| Answer | Marks | Guidance |
|---|---|---|
| \(25\text{ N}\) | B1 | Condone no units. Do not accept \(-25\text{ N}\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(50\cos25 = 45.31538\ldots \approx 45.3\text{ N}\) | M1, A1 | Attempt to resolve \(50\text{ N}\). Accept \(\sin \leftrightarrow \cos\). No extra forces. cao but accept \(-45.3\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 57.26908\ldots \approx 57.3\text{ N}\) | M1, A1, A1 | All relevant forces with resolution of \(50\text{ N}\). No extras. Accept \(\sin \leftrightarrow \cos\). All correct. |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 1.4064105\ldots \approx 1.41\text{ m s}^{-2}\) | M1, A1, A1 | N2L with \(m=18\). Accept \(F=mga\). Attempt at resolving \(50\text{ N}\). Allow \(20\text{ N}\) omitted and \(\sin\leftrightarrow\cos\). No extra forces. Allow only sign error and \(\sin\leftrightarrow\cos\). cao. |
| Answer | Marks | Guidance |
|---|---|---|
| Resolution of weight down the slope | B1 | \(mg\sin5°\) where \(m = 8\) or \(10\) or \(18\), wherever first seen. |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = -3.888\ldots \approx -3.89\text{ N}\), the force is a thrust | M1, A1, M1, F1, A1 | \(F=ma\). Must have \(20\text{ N}\) and \(m=18\). Allow weight not resolved, accept \(\sin\leftrightarrow\cos\) and sign errors. cao. \(F=ma\) must consider motion of either C or D including weight component, resistance and \(T\). No extra forces. Condone sign errors and \(\sin\leftrightarrow\cos\). FT only applies to \(a\) and if direction is consistent. \('+T'\) if \(T\) taken as thrust, \('-T'\) if \(T\) taken as thrust. If \(T\) taken as thrust, \(T = +3.89\). Dependent on \(T\) correct. |
| Answer | Marks | Guidance |
|---|---|---|
| The force is a thrust | M1, M1, A1, A1, F1, A1 | \(F=ma\) for C including weight component, resistance and \(T\). \(F=ma\) for D including weight component, resistance and \(T\). No extra forces. Condone sign errors and \(\sin\leftrightarrow\cos\). Award for either C or D equation correct (\('-T'\) if \(T\) taken as thrust). First of \(a\) and \(T\) found is correct. If \(T\) taken as thrust, \(T=+3.89\). The second of \(a\) and \(T\) found is FT. Dependent on \(T\) correct. |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 2.4860303\ldots \approx 2.49\text{ m s}^{-1}\) | M1, F1 | Allow sign of \(a\) not followed. FT their value of \(a\). Allow change to correct sign of \(a\) at this stage. FT from magnitude of their \(a\) but must be consistent with its direction. |
## Question 2:
### Part (i)
$25\text{ N}$ | B1 | Condone no units. Do not accept $-25\text{ N}$.
### Part (ii)
$50\cos25 = 45.31538\ldots \approx 45.3\text{ N}$ | M1, A1 | Attempt to resolve $50\text{ N}$. Accept $\sin \leftrightarrow \cos$. No extra forces. cao but accept $-45.3$.
### Part (iii)
Resolving vertically: $R + 50\sin25 - 8 \times 9.8 = 0$
$R = 57.26908\ldots \approx 57.3\text{ N}$ | M1, A1, A1 | All relevant forces with resolution of $50\text{ N}$. No extras. Accept $\sin \leftrightarrow \cos$. All correct.
### Part (iv)
Newton's 2nd Law in direction DC:
$50\cos25 - 20 = 18a$
$a = 1.4064105\ldots \approx 1.41\text{ m s}^{-2}$ | M1, A1, A1 | N2L with $m=18$. Accept $F=mga$. Attempt at resolving $50\text{ N}$. Allow $20\text{ N}$ omitted and $\sin\leftrightarrow\cos$. No extra forces. Allow only sign error and $\sin\leftrightarrow\cos$. cao.
### Part (v)
Resolution of weight down the slope | B1 | $mg\sin5°$ where $m = 8$ or $10$ or $18$, wherever first seen.
**Either method:**
N2L down slope overall:
$18 \times 9.8 \times \sin5 - 20 = 18a$
$a = -0.2569\ldots$
N2L down slope for D (taking rod force as tension $T$):
$10 \times 9.8 \times \sin5 - 15 - T = 10a$
(For C: $8 \times 9.8 \times \sin5 - 5 + T = 8a$)
$T = -3.888\ldots \approx -3.89\text{ N}$, the force is a thrust | M1, A1, M1, F1, A1 | $F=ma$. Must have $20\text{ N}$ and $m=18$. Allow weight not resolved, accept $\sin\leftrightarrow\cos$ and sign errors. cao. $F=ma$ must consider motion of either C or D including weight component, resistance and $T$. No extra forces. Condone sign errors and $\sin\leftrightarrow\cos$. FT only applies to $a$ and if direction is consistent. $'+T'$ if $T$ taken as thrust, $'-T'$ if $T$ taken as thrust. If $T$ taken as thrust, $T = +3.89$. Dependent on $T$ correct.
**Or method:**
For C: $8\times9.8\times\sin5 - 5 + T = 8a$
For D: $10\times9.8\times\sin5 - 15 - T = 10a$
$a = -0.2569\ldots$, $T = -3.888\ldots \approx -3.89\text{ N}$
The force is a thrust | M1, M1, A1, A1, F1, A1 | $F=ma$ for C including weight component, resistance and $T$. $F=ma$ for D including weight component, resistance and $T$. No extra forces. Condone sign errors and $\sin\leftrightarrow\cos$. Award for either C or D equation correct ($'-T'$ if $T$ taken as thrust). First of $a$ and $T$ found is correct. If $T$ taken as thrust, $T=+3.89$. The second of $a$ and $T$ found is FT. Dependent on $T$ correct.
**Then:**
After $2\text{ s}$: $v = 3 + 2 \times a$
$v = 2.4860303\ldots \approx 2.49\text{ m s}^{-1}$ | M1, F1 | Allow sign of $a$ not followed. FT their value of $a$. Allow change to correct sign of $a$ at this stage. FT from magnitude of **their** $a$ but must be consistent with its direction.
---2 In this question, positions are given relative to a fixed origin, O. The $x$-direction is east and the $y$-direction north; distances are measured in kilometres.
Two boats, the Rosemary and the Sage, are having a race between two points A and B.\\
The position vector of the Rosemary at time $t$ hours after the start is given by
$$\mathbf { r } = \binom { 3 } { 2 } + \binom { 6 } { 8 } t , \text { where } 0 \leqslant t \leqslant 2 .$$
The Rosemary is at point A when $t = 0$, and at point B when $t = 2$.\\
(i) Find the distance AB .\\
(ii) Show that the Rosemary travels at constant velocity.
The position vector of the Sage is given by
$$\mathbf { r } = \binom { 3 ( 2 t + 1 ) } { 2 \left( 2 t ^ { 2 } + 1 \right) }$$
(iii) Plot the points A and B .
Draw the paths of the two boats for $0 \leqslant t \leqslant 2$.\\
(iv) What can you say about the result of the race?\\
(v) Find the speed of the Sage when $t = 2$. Find also the direction in which it is travelling, giving your answer as a compass bearing, to the nearest degree.\\
(vi) Find the displacement of the Rosemary from the Sage at time $t$ and hence calculate the greatest distance between the boats during the race.
\hfill \mbox{\textit{OCR MEI M1 Q2 [18]}}