| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Energy methods in projectiles |
| Difficulty | Standard +0.3 This is a standard M2 projectiles question requiring energy conservation for part (a) and standard kinematic equations for part (b). The energy method is straightforward (KE loss = PE gain), and part (b) involves routine application of projectile formulas with a given angle. Slightly above average difficulty due to the two-part structure and need to recognize the energy approach, but all techniques are standard M2 material with no novel problem-solving required. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}m \times 14^2 = \frac{1}{2}m \times 10^2 + mgh\) | M1, A2 | All terms required. Correct form but condone sign errors. \(-1\) each error in unsimplified equation |
| \(h = \frac{48}{g} = 4.90\) | A1 | Accept \(\frac{48}{g}\). Maximum 3 s.f. if they go into decimals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Initial \(v_y = 14\sin\alpha\), Final \(v_y = \sqrt{100 - 14^2\cos^2\alpha}\) | ||
| \(100 - 196\cos^2\alpha = 196\sin^2\alpha - 2gh\) | M1A2 | Using \(v^2 = u^2 + 2as\) on vertical components of speed. \(-1\) each error |
| \(h = \frac{48}{g} = 4.90\) | A1 | Accept in exact form. Maximum 3 s.f. if decimals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vertical distance: \(h = 14\sin\alpha t - \frac{1}{2} \times 9.8t^2\) | M1 | Complete method for equation in \(t\). Must involve trig, condone sin/cos confusion |
| \(4.9t^2 - 11.9t + h = 0\) | A2 | Correct in \(h\) or their \(h\). \(-1\) each error |
| \(t = \frac{11.9 \pm \sqrt{11.9^2 - 4 \times 4.9^2}}{9.8}\) | DM1 | Solve 3-term quadratic for \(t\). Needs their value of \(h\) |
| \(t = 1.903...\) | A1 | 1.9 or better |
| Horizontal distance: \(x = 14\cos\alpha \times t\) | M1 | Method for horizontal distance. Condone consistent sin/cos confusion |
| \(= 14.0\) (m) | A1, A1 | Correct for their positive \(t\). Accept 14 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vertical speed \(= \sqrt{100 - (14\cos\alpha)^2}\) (\(= 6.75\)) | M1, A2 | Complete method for vertical component of speed at \(B\). Correct unsimplified. \(-1\) each error |
| \(v = u + at = 14 \times 0.85 - 9.8t \quad (-6.75 = 11.9 - 9.8t)\) | DM1 | Use vertical component to find \(t\) |
| \(t = 1.903...\) | A1 | 1.9 or better |
| Horizontal distance: \(x = 14\cos\alpha \times t = 14.0\) (m) | M1, A1, A1 | Method for horizontal distance. Correct for positive \(t\). Accept 14 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}m \times 14^2 = \frac{1}{2}m \times 10^2 + mgh$ | M1, A2 | All terms required. Correct form but condone sign errors. $-1$ each error in unsimplified equation |
| $h = \frac{48}{g} = 4.90$ | A1 | Accept $\frac{48}{g}$. Maximum 3 s.f. if they go into decimals |
**Alt(a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Initial $v_y = 14\sin\alpha$, Final $v_y = \sqrt{100 - 14^2\cos^2\alpha}$ | | |
| $100 - 196\cos^2\alpha = 196\sin^2\alpha - 2gh$ | M1A2 | Using $v^2 = u^2 + 2as$ on vertical components of speed. $-1$ each error |
| $h = \frac{48}{g} = 4.90$ | A1 | Accept in exact form. Maximum 3 s.f. if decimals |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vertical distance: $h = 14\sin\alpha t - \frac{1}{2} \times 9.8t^2$ | M1 | Complete method for equation in $t$. Must involve trig, condone sin/cos confusion |
| $4.9t^2 - 11.9t + h = 0$ | A2 | Correct in $h$ or their $h$. $-1$ each error |
| $t = \frac{11.9 \pm \sqrt{11.9^2 - 4 \times 4.9^2}}{9.8}$ | DM1 | Solve 3-term quadratic for $t$. Needs their value of $h$ |
| $t = 1.903...$ | A1 | 1.9 or better |
| Horizontal distance: $x = 14\cos\alpha \times t$ | M1 | Method for horizontal distance. Condone consistent sin/cos confusion |
| $= 14.0$ (m) | A1, A1 | Correct for their positive $t$. Accept 14 |
**Alt (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vertical speed $= \sqrt{100 - (14\cos\alpha)^2}$ ($= 6.75$) | M1, A2 | Complete method for vertical component of speed at $B$. Correct unsimplified. $-1$ each error |
| $v = u + at = 14 \times 0.85 - 9.8t \quad (-6.75 = 11.9 - 9.8t)$ | DM1 | Use vertical component to find $t$ |
| $t = 1.903...$ | A1 | 1.9 or better |
| Horizontal distance: $x = 14\cos\alpha \times t = 14.0$ (m) | M1, A1, A1 | Method for horizontal distance. Correct for positive $t$. Accept 14 |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{47420c50-c232-41e9-8c4d-a890d86ea933-10_645_1196_125_351}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A small ball is projected with speed $14 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $A$ on horizontal ground. The angle of projection is $\alpha$ above the horizontal. A horizontal platform is at height $h$ metres above the ground. The ball moves freely under gravity until it hits the platform at the point B, as shown in Figure 2. The speed of the ball immediately before it hits the platform at $B$ is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $h$.
Given that $\sin \alpha = 0.85$,
\item find the horizontal distance from $A$ to $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q6 [12]}}