:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $ct^{\frac{1}{2}} = 2c$ **and** $(-)\ \frac{3t^2}{8} = (-)6$ when $t=4$, seen or implied | B1 | |
| **ALT 3:** Diagram with $ct^{\frac{1}{2}}$ and $\frac{3t^2}{8}$ at $45°$ **AND either** $\tan 45° = \frac{\frac{3t^2}{8}}{ct^{\frac{1}{2}}} \Rightarrow 2c=6$ when $t=4$ **or states** isosceles triangle so $ct^{\frac{1}{2}} = \frac{3t^2}{8} \Rightarrow 2c=6$ when $t=4$ | M1 | N.B. M0 if using wrong bearing |
| $c=3$ | A1* | |
| **SC3 M1A0:** $(ct^{\frac{1}{2}}\mathbf{i} - \frac{3t^2}{8}\mathbf{j}) = k(\mathbf{i}-\mathbf{j}) \Rightarrow 2c=6$ when $t=4$ **or** $\mathbf{i}$-cpt $= -\mathbf{j}$-cpt $\Rightarrow ct^{\frac{1}{2}} = \frac{3t^2}{8} \Rightarrow 2c=6$ when $t=4$ (must see justification) | | |
| **SC4 M1A0:** $\tan 45° = \frac{\frac{3t^2}{8}}{ct^{\frac{1}{2}}} \Rightarrow 2c=6$ when $t=4$ (must see justification) | | |
## Question 4a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2c\mathbf{i} - 6\mathbf{j}$ seen or implied | B1 | B0 for $r = 2c - 6$ if no evidence of components |
| Use bearing to obtain correct diagram showing right-angled triangle with at least one 45° angle marked; **ALT 2:** Use $\tan 135° = \frac{2c}{-6} \Rightarrow 2c = 6$ | M1 | Must see justification for equation |
| bearing $= 45° + 90° = 135°$ | A1* | Given answer correctly obtained |
---
## Question 4b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate $\mathbf{r}$ wrt $t$ to obtain $\mathbf{v}$ | M1 | Both powers of $t$ decreasing by 1; M0 if $\mathbf{i}$ or $\mathbf{j}$ missing |
| $\mathbf{v} = \frac{3}{2}t^{-\frac{1}{2}}\mathbf{i} - \frac{3}{4}t\mathbf{j}$ | A1 | Correct unsimplified derivative or two correct components |
| Put $t=4$ into both components and use Pythagoras: $\sqrt{\left(\frac{3}{4}\right)^2 + (-3)^2}$ | M1 | Must use attempted derivative of $\mathbf{r}$; allow missing $-$ sign |
| $\sqrt{\frac{153}{16}}$ or $\frac{\sqrt{153}}{4}$ or $\frac{3\sqrt{17}}{4}$ or $3\sqrt{\frac{17}{16}} = 3.0923\ldots$ (m s$^{-1}$) | A1 | Accept 3.1 or better |
---
## Question 4c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate $\mathbf{v}$ wrt $t$ to obtain $\mathbf{a}$ | M1 | Both powers of $t$ decreasing by 1; M0 if $\mathbf{i}$ or $\mathbf{j}$ missing |
| $\mathbf{a} = -\frac{3}{4}t^{-\frac{3}{2}}\mathbf{i} - \frac{3}{4}\mathbf{j}$ | A1 | Correct unsimplified derivative |
| $\dfrac{-\frac{3}{4}T^{-\frac{1}{2}}}{-\frac{3}{4}} = \frac{-1}{-27}$ | M1 | Use of appropriate ratio using attempted derivative of $\mathbf{v}$; condone sign error and reciprocal |
| $(T =)\ 9$ | A1 | cao (allow $t$ instead of $T$) |
---
## Question 5a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using horizontal motion $s = ut$, with 35 resolved: $x = 35\cos\alpha \times t$ | M1, A1 | Correct terms, condone sin/cos confusion and sign errors |
| Using vertical motion $s = ut + \frac{1}{2}at^2$, with 35 resolved: $y = 35\sin\alpha \times t - \frac{1}{2}gt^2$ | M1, A1 | Available if different letter used for $y$ provided not same as $x$ |
| Eliminate $t$: $y = 35\sin\alpha \times \frac{x}{35\cos\alpha} - \frac{1}{2}g\left(\frac{x}{35\cos\alpha}\right)^2$ | DM1 | Dependent on both previous M marks |
| $y = \frac{3}{4}x - \frac{1}{160}x^2$ | A1* | Given answer; must show at least one further line with trig ratios and $g = 9.8$ explicitly seen |
---
## Question 5b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **ALT 1:** $0 = \frac{3}{4}x - \frac{1}{160}x^2$ and solve for $x$ | M1 | **ALT 2:** $\frac{dy}{dx} = \frac{3}{4} - \frac{x}{80} = 0 \Rightarrow x = 60$ and $OA = 2 \times 60$; **ALT 3:** complete suvat method |
| $(OA =)\ 120$ (m) | A1 | cao |
---
## Question 5c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **ALT 1:** $H = \frac{3}{4}\times 60 - \frac{1}{160}\times 60^2$ | M1 | **ALT 2:** complete square; **ALT 3:** $\frac{dy}{dx}=0 \Rightarrow x=60$; **ALT 4:** suvat $H = \frac{(35\sin\alpha)^2}{2g}$ |
| $(H =)\ 22.5$ | A1 | Accept 23 |
---
## Question 5d:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H$ is greater (or $K$ is smaller), as air resistance would slow the particle down | B1 | oe |
---
## Question 5e:
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. the inaccuracy of using $9.8\ \text{m s}^{-2}$ for $g$ | B1 | |
## Question 5d:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Justification for $H$ greater (or $K$ smaller): air resistance will provide an extra force acting against the stone / opposing the motion | B1 | Also accept: air resistance would take away energy from the stone; work would be done against air resistance; air resistance will mean the acceleration will be less (or deceleration greater); air resistance would reduce the velocity/speed. Allow 'it' for air resistance. B0 if no justification given. B0 for air resistance would reduce the **initial** velocity. B0 for air resistance will limit the vertical height. |
---
## Question 5e:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Any single correct answer e.g. the inaccuracy of using $9.8\ \text{ms}^{-2}$ for $g$; wind or weather effects; the spin of the stone; the size (or shape or surface area) of the stone; the stone is still modelled as a particle | B1 | Allow if stone referred to as e.g. a ball. B0 if any incorrect extras. Unacceptable: mass or weight of stone; air resistance; ground may not be horizontal. B0 for consequences of air resistance e.g. path won't be a parabola / path won't be symmetrical. |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
[In this question, $\mathbf { i }$ is a unit vector due east and $\mathbf { j }$ is a unit vector due north.\\
Position vectors are given relative to a fixed origin $O$.]
At time $t$ seconds, $t \geqslant 1$, the position vector of a particle $P$ is $\mathbf { r }$ metres, where
$$\mathbf { r } = c t ^ { \frac { 1 } { 2 } } \mathbf { i } - \frac { 3 } { 8 } t ^ { 2 } \mathbf { j }$$
and $c$ is a constant.\\
When $t = 4$, the bearing of $P$ from $O$ is $135 ^ { \circ }$\\
(a) Show that $c = 3$\\
(b) Find the speed of $P$ when $t = 4$
When $t = T , P$ is accelerating in the direction of ( $\mathbf { - i } - \mathbf { 2 7 j }$ ).\\
(c) Find the value of $T$.
\hfill \mbox{\textit{Edexcel Paper 3 2024 Q4 [11]}}