| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Direct collision, find impulse magnitude |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring standard techniques: impulse calculation from velocity change (vector subtraction and magnitude), energy conservation with projectile motion (KE + PE), and angle calculation using velocity components. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = 0.4(15i + 16j + 20i - 4j) \,(= 0.4(35i + 12j) = 14i + 4.8j)\) | M1 | |
| \( | I | = \sqrt{(14^2 + 4.8^2)}\) or \(0.4\sqrt{(35^2 + 12^2)}\) |
| \(= 14.8\) (Ns) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Initial K.E. \(= \frac{1}{2}m(15^2 + 16^2) \,(= 240.5m = 96.2 \text{ J})\) | M1 | |
| \(\frac{1}{2}mv^2 = \frac{1}{2}m(15^2 + 16^2) = m \times 9.8 \times 1.2\) | M1 A2, 1.0 | (−1 each incorrect term) |
| \(v^2 = 504.52\) | M1 | |
| \(v = 22\) (m s\(^{-1}\)) accept 22.5 | A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\arccos \frac{15}{22.5} = 48°\) accept 48.1° | M1 A1 A1 A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Any 2 of: Air resistance; Wind (problem not 2 dimensional); Rotation of ball (ball is not a particle) | B1, B1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Resolve ↑ with 16 and 9.8 | M1 | |
| (↑) \(v_y^2 = 16^2 + 2 \times (-9.8) \times (-1.2)\) | M1 A1 | |
| \((v_y^2 = 279.52, v_y \approx 16.7.....)\) | ||
| \(v^2 = 15^2 + 279.52\) | M1 A1 | |
| \(v = 22\) (m s\(^{-1}\)) accept 22.5 | A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\arctan \frac{16.7}{15} = 48°\) | M1 A1 A1 A1 | (4 marks) |
## Part (a)
$I = 0.4(15i + 16j + 20i - 4j) \,(= 0.4(35i + 12j) = 14i + 4.8j)$ | M1 |
$|I| = \sqrt{(14^2 + 4.8^2)}$ or $0.4\sqrt{(35^2 + 12^2)}$ | M1 for any magnitude | M1 A1 |
$= 14.8$ (Ns) | A1 | **(4 marks)**
## Part (b)
Initial K.E. $= \frac{1}{2}m(15^2 + 16^2) \,(= 240.5m = 96.2 \text{ J})$ | M1 |
$\frac{1}{2}mv^2 = \frac{1}{2}m(15^2 + 16^2) = m \times 9.8 \times 1.2$ | M1 A2, 1.0 | (−1 each incorrect term) |
$v^2 = 504.52$ | M1 |
$v = 22$ (m s$^{-1}$) accept 22.5 | A1 | **(6 marks)**
## Part (c)
$\arccos \frac{15}{22.5} = 48°$ accept 48.1° | M1 A1 A1 A1 | **(4 marks)**
## Part (d)
Any 2 of: Air resistance; Wind (problem not 2 dimensional); Rotation of ball (ball is not a particle) | B1, B1 | **(2 marks)**
**(16 marks total)**
---
# Alternative Methods
## Alt (b)
Resolve ↑ with 16 and 9.8 | M1 |
(↑) $v_y^2 = 16^2 + 2 \times (-9.8) \times (-1.2)$ | M1 A1 |
$(v_y^2 = 279.52, v_y \approx 16.7.....)$ | |
$v^2 = 15^2 + 279.52$ | M1 A1 |
$v = 22$ (m s$^{-1}$) accept 22.5 | A1 | **(6 marks)**
## Alt (c)
$\arctan \frac{16.7}{15} = 48°$ | M1 A1 A1 A1 | **(4 marks)**\includegraphics{figure_3}
A ball $B$ of mass 0.4 kg is struck by a bat at a point $O$ which is 1.2 m above horizontal ground. The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are respectively horizontal and vertical. Immediately before being struck, $B$ has velocity $(-20\mathbf{i} + 4\mathbf{j})$ m s$^{-1}$. Immediately after being struck it has velocity $(15\mathbf{i} + 16\mathbf{j})$ m s$^{-1}$.
After $B$ has been struck, it moves freely under gravity and strikes the ground at the point $A$, as shown in Fig. 3. The ball is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Calculate the magnitude of the impulse exerted by the bat on $B$. [4]
\item By using the principle of conservation of energy, or otherwise, find the speed of $B$ when it reaches $A$. [6]
\item Calculate the angle which the velocity of $B$ makes with the ground when $B$ reaches $A$. [4]
\item State two additional physical factors which could be taken into account in a refinement of the model of the situation which would make it more realistic. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2003 Q7 [16]}}