| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2001 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Vector impulse: find velocity or speed after impulse |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard application of impulse-momentum theorem (part a), projectile motion equations for maximum height (part b), and range calculation (part c). All three parts follow textbook methods with no novel problem-solving required, though the multi-step nature and vector notation place it slightly below average difficulty for A-level overall. |
| Spec | 3.02i Projectile motion: constant acceleration model6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: \(\mathbf{u} + \mathbf{v}\mathbf{j} = (-25\mathbf{i} - 5\mathbf{j})\) ms\(^{-1}\) | M1A1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: 32.9 m or \(\frac{625}{19}\)m | M1 A1 A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Horizontal distance = \(10 \times t = 51\) m [N to 10 or 20+] | M1A1 M1A1√ | (4) |
| Notes: If i and j interchanged, then can score Ms in (b) and (c); allow \(\sqrt{\text{for}} 25 \times 2.04 = 51\). | ||
| Use of answer in (a) can score M marks in (b)(c) only | ||
| Use of \(\frac{V^2 \sin\theta}{2g}\) and \(\frac{V^2 \sin 2\theta}{g}\): M1 method for V or \(\theta\), A1 both correct for first two marks |
**(a)** Impulse = change in momentum
$3.5\mathbf{i} + 3\mathbf{j} = m[\mathbf{(}10\mathbf{i} + 25\mathbf{j}) - (\mathbf{u} + \mathbf{v}\mathbf{j})]$
Answer: $\mathbf{u} + \mathbf{v}\mathbf{j} = (-25\mathbf{i} - 5\mathbf{j})$ ms$^{-1}$ | M1A1 A1 | (3)
**(b)** Complete method to find height $s$ above hit position
Correct equation in $s$ only: $0 = 625 - 2(9.8)s$; $s = 25(25/g) - \frac{1}{2}g(25/g)^2$
Answer: **32.9 m** or **$\frac{625}{19}$m** | M1 A1 A1 | (3)
**(c)** Method for total time: $0 = 25t - 4.9t^2$ $\Rightarrow$ $t = 5.10$ s
or "half time" $0 = 25 - 9.8 \cdot t'$ $\Rightarrow$ $t' = 2.55$ s
Horizontal distance = $10 \times t = 51$ m [N to 10 or 20+] | M1A1 M1A1√ | (4)
**Notes:** If i and j interchanged, then can score Ms in (b) and (c); allow $\sqrt{\text{for}} 25 \times 2.04 = 51$. | |
Use of answer in (a) can score M marks in (b)(c) only | |
Use of $\frac{V^2 \sin\theta}{2g}$ and $\frac{V^2 \sin 2\theta}{g}$: M1 method for V or $\theta$, A1 both correct for first two marks | |
---The unit vectors $\mathbf{i}$ and $\mathbf{j}$ lie in a vertical plane, $\mathbf{i}$ being horizontal and $\mathbf{j}$ vertical. A ball of mass 0.1 kg is hit by a bat which gives it an impulse of $(3.5\mathbf{i} + 3\mathbf{j})$ Ns. The velocity of the ball immediately after being hit is $(10\mathbf{i} + 25\mathbf{j})$ m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of the ball immediately before it is hit.
[3]
\end{enumerate}
In the subsequent motion the ball is modelled as a particle moving freely under gravity. When it is hit the ball is 1 m above horizontal ground.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the greatest height of the ball above the ground in the subsequent motion.
[3]
\end{enumerate}
The ball is caught when it is again 1 m above the ground.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the distance from the point where the ball is hit to the point where it is caught.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2001 Q4 [10]}}