| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Oblique collision of spheres |
| Difficulty | Challenging +1.2 This is a standard oblique collision problem requiring conservation of momentum along the line of centres, coefficient of restitution, and perpendicular component preservation. While it involves multiple steps and vector decomposition, the techniques are routine for Further Maths mechanics students. The 'show that' structure and energy loss calculation are typical exam questions, though the three-dimensional reasoning elevates it slightly above average difficulty. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (i) | Perpendicular to line of centres, momentum of |
| Answer | Marks |
|---|---|
| 8 | B1 |
| Answer | Marks |
|---|---|
| [6] | I |
| Answer | Marks |
|---|---|
| 2.1 | Use of NEL. Allow sign errors. |
| Answer | Marks |
|---|---|
| Use of PCLM. Allow sign errors | NEL is Newton's experimental law |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (ii) | Perpendicular to line of centres, momentum |
| Answer | Marks |
|---|---|
| 64 | M1 |
| Answer | Marks |
|---|---|
| [7] | 3.4 |
| Answer | Marks |
|---|---|
| 2.1 | Use of conservation of linear |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (iii) | The component of linear momentum of A |
| Answer | Marks |
|---|---|
| change | E |
| Answer | Marks |
|---|---|
| [1] | 3.5b |
Question 9:
9 | (i) | Perpendicular to line of centres, momentum of
A unchanged, so A has no component of
velocity, so A moves parallel to BA
Taking L to R as positive
NEL along line of centres:
V (cid:16)V (cid:32)e(v(cid:14)vcos60 ) P
A B
3v
V (cid:16)V (cid:32) S
A B 4
v
PCLM V (cid:14)V (cid:32)(cid:16)v(cid:14)vcos60 (cid:32)(cid:16)
A B 2
1 5
Solve: V (cid:32) v (V (cid:32)(cid:16) v) so speed of A
A 8 B 8
1
is vtowards right
8 | B1
E
M1
A1
M1
A1
A1
[6] | I
C3.4
3.3
1.1
3.4
1.1
2.1 | Use of NEL. Allow sign errors.
speed of separation
Must be
speed of approach
Use of PCLM. Allow sign errors | NEL is Newton's experimental law
PCLM is Principle of conservation
of linear momentum
9 | (ii) | Perpendicular to line of centres, momentum
and so velocity of B unchanged:
3
vsin60 (cid:32) v
2
2
1 1 (cid:167) 3v(cid:183)
Final KE of B = mV 2 (cid:14) m(cid:168) (cid:184) =
2 B 2 (cid:168) 2 (cid:184)
(cid:169) (cid:185)
73
mv2
128
1 1 (cid:167)v(cid:183) 2 73
Loss in KE = 2(cid:117) (cid:117)mv2 (cid:16) m (cid:168) (cid:184) (cid:16) mv2
2 2 (cid:169)8(cid:185) 128
27
= mv2
64
27
Percentage loss = (cid:117)100(cid:32)42%
64 | M1
A1
M1
A1
M1
A1
A1
[7] | 3.4
1.1
3.1b
1.1
1.1
I
1.1
C
2.1 | Use of conservation of linear
momentum for B
N
Use of KE using both velocity
components
E
M
All KE terms present
9 | (iii) | The component of linear momentum of A
perpendicular to line of centres does not
change | E
B1
[1] | 3.5b\includegraphics{figure_9}
Fig. 9 shows the instant of impact of two identical uniform smooth spheres, A and B, each with mass $m$. Immediately before they collide, the spheres are sliding towards each other on a smooth horizontal table in the directions shown in the diagram, each with speed $v$. The coefficient of restitution between the spheres is $\frac{1}{2}$.
\begin{enumerate}[label=(\roman*)]
\item Show that, immediately after the collision, the speed of A is $\frac{1}{8}v$. Find its direction of motion. [6]
\item Find the percentage of the original kinetic energy that is lost in the collision. [7]
\item State where in your answer to part (i) you have used the assumption that the contact between the spheres is smooth. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major Q9 [14]}}