| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Collision with vector velocities |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard calculus operations (differentiation for acceleration, integration for position) and basic vector manipulation. All parts follow directly from definitions: momentum = mass × velocity, F = ma, and position from integrating velocity. While it involves multiple steps and vector components, it requires no problem-solving insight—just systematic application of learned formulas. |
| Spec | 1.08d Evaluate definite integrals: between limits3.03d Newton's second law: 2D vectors6.03a Linear momentum: p = mv |
| Answer | Marks | Guidance |
|---|---|---|
| Momentum = \(0.25 \times ((2-12)i + (9+6)j)\) | M1 | Substitution of \(m = 0.25\) and \(t = 3\) |
| \(= -2.5i + 3.75j\) | A1 | or simplified equivalent \(-\frac{5}{2}i + \frac{15}{4}j\) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = -4i + (2t+2)j\) | M1 | Differentiate v |
| A2 | A1 for each term | |
| Use of \(F = ma\) | M1 | |
| Use of Pythagoras on F or a | M1 | |
| \(\ | F\ | = 0.25 \times \sqrt{16 + (6+2)^2} = \sqrt{5}\) |
| (6) |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = (2t - 2t^2)i + \left(\frac{t^3}{3} + t^2\right)j (+ r_0)\) | M1 | Integrate to find displacement |
| A2 | A1 for each component. \(r_0\) not required. | |
| Use of \(r_0\) and \(t = 3\) | M1 | |
| Position vector of A is \(-10i + 14j\) | A1 | |
| (5) | ||
| [13] |
**(a)**
| Momentum = $0.25 \times ((2-12)i + (9+6)j)$ | M1 | Substitution of $m = 0.25$ and $t = 3$ |
| $= -2.5i + 3.75j$ | A1 | or simplified equivalent $-\frac{5}{2}i + \frac{15}{4}j$ |
| | (2) | |
**(b)**
| $a = -4i + (2t+2)j$ | M1 | Differentiate v |
| | A2 | A1 for each term |
| Use of $F = ma$ | M1 | |
| Use of Pythagoras on F or **a** | M1 | |
| $\|F\| = 0.25 \times \sqrt{16 + (6+2)^2} = \sqrt{5}$ | A1 | 2.24 or better |
| | (6) | |
**(c)**
| $s = (2t - 2t^2)i + \left(\frac{t^3}{3} + t^2\right)j (+ r_0)$ | M1 | Integrate to find displacement |
| | A2 | A1 for each component. $r_0$ not required. |
| Use of $r_0$ and $t = 3$ | M1 | |
| Position vector of A is $-10i + 14j$ | A1 | |
| | (5) |
| | [13] | |A particle $P$ of mass 0.25 kg moves under the action of a single force $\mathbf{F}$ newtons. At time $t$ seconds, the velocity of $P$ is $\mathbf{v}$ m s$^{-1}$, where
$$\mathbf{v} = (2 - 4t)\mathbf{i} + (t^2 + 2t)\mathbf{j}$$
When $t = 0$, $P$ is at the point with position vector $(2\mathbf{i} - 4\mathbf{j})$ m with respect to a fixed origin $O$. When $t = 3$, $P$ is at the point $A$. Find
\begin{enumerate}[label=(\alph*)]
\item the momentum of $P$ when $t = 3$, [2]
\item the magnitude of $\mathbf{F}$ when $t = 3$, [6]
\item the position vector of $A$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q3 [13]}}