| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Moderate -0.3 This is a standard M2 projectiles question requiring routine application of SUVAT equations and trajectory formulas. Parts (a) and (b) involve straightforward substitution into well-practiced methods (max height and finding y at given x), while part (c) is a standard modelling assumption recall. Slightly easier than average due to being a textbook-style multi-part question with no novel problem-solving required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Initial vertical speed \(= u\sin\alpha = 25\frac{5}{13}\) ms\(^{-1}\) | B1 | |
| \(v^2 = u^2 + 2as\), \(\quad 100 = 2gh\) | M1 | |
| \(h = \frac{100}{2g} \approx 5.1\) m | A1 | |
| \(\therefore\) Ht \(+ 5.1 + 0.8 = 5.9\) m | A1 ft | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Horizontal speed \(= u\cos\alpha = 24\) ms\(^{-1}\) | B1 | |
| Time to window \(36 = 24t \Rightarrow t = 1.5\) s | M1 A1 | |
| \(h = 0.8 + 10 \times 1.5 - \frac{1}{2} \times 9.8 \times 1.5^2\) | M1 A1 A1 ft | |
| \(\approx 4.8\) m | A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| One of, e.g., air resistance; spin of ball; variation in \(g\); wind | B1 | (1) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| Initial vertical speed $= u\sin\alpha = 25\frac{5}{13}$ ms$^{-1}$ | B1 | |
| $v^2 = u^2 + 2as$, $\quad 100 = 2gh$ | M1 | |
| $h = \frac{100}{2g} \approx 5.1$ m | A1 | |
| $\therefore$ Ht $+ 5.1 + 0.8 = 5.9$ m | A1 ft | **(4)** |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| Horizontal speed $= u\cos\alpha = 24$ ms$^{-1}$ | B1 | |
| Time to window $36 = 24t \Rightarrow t = 1.5$ s | M1 A1 | |
| $h = 0.8 + 10 \times 1.5 - \frac{1}{2} \times 9.8 \times 1.5^2$ | M1 A1 A1 ft | |
| $\approx 4.8$ m | A1 | **(7)** |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| One of, e.g., air resistance; spin of ball; variation in $g$; wind | B1 | **(1)** |
---6. A cricket ball is hit from a height of 0.8 m above horizontal ground with a speed of $26 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ above the horizontal, where $\tan \alpha = \frac { 5 } { 12 }$. The motion of the ball is modelled as that of a particle moving freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Find, to 2 significant figures, the greatest height above the ground reached by the ball.
When the ball has travelled a horizontal distance of 36 m , it hits a window.
\item Find, to 2 significant figures, the height above the ground at which the ball hits the window.
\item State one physical factor which could be taken into account in any refinement of the model which would make it more realistic.
Figure 2
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [12]}}