| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Projectile energy - finding speed or height |
| Difficulty | Standard +0.3 This is a standard M2 projectile motion question with straightforward energy methods. Part (a) uses basic conservation of energy (one equation), parts (b-c) apply standard projectile formulas with a given angle, and part (d) requires combining vertical and horizontal motion equations. All techniques are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m\times 15^2 = 47.5mg\) | M1 | Need all 3 terms. Condone sign errors. Must be dimensionally correct. |
| A1 | Unsimplified equation with at most one error | |
| A1 | Correct unsimplified equation | |
| \(v = 34\ \text{m s}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = 15\times\frac{3}{5}\ \text{m s}^{-1},\ a=-9.8\ \text{m s}^{-1},\ v=0\) | ||
| \(0 = 9^2 - 2\times 9.8s\) | M1 | Complete method using suvat to reach equation in \(s\) |
| \(s = 4.1326...\) | A1 | |
| ht above beach \(= 51.63... = 52\) (m) | A1ft | Or \(51.6\)(m). Their \(s + 47.5\). Max 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| least speed \(= 15\times\dfrac{4}{5} = 12\ \text{m s}^{-1}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u = -15\times\frac{3}{5}\ \text{m s}^{-1},\ a=9.8\ \text{m s}^{-2},\ s=47.5\) | ||
| \(47.5 = -9t + \frac{1}{2}\times 9.8t^2\) | M1 | Complete method using suvat to reach equation in \(t\) |
| \(4.9t^2 - 9t - 47.5 = 0\) | A1 | Correct equation (any form) |
| \(t = \dfrac{9\pm\sqrt{9^2 + 4\times 4.9\times 47.5}}{9.8}\) | dM1 | Solve for \(t\). Dependent on preceding M |
| \(t = 4.16448...\) | A1 | Only. \(-\)ve value must be rejected if seen. |
| Horiz dist \(= 15\times\dfrac{4}{5}\times 4.16448... (= 49.9738...\)m\()\) | M1 | Complete method using suvat and their \(t\) to find distance. Independent |
| \(= 50\) or \(50.0\) (m) | A1 | Max 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to find vertical component of speed on impact with the ground | M1 | Or use their \(\sqrt{(a)^2-(c)^2}\) provided \((c)\neq 0\) |
| \(v = \sqrt{1012}\ (=31.8...)\) | A1 | |
| \(\sqrt{1012} = -9 + gt\) | M1 | Use suvat to find \(t\). Condone sign error(s) |
| \(t = 4.16448...\) | A1 |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - \frac{1}{2}m\times 15^2 = 47.5mg$ | M1 | Need all 3 terms. Condone sign errors. Must be dimensionally correct. |
| | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $v = 34\ \text{m s}^{-1}$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = 15\times\frac{3}{5}\ \text{m s}^{-1},\ a=-9.8\ \text{m s}^{-1},\ v=0$ | | |
| $0 = 9^2 - 2\times 9.8s$ | M1 | Complete method using suvat to reach equation in $s$ |
| $s = 4.1326...$ | A1 | |
| ht above beach $= 51.63... = 52$ (m) | A1ft | Or $51.6$(m). Their $s + 47.5$. Max 3 s.f. |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| least speed $= 15\times\dfrac{4}{5} = 12\ \text{m s}^{-1}$ | B1 | |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = -15\times\frac{3}{5}\ \text{m s}^{-1},\ a=9.8\ \text{m s}^{-2},\ s=47.5$ | | |
| $47.5 = -9t + \frac{1}{2}\times 9.8t^2$ | M1 | Complete method using suvat to reach equation in $t$ |
| $4.9t^2 - 9t - 47.5 = 0$ | A1 | Correct equation (any form) |
| $t = \dfrac{9\pm\sqrt{9^2 + 4\times 4.9\times 47.5}}{9.8}$ | dM1 | Solve for $t$. Dependent on preceding M |
| $t = 4.16448...$ | A1 | Only. $-$ve value must be rejected if seen. |
| Horiz dist $= 15\times\dfrac{4}{5}\times 4.16448... (= 49.9738...$m$)$ | M1 | Complete method using suvat and their $t$ to find distance. Independent |
| $= 50$ or $50.0$ (m) | A1 | Max 3 s.f. |
**Alternative for first 4 marks in (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find vertical component of speed on impact with the ground | M1 | Or use their $\sqrt{(a)^2-(c)^2}$ provided $(c)\neq 0$ |
| $v = \sqrt{1012}\ (=31.8...)$ | A1 | |
| $\sqrt{1012} = -9 + gt$ | M1 | Use suvat to find $t$. Condone sign error(s) |
| $t = 4.16448...$ | A1 | |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ef8231d-5b95-4bbb-a8e2-788c708fa078-24_711_1009_251_479}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A small ball $P$ is projected with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $A$ which is 47.5 m above a horizontal beach. The ball moves freely under gravity and hits the beach at the point $B$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item By considering energy, find the speed of $P$ immediately before it hits the beach.
The ball was projected from $A$ at an angle $\theta$ above the horizontal, where $\sin \theta = \frac { 3 } { 5 }$
\item Find the greatest height above the beach of $P$ as it moved from $A$ to $B$.
\item Find the least speed of $P$ as it moved between $A$ and $B$.
\item Find the horizontal distance from $A$ to $B$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2017 Q7 [14]}}