| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Standard +0.3 This is a standard M2 work-energy and projectile motion question with straightforward application of well-practiced techniques. Part (a) uses work-energy principle with given values, parts (b-d) are routine projectile motion calculations with a given angle. All steps are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Work-Energy: \(R \times 60 = 80 \times 9.8 \times 24.4 - \frac{1}{2} \times 80 \times 20^2\) | M1 A2(1,0) | |
| \((= 19129.6 - 16000 = 3129.6)\) | ||
| \(R = 52\) (N) | M1 A1 | accept 52.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-8.1 = 20\sin\alpha \times t - \frac{1}{2}gt^2\) | M1 A2(1,0) | |
| \(4.9t^2 - 12t - 8.1 = 0\) | M1 | |
| \(t = 3\) (s) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(20\cos\alpha \times 3 = 16 \times 3 = 48\) (m) | M1 A1ft | ft their \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Energy: \(\frac{1}{2}mv^2 - \frac{1}{2}m \times 20^2 = m \times 9.8 \times 8.1\) | M1 A2(1,0) | |
| \(v = \sqrt{558.56} \approx 24\) \((\text{ms}^{-1})\) | M1 A1 | accept 23.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\uparrow\) \(v_y = 12 - 3g = -17.4\); \(\rightarrow\) \(v_x = 16\) | M1 A1, A1 | |
| \(v = \sqrt{17.4^2 + 16^2} \approx 24\) \((\text{ms}^{-1})\) | M1 A1 | accept 23.6 |
## Question 7:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Work-Energy: $R \times 60 = 80 \times 9.8 \times 24.4 - \frac{1}{2} \times 80 \times 20^2$ | M1 A2(1,0) | |
| $(= 19129.6 - 16000 = 3129.6)$ | | |
| $R = 52$ (N) | M1 A1 | accept 52.2 |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-8.1 = 20\sin\alpha \times t - \frac{1}{2}gt^2$ | M1 A2(1,0) | |
| $4.9t^2 - 12t - 8.1 = 0$ | M1 | |
| $t = 3$ (s) | A1 | |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $20\cos\alpha \times 3 = 16 \times 3 = 48$ (m) | M1 A1ft | ft their $t$ |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Energy: $\frac{1}{2}mv^2 - \frac{1}{2}m \times 20^2 = m \times 9.8 \times 8.1$ | M1 A2(1,0) | |
| $v = \sqrt{558.56} \approx 24$ $(\text{ms}^{-1})$ | M1 A1 | accept 23.6 |
**Alternative for (d):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\uparrow$ $v_y = 12 - 3g = -17.4$; $\rightarrow$ $v_x = 16$ | M1 A1, A1 | |
| $v = \sqrt{17.4^2 + 16^2} \approx 24$ $(\text{ms}^{-1})$ | M1 A1 | accept 23.6 |7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{8e694174-b9a9-4018-8896-31a3b4f0d344-5_424_1324_264_383}
\end{center}
\end{figure}
In a ski-jump competition, a skier of mass 80 kg moves from rest at a point $A$ on a ski-slope. The skier's path is an arc $A B$. The starting point $A$ of the slope is 32.5 m above horizontal ground. The end $B$ of the slope is 8.1 m above the ground. When the skier reaches $B$, she is travelling at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and moving upwards at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$, as shown in Fig. 2. The distance along the slope from $A$ to $B$ is 60 m . The resistance to motion while she is on the slope is modelled as a force of constant magnitude $R$ newtons. By using the work-energy principle,
\begin{enumerate}[label=(\alph*)]
\item find the value of $R$.
On reaching $B$, the skier then moves through the air and reaches the ground at the point $C$. The motion of the skier in moving from $B$ to $C$ is modelled as that of a particle moving freely under gravity.
\item Find the time for the skier to move from $B$ to $C$.
\item Find the horizontal distance from $B$ to $C$.
\item Find the speed of the skier immediately before she reaches $C$.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2004 Q7 [17]}}