| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Moderate -0.3 This is a straightforward two-part mechanics question requiring standard application of impulse-momentum theorem and kinetic energy formula. Part (a) involves vector subtraction and magnitude calculation, while part (b) requires computing KE before and after using ½mv². No problem-solving insight needed, just routine application of formulas with given values. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03c Momentum in 2D: vector form6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(\mathbf{I} = m\mathbf{v} - m\mathbf{u}\) | M1 | Used |
| \(= 0.6(2\mathbf{i}+3\mathbf{j}-4\mathbf{i}+2\mathbf{j})\) | A1 | Correct unsimplified; allow subtraction wrong way round, or simplified equivalent |
| \(= 0.6(-2\mathbf{i}+5\mathbf{j})\) \(\quad (-1.2\mathbf{i}+3\mathbf{j})\) | ||
| \( | \mathbf{I} | = 0.6\sqrt{4+25}\) |
| \(= 3.23\) | A1 | Or better \(\frac{3}{5}\sqrt{29}\), 3.231098... condone \(0.6\sqrt{29}\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{KE lost} = \frac{1}{2}\times0.6\times\left(\ | 4\mathbf{i}-2\mathbf{j}\ | ^2 - \ |
| \(= \frac{1}{2}\times0.6\times(20-13)\) | A1 | Correct unsimplified; accept \(\pm\); allow after approximation e.g. \(\frac{1}{2}\times0.6\times(4.5^2-3.6^2)\) |
| \(= 2.1\) (J) | A1 | CAO |
| [3] |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{I} = m\mathbf{v} - m\mathbf{u}$ | M1 | Used |
| $= 0.6(2\mathbf{i}+3\mathbf{j}-4\mathbf{i}+2\mathbf{j})$ | A1 | Correct unsimplified; allow subtraction wrong way round, or simplified equivalent |
| $= 0.6(-2\mathbf{i}+5\mathbf{j})$ $\quad (-1.2\mathbf{i}+3\mathbf{j})$ | | |
| $|\mathbf{I}| = 0.6\sqrt{4+25}$ | M1 | Use of Pythagoras on their impulse |
| $= 3.23$ | A1 | Or better $\frac{3}{5}\sqrt{29}$, 3.231098... condone $0.6\sqrt{29}$ |
| **[4]** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{KE lost} = \frac{1}{2}\times0.6\times\left(\|4\mathbf{i}-2\mathbf{j}\|^2 - \|2\mathbf{i}+3\mathbf{j}\|^2\right)$ | M1 | Change in KE; terms of correct structure; subtract in either order |
| $= \frac{1}{2}\times0.6\times(20-13)$ | A1 | Correct unsimplified; accept $\pm$; allow after approximation e.g. $\frac{1}{2}\times0.6\times(4.5^2-3.6^2)$ |
| $= 2.1$ (J) | A1 | CAO |
| **[3]** | | |
---\begin{enumerate}
\item A particle $P$ of mass 0.6 kg is moving with velocity ( $4 \mathbf { i } - 2 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$ when it receives an impulse $\mathbf { I } \mathrm { N }$ s. Immediately after receiving the impulse, $P$ has velocity ( $2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\end{enumerate}
Find\\
(a) the magnitude of $\mathbf { I }$,\\
(b) the kinetic energy lost by $P$ as a result of receiving the impulse.\\
\hfill \mbox{\textit{Edexcel M2 2015 Q1 [7]}}