| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Velocity after impulse (direct calculation) |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring integration of force to find velocity (using F=ma), then applying impulse-momentum theorem. Both parts use standard techniques with no conceptual challenges—slightly easier than average due to routine calculus and vector arithmetic. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| N2L: \((6t - 5)i + (t^2 - 2t)j = 0.5a\) | M1 | |
| \(a = (12t - 10)i + (2t^2 - 4t)j\) | A1 | |
| \(v = (6t^2 - 10t)i + \left(\frac{2}{3}t^3 - 2t^2\right)j\) (+C) | M1 A1 ft + A1 ft | |
| ft their \(a\) | ||
| \(v = (6t^2 - 10t + 1)i + \left(\frac{2}{3}t^3 - 2t^2 - 4\right)j\) | A1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| When \(t = 3\): \(v_3 = 25i - 4j\) | M1 | |
| \(-5i + 12j = 0.5(v - (25i - 4j))\) | ft their \(v_3\) | |
| \(v = 15i + 20j\) | M1 A1 ft | |
| \( | v | = \sqrt{(15^2 + 20^2)} = 25\) (ms\(^{-1}\)) |
**Part (a):**
N2L: $(6t - 5)i + (t^2 - 2t)j = 0.5a$ | M1 |
$a = (12t - 10)i + (2t^2 - 4t)j$ | A1 |
$v = (6t^2 - 10t)i + \left(\frac{2}{3}t^3 - 2t^2\right)j$ (+C) | M1 A1 ft + A1 ft |
| | ft their $a$
$v = (6t^2 - 10t + 1)i + \left(\frac{2}{3}t^3 - 2t^2 - 4\right)j$ | A1 | [6]
**Part (b):**
When $t = 3$: $v_3 = 25i - 4j$ | M1 |
$-5i + 12j = 0.5(v - (25i - 4j))$ | | ft their $v_3$
$v = 15i + 20j$ | M1 A1 ft |
$|v| = \sqrt{(15^2 + 20^2)} = 25$ (ms$^{-1}$) | cso M1 A1 | [6] [12]\begin{enumerate}
\item A particle $P$ of mass 0.5 kg is moving under the action of a single force $\mathbf { F }$ newtons. At time $t$ seconds,
\end{enumerate}
$$\mathbf { F } = ( 6 t - 5 ) \mathbf { i } + \left( t ^ { 2 } - 2 t \right) \mathbf { j }$$
The velocity of $P$ at time $t$ seconds is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = 0 , \mathbf { v } = \mathbf { i } - 4 \mathbf { j }$.\\
(a) Find $\mathbf { v }$ at time $t$ seconds.
When $t = 3$, the particle $P$ receives an impulse ( $- 5 \mathbf { i } + 12 \mathbf { j }$ ) N s.\\
(b) Find the speed of $P$ immediately after it receives the impulse.\\
\hfill \mbox{\textit{Edexcel M2 2008 Q4 [12]}}