| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Impulse from velocity change |
| Difficulty | Standard +0.3 This is a standard M3 impulse-momentum question requiring vector resolution in two perpendicular directions and Pythagoras' theorem to find magnitude and direction. The setup is straightforward with given speeds and angle, requiring routine application of impulse-momentum principles without novel insight. Slightly above average difficulty due to the vector component work, but well within typical M3 scope. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pm(5.4\cos45° - 8.7)\) | M1 | For attempting to find \(\Delta v\) in i direction |
| \(I\cos\theta = \pm 0.4(5.4\cos45° - 8.7)\) | M1, A1 | For using \(I = m(\Delta v)\) in i direction \((= \mp 1.953)\) |
| \(I\sin\theta = 0.4 \times 5.4\sin45\) | B1 | \((= 1.527)\) |
| \(I = \sqrt{(1.527^2 + 1.953^2)}\) or \(\theta = \tan^{-1}[1.527/(-1.953)]\) | M1 | For using Pythagoras or trig. |
| Magnitude is \(2.48\) kgms\(^{-1}\) | A1 | |
| Direction is \(142°\) to original dir'n | A1 | [7] Accept \(\theta = 38.0°\) with \(\theta\) shown appropriately |
| OR: \(I = 0.4(5.4^2 + 8.7^2 - 2\times5.4\times8.7\cos45°)^{1/2}\) | M1, M1 | For using Impulse \(=\) mass \(\times \Delta v\); For appropriate use of cosine rule |
| Magnitude is \(2.48\) kgms\(^{-1}\) | A1 | |
| \(\sin\theta/5.4 = \sin45°/6.1976\) | A1, M1, A1, A1 | For appropriate use of sine rule |
| \(\theta = 38.0°\) |
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pm(5.4\cos45° - 8.7)$ | M1 | For attempting to find $\Delta v$ in **i** direction |
| $I\cos\theta = \pm 0.4(5.4\cos45° - 8.7)$ | M1, A1 | For using $I = m(\Delta v)$ in **i** direction $(= \mp 1.953)$ |
| $I\sin\theta = 0.4 \times 5.4\sin45$ | B1 | $(= 1.527)$ |
| $I = \sqrt{(1.527^2 + 1.953^2)}$ or $\theta = \tan^{-1}[1.527/(-1.953)]$ | M1 | For using Pythagoras or trig. |
| Magnitude is $2.48$ kgms$^{-1}$ | A1 | |
| Direction is $142°$ to original dir'n | A1 | [7] Accept $\theta = 38.0°$ with $\theta$ shown appropriately |
| **OR:** $I = 0.4(5.4^2 + 8.7^2 - 2\times5.4\times8.7\cos45°)^{1/2}$ | M1, M1 | For using Impulse $=$ mass $\times \Delta v$; For appropriate use of cosine rule |
| Magnitude is $2.48$ kgms$^{-1}$ | A1 | |
| $\sin\theta/5.4 = \sin45°/6.1976$ | A1, M1, A1, A1 | For appropriate use of sine rule |
| $\theta = 38.0°$ | | |
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\includegraphics[max width=\textwidth, alt={}, center]{5bb3bd29-a2eb-4124-802c-fb17b68c50e4-2_246_693_278_731}
A particle $P$ of mass 0.4 kg moving in a straight line has speed $8.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. An impulse applied to $P$ deflects it through $45 ^ { \circ }$ and reduces its speed to $5.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see diagram). Calculate the magnitude and direction of the impulse exerted on $P$.\\
$2 \quad O$ is a fixed point on a horizontal straight line. A particle $P$ of mass 0.5 kg is released from rest at $O$. At time $t$ seconds after release the only force acting on $P$ has magnitude $\left( 1 + k t ^ { 2 } \right) \mathrm { N }$ and acts horizontally and away from $O$ along the line, where $k$ is a positive constant.\\
(i) Find the speed of $P$ in terms of $k$ and $t$.\\
(ii) Given that $P$ is 2 m from $O$ when $t = 1$, find the value of $k$ and the time taken by $P$ to travel 20 m from $O$.
\hfill \mbox{\textit{OCR M3 2006 Q1 [7]}}