| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| 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 using energy conservation and kinematic equations. Part (a) is straightforward energy calculation, part (b) requires resolving vertical motion with given conditions, and part (c) uses standard velocity components. 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 model6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Energy at A = Energy at B | M1 | Equation requires all 3 terms. Condone sign errors |
| \(\frac{1}{2}m \times 15^2 = \frac{1}{2}m \times 12^2 + mgh\) | A2 | -1 each error |
| \(h = \frac{27 \times 3}{2g} \approx 4.1\) | A1 | Allow 4.13; \(\frac{405}{98}\) is A0 |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(s = ut + \frac{1}{2}gt^2\) | M1 | or equivalent complete method to an equation in \(a\) |
| \(4 = 12 \sin a \times 1.5 - \frac{g}{2} \times 1.5^2\) | A1 | Correct unsimplified equation |
| \((\sin a = 0.834.....)\) \(a = 56.6°\) | A1 | Or \(57°\) |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos \theta = \frac{12 \cos a}{15}\) | M1 | Use of horizontal component and speed to find direction, or equivalent. Follow their \(a\): 6.606, 13.47, 15 |
| \(\theta = 64°\) | A1 | 63.9 from calculator values, 64.2 from 57 |
| (3) | ||
| [10] |
**(a)**
| Energy at A = Energy at B | M1 | Equation requires all 3 terms. Condone sign errors |
| $\frac{1}{2}m \times 15^2 = \frac{1}{2}m \times 12^2 + mgh$ | A2 | -1 each error |
| $h = \frac{27 \times 3}{2g} \approx 4.1$ | A1 | Allow 4.13; $\frac{405}{98}$ is A0 |
| | (4) | |
**(b)**
| Use of $s = ut + \frac{1}{2}gt^2$ | M1 | or equivalent complete method to an equation in $a$ |
| $4 = 12 \sin a \times 1.5 - \frac{g}{2} \times 1.5^2$ | A1 | Correct unsimplified equation |
| $(\sin a = 0.834.....)$ $a = 56.6°$ | A1 | Or $57°$ |
| | (3) | |
**(c)**
| $\cos \theta = \frac{12 \cos a}{15}$ | M1 | Use of horizontal component and speed to find direction, or equivalent. Follow their $a$: 6.606, 13.47, 15 |
| $\theta = 64°$ | A1 | 63.9 from calculator values, 64.2 from 57 |
| | (3) |
| | [10] | |\includegraphics{figure_1}
The points $O$ and $B$ are on horizontal ground. The point $A$ is $h$ metres vertically above $O$. A particle $P$ is projected from $A$ with speed 12 m s$^{-1}$ at an angle $\alpha°$ to the horizontal. The particle moves freely under gravity and hits the ground at $B$, as shown in Figure 1. The speed of $P$ immediately before it hits the ground is 15 m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item By considering energy, find the value of $h$. [4]
\end{enumerate}
Given that 1.5 s after it is projected from $A$, $P$ is at a point 4 m above the level of $A$, find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item the value of $\alpha$, [3]
\item the direction of motion of $P$ immediately before it reaches $B$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q4 [10]}}