| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Projectile motion: trajectory equation |
| Difficulty | Moderate -0.8 This is a straightforward two-part projectile motion question requiring standard SUVAT application. Part (i) asks for a basic kinematic equation (y = ut - ½gt²) with given components, and part (ii) requires finding range by setting height to zero—both are routine textbook exercises with no problem-solving insight needed. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(40\times0.6t - 5t^2 = 24t - 5t^2\) | M1 | Use of \(s = ut + 0.5at^2\) with \(a = \pm9.8, \pm10\). Accept 40 or \(40\times0.8\) for \(u\) |
| A1 | Any form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Need zero vertical distance so \(24t - 5t^2 = 0\) | M1 | Equate their \(y\) to zero. With fresh start must have correct \(y\) |
| so \(t = 0\) or \(t = 4.8\) | A1 | Accept no reference to \(t=0\) and the other root in any form. FT their \(y\) if gives \(t > 0\) |
| or Time to highest point \(T\): \(0 = 40\times0.6 - 10T\) so \(T = 2.4\) and time of flight is 4.8 | M1, A1 | Allow use of \(u = 40\) and \(40\times0.8\). May be awarded for doubling half range later |
| Range is \(40\times0.8\times4.8 = 153.6\) | M1 | Horiz cpt. Accept 0.6 instead of 0.8 only if consistent with expression in (i). FT their \(t\) |
| so 154 m (3 s.f.) | A1 | cao. NB use of half range or half time to get 76.8… (\(g=10\)) or 78.36… (\(g=9.8\)) scores 2. If range formula used: M1 sensible attempt at substitution; allow \(\sin2\alpha\) wrong. B1 \(\sin2\alpha\) correct, A1 all correct, A1 cao |
# Question 6:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $40\times0.6t - 5t^2 = 24t - 5t^2$ | M1 | Use of $s = ut + 0.5at^2$ with $a = \pm9.8, \pm10$. Accept 40 or $40\times0.8$ for $u$ |
| | A1 | Any form |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Need zero vertical distance so $24t - 5t^2 = 0$ | M1 | Equate their $y$ to zero. With fresh start must have correct $y$ |
| so $t = 0$ or $t = 4.8$ | A1 | Accept no reference to $t=0$ and the other root in any form. FT their $y$ if gives $t > 0$ |
| **or** Time to highest point $T$: $0 = 40\times0.6 - 10T$ so $T = 2.4$ and time of flight is 4.8 | M1, A1 | Allow use of $u = 40$ and $40\times0.8$. May be awarded for doubling half range later |
| Range is $40\times0.8\times4.8 = 153.6$ | M1 | Horiz cpt. Accept 0.6 instead of 0.8 only if consistent with expression in (i). FT their $t$ |
| so 154 m (3 s.f.) | A1 | cao. NB use of half range or half time to get 76.8… ($g=10$) or 78.36… ($g=9.8$) scores 2. If range formula used: M1 sensible attempt at substitution; allow $\sin2\alpha$ wrong. B1 $\sin2\alpha$ correct, A1 all correct, A1 cao |
---6 In this question take $\boldsymbol { g } = \mathbf { 1 0 }$.\\
A golf ball is hit from ground level over horizontal ground. The initial velocity of the ball is $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.6$ and $\cos \alpha = 0.8$. Air resistance may be neglected.\\
(i) Find an expression for the height of the ball above the ground $t$ seconds after projection.\\
(ii) Calculate the horizontal range of the ball.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{170edb27-324e-44df-8dc1-7d8fbad680fe-4_358_447_360_849}
\captionsetup{labelformat=empty}
\caption{Fig. 7.1}
\end{center}
\end{figure}
A box of mass 8 kg is supported by a continuous light string ACB that is fixed at A and at B and passes through a smooth ring on the box at C, as shown in Fig. 7.1. The box is in equilibrium and the tension in the string section AC is 60 N .\\
\hfill \mbox{\textit{OCR MEI M1 2008 Q6 [6]}}