| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projection from elevated point - angle above horizontal |
| Difficulty | Moderate -0.3 This is a standard M2 projectile question requiring application of SUVAT equations and energy conservation or kinematic equations. All three parts use routine methods: (a) energy conservation or vertical motion equation, (b) resolving velocity components, (c) vertical motion with known displacement. The elevated starting point adds minimal complexity to what are otherwise textbook exercises. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Conservation of energy: \(\frac{1}{2}m\times49+10mg=\frac{1}{2}mv^2\) | M1 | Equation must include all three terms |
| A2 | -1 each error | |
| \(v=15.7\ \text{m s}^{-1}\) | A1 | Accept 16. Not 15.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Find horizontal and vertical components of speed at \(B\) | M1 | Use of suvat for both components and combine |
| \(v_x=7\cos55\) | A1 | |
| \((v_y)^2=(7\sin55)^2+20g\) | A1 | |
| \(v=15.7\ \text{m s}^{-1}\) | A1 | Accept 16. Not 15.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\cos\theta=\frac{7\cos55}{\text{their }v}\), \(\tan\theta=\frac{\text{their }v_y}{7\cos55}\) | M1 | Correct trig to form equation in a relevant angle |
| \(\cos\theta=\frac{7\cos55}{\sqrt{49+20g}}\), \(\tan\theta=\frac{\sqrt{(7\sin55)^2+20g}}{7\cos55}\) | A1 | |
| \(\theta=75.1°\) to the horizontal (75) (75.2 from 15.7) | A1 | \(14.9°\) to the vertical (direction seen or implied). A0 if magnitude and direction contradict |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vertical distance: \(-10=(7\sin55)t-4.9t^2\) | M1 | Use of suvat - condone sign errors |
| A2 | -1 each error | |
| \(t=\frac{7\sin55+\sqrt{(7\sin55)^2+40\times4.9}}{9.8}\) | DM1 | Solve for \(t\). Incorrect answers must be supported by working |
| \(=2.13\ \text{(s)}\) | A1 | Accept 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy to find \(t\) | M1 | Complete strategy to find \(t\) |
| Vertical component of speed at \(B\) \(= 15.7 \times \sin(75.1)\) | A1 | |
| \(v = u + at\) : \(15.1... = (-7\sin 55) + gt\) | DM1 | Use suvat. Condone sign error(s) |
| A1 | ||
| \(t = 2.13\) s \((2.1)\) | A1[5] | Accept 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy: time to top + top to ground | M1 | Complete strategy for time to top + Top to ground |
| Time to top: \(0 = 7\sin 55 - gt\), \(t_1 = 0.5851...\) | A1 | |
| Distance to top: \(0 = (7\sin 55)^2 - 2 \times 9.8s\), \(s = 1.6775...\) | DM1 | |
| Time to fall 11.68: \(11.68 = \frac{1}{2}gt^2\), \(t_2 = 1.5437...\) | A1 | |
| Total time \(= t_1 + t_2 = 2.13\) s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy: time to level + time to fall 10 m | M1 | Complete strategy for time to level + time to fall 10 m |
| \(-7\sin 55 = 7\sin 55 - gt\), \(t_1 = 1.17...\) | A1 | |
| Time to fall 10 m: \(-10 = -7\sin 55t - 9.8t^2\) | DM1 | |
| \(t_2 = 0.959...\) | A1 | |
| Total time \(= t_1 + t_2 = 2.13\) s | A1[5] |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Conservation of energy: $\frac{1}{2}m\times49+10mg=\frac{1}{2}mv^2$ | M1 | Equation must include all three terms |
| | A2 | -1 each error |
| $v=15.7\ \text{m s}^{-1}$ | A1 | Accept 16. Not 15.6 |
**Alt 6(a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Find horizontal and vertical components of speed at $B$ | M1 | Use of suvat for both components and combine |
| $v_x=7\cos55$ | A1 | |
| $(v_y)^2=(7\sin55)^2+20g$ | A1 | |
| $v=15.7\ \text{m s}^{-1}$ | A1 | Accept 16. Not 15.6 |
---
## Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\theta=\frac{7\cos55}{\text{their }v}$, $\tan\theta=\frac{\text{their }v_y}{7\cos55}$ | M1 | Correct trig to form equation in a relevant angle |
| $\cos\theta=\frac{7\cos55}{\sqrt{49+20g}}$, $\tan\theta=\frac{\sqrt{(7\sin55)^2+20g}}{7\cos55}$ | A1 | |
| $\theta=75.1°$ to the horizontal (75) (75.2 from 15.7) | A1 | $14.9°$ to the vertical (direction seen or implied). A0 if magnitude and direction contradict |
---
## Question 6(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vertical distance: $-10=(7\sin55)t-4.9t^2$ | M1 | Use of suvat - condone sign errors |
| | A2 | -1 each error |
| $t=\frac{7\sin55+\sqrt{(7\sin55)^2+40\times4.9}}{9.8}$ | DM1 | Solve for $t$. Incorrect answers must be supported by working |
| $=2.13\ \text{(s)}$ | A1 | Accept 2.1 |
## Question 6(c) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy to find $t$ | M1 | Complete strategy to find $t$ |
| Vertical component of speed at $B$ $= 15.7 \times \sin(75.1)$ | A1 | |
| $v = u + at$ : $15.1... = (-7\sin 55) + gt$ | DM1 | Use suvat. Condone sign error(s) |
| | A1 | |
| $t = 2.13$ s $(2.1)$ | A1[5] | Accept 2.1 |
## Question 6(c) alt 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy: time to top + top to ground | M1 | Complete strategy for time to top + Top to ground |
| Time to top: $0 = 7\sin 55 - gt$, $t_1 = 0.5851...$ | A1 | |
| Distance to top: $0 = (7\sin 55)^2 - 2 \times 9.8s$, $s = 1.6775...$ | DM1 | |
| Time to fall 11.68: $11.68 = \frac{1}{2}gt^2$, $t_2 = 1.5437...$ | A1 | |
| Total time $= t_1 + t_2 = 2.13$ s | A1 | |
## Question 6(c) alt 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy: time to level + time to fall 10 m | M1 | Complete strategy for time to level + time to fall 10 m |
| $-7\sin 55 = 7\sin 55 - gt$, $t_1 = 1.17...$ | A1 | |
| Time to fall 10 m: $-10 = -7\sin 55t - 9.8t^2$ | DM1 | |
| $t_2 = 0.959...$ | A1 | |
| Total time $= t_1 + t_2 = 2.13$ s | A1[5] | |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{25c503ad-94c7-4137-83b5-c3e0aea62f0c-11_452_865_264_495}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A small ball $P$ is projected with speed $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $A 10 \mathrm {~m}$ above horizontal ground. The angle of projection is $55 ^ { \circ }$ above the horizontal. The ball moves freely under gravity and hits the ground at the point $B$, as shown in Figure 3.
Find
\begin{enumerate}[label=(\alph*)]
\item the speed of $P$ as it hits the ground at $B$,
\item the direction of motion of $P$ as it hits the ground at $B$,
\item the time taken for $P$ to move from $A$ to $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2015 Q6 [12]}}