| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Impulse and momentum, vector form |
| Difficulty | Standard +0.3 This is a straightforward application of impulse-momentum theorem and kinetic energy formulas with vector components. Students need to find change in momentum (impulse), calculate magnitude and angle using standard vector methods, then compute KE before and after. All steps are routine for Further Mechanics with no conceptual challenges or novel problem-solving required—slightly easier than average A-level due to direct application of formulas. |
| Spec | 1.10c Magnitude and direction: of vectors6.02f KE with vectors: using scalar product6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (a) | I = mv – mu = 2(–3i + j – (5i + 16j)) |
| Answer | Marks |
|---|---|
| 17 | M1 |
| Answer | Marks |
|---|---|
| [6] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Correct use of formula (award if |
| Answer | Marks | Guidance |
|---|---|---|
| required angle | or using the cosine rule on vectors | |
| u,v,I to reach | I | = 34 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (b) | Init KE = 1 ×2× ( 52 +162) |
| Answer | Marks |
|---|---|
| Loss = 281 – 10 = 271 J | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | 281 J |
Question 2:
2 | (a) | I = mv – mu = 2(–3i + j – (5i + 16j))
= 2(–8i – 15j)
I =2 (−8)2+(−15)2
=2 289 =34
I.i −16×1
cosθ= =
I i 34×1
−8
θ=cos−1 =118.1° or 2.06 rad
17 | M1
A1
M1
A1
M1
A1
[6] | 1.1
1.1
1.1
1.1
1.1
1.1 | Correct use of formula (award if
mu – mv)
Allow 16i + 30j
or (−16)2+(−30)2 oe
Attempting to use the dot
product of I and i to find the
required angle | or using the cosine rule on vectors
u,v,I to reach |I| = 34
or use of ordinary trigonometry eg
−30
tanθ=
−16
2 | (b) | Init KE = 1 ×2× ( 52 +162)
2
Final KE = 1 ×2× ( (−3)2 +12)
2
Loss = 281 – 10 = 271 J | M1
M1
A1
[3] | 1.1
1.1
1.1 | 281 J
10 J2 A particle $P$ of mass 2 kg is moving on a large smooth horizontal plane when it collides with a fixed smooth vertical wall. Before the collision its velocity is $( 5 \mathbf { i } + 16 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and after the collision its velocity is $( - 3 \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item The impulse imparted on $P$ by the wall is denoted by INs.
Find the following.
\begin{itemize}
\item The magnitude of $\mathbf { I }$
\item The angle between I and i
\item Find the loss of kinetic energy of $P$ as a result of the collision.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2021 Q2 [9]}}