| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question testing Newton's second law and kinematics with constant acceleration. Part (i) uses basic SUVAT equations, parts (ii) and (iii) apply F=ma in straightforward contexts, and part (iv) requires considering two connected bodies but with clear force diagrams. All techniques are routine for M1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| \(3.5 = 0.5 + 1.5T^2\) so \(T = 2\) so \(2\) s | M1, A1 | Suitable uvast, condone sign errors. cao |
| \(s = \frac{3.5 + 0.5}{\sqrt{2}} \times 2\) so \(s = 4\) so \(4\) m | M1, F1 | Suitable uvast, condone sign errors. FT their \(T\). [If \(s\) found first then it is cao. In this case when finding \(T\), FT their \(s\), if used.] |
| Sub: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| N2L \(\downarrow: 80 \times 9.8 - T = 80 \times 1.5\) | M1 | Use of N2L. Allow weight omitted and use of \(F = ma\). Condone errors in sign but do not allow extra forces. |
| \(T = 664\) so \(664\) N | B1, A1 | weight correct (seen in (A) or (B)). cao |
| Sub: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| N2L \(\downarrow: 80 \times 9.8 - T = 80 \times (-1.5)\) | M1 | N2L with all forces and using \(F = ma\). Condone errors in sign but do not allow extra forces. |
| \(T = 904\) so \(904\) N | A1 | cao [Accept 904 N seen for M1 A1] |
| Sub: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| N2L \(\uparrow: 2500 - 80 - 9.8 - 116 = 80a\) | M1 | Use of N2L with \(F = ma\). Allow 1 force missing. No extra forces. Condone errors in sign. |
| \(a = 20\) so \(20\) m s\(^{-2}\) upwards. | A1, A1, A1 | \(\pm20\), accept direction wrong or omitted. upwards made clear (accept diagram) |
| Sub: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| N2L \(\uparrow\) on equipment: \(80 - 10 \times 9.8 = 10a\) | M1 | Use of N2L on equipment. All forces. \(F = ma\). No extra forces. Allow sign errors. Allow \(\pm1.8\) |
| \(a = -1.8\) | A1 | |
| N2L \(\uparrow\) | M1 | N2L for system or for man alone. Forces correct (with no extras); accept sign errors; their \(\pm1.8\) used |
| either all: \(T - (80 + 10) \times 9.8 - 116 = 90 \times (-1.8)\) | ||
| or on man: \(T - (80 \times 9.8) - 116 - 80 = 80 \times (-1.8)\) \(T = 836\) so \(836\) N | A1 | cao [NB: The answer 836 N is independent of the value taken for \(g\) and hence may be obtained if all weights are omitted.] |
| Sub: 4 |
### (i)
$3.5 = 0.5 + 1.5T^2$ so $T = 2$ so $2$ s | M1, A1 | Suitable uvast, condone sign errors. cao
$s = \frac{3.5 + 0.5}{\sqrt{2}} \times 2$ so $s = 4$ so $4$ m | M1, F1 | Suitable uvast, condone sign errors. FT their $T$. [If $s$ found first then it is cao. In this case when finding $T$, FT their $s$, if used.]
| Sub: 4
### (ii)(A)
N2L $\downarrow: 80 \times 9.8 - T = 80 \times 1.5$ | M1 | Use of N2L. Allow weight omitted and use of $F = ma$. Condone errors in sign but do not allow extra forces.
$T = 664$ so $664$ N | B1, A1 | weight correct (seen in (A) or (B)). cao
| Sub: 5
### (ii)(B)
N2L $\downarrow: 80 \times 9.8 - T = 80 \times (-1.5)$ | M1 | N2L with all forces and using $F = ma$. Condone errors in sign but do not allow extra forces.
$T = 904$ so $904$ N | A1 | cao [Accept 904 N seen for M1 A1]
| Sub: 5
### (iii)
N2L $\uparrow: 2500 - 80 - 9.8 - 116 = 80a$ | M1 | Use of N2L with $F = ma$. Allow 1 force missing. No extra forces. Condone errors in sign.
$a = 20$ so $20$ m s$^{-2}$ upwards. | A1, A1, A1 | $\pm20$, accept direction wrong or omitted. upwards made clear (accept diagram)
| Sub: 4
### (iv)
N2L $\uparrow$ on equipment: $80 - 10 \times 9.8 = 10a$ | M1 | Use of N2L on equipment. All forces. $F = ma$. No extra forces. Allow sign errors. Allow $\pm1.8$
$a = -1.8$ | A1 |
N2L $\uparrow$ | M1 | N2L for system or for man alone. Forces correct (with no extras); accept sign errors; their $\pm1.8$ used
either all: $T - (80 + 10) \times 9.8 - 116 = 90 \times (-1.8)$ | |
or on man: $T - (80 \times 9.8) - 116 - 80 = 80 \times (-1.8)$ $T = 836$ so $836$ N | A1 | cao [NB: The answer 836 N is independent of the value taken for $g$ and hence may be obtained if all weights are omitted.]
| Sub: 4
---A helicopter rescue activity at sea is modelled as follows. The helicopter is stationary and a man is suspended from it by means of a vertical, light, inextensible wire that may be raised or lowered, as shown in Fig. 6.1.
\includegraphics{figure_6_1}
\begin{enumerate}[label=(\roman*)]
\item When the man is descending with an acceleration 1.5 m s$^{-2}$ downwards, how much time does it take for his speed to increase from 0.5 m s$^{-1}$ downwards to 3.5 m s$^{-1}$ downwards?
How far does he descend in this time? [4]
\end{enumerate}
The man has a mass of 80 kg. All resistances to motion may be neglected.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the tension in the wire when the man is being lowered
\begin{enumerate}[label=(\Alph*)]
\item with an acceleration of 1.5 m s$^{-2}$ downwards,
\item with an acceleration of 1.5 m s$^{-2}$ upwards. [5]
\end{enumerate}
\end{enumerate}
Subsequently, the man is raised and this situation is modelled with a constant resistance of 116 N to his upward motion.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item For safety reasons, the tension in the wire should not exceed 2500 N. What is the maximum acceleration allowed when the man is being raised? [4]
\end{enumerate}
At another stage of the rescue, the man has equipment of mass 10 kg at the bottom of a vertical rope which is hanging from his waist, as shown in Fig. 6.2. The man and his equipment are being raised; the rope is light and inextensible and the tension in it is 80 N.
\includegraphics{figure_6_2}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item Assuming that the resistance to the upward motion of the man is still 116 N and that there is negligible resistance to the motion of the equipment, calculate the tension in the wire. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 2008 Q6 [17]}}