| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Loss of kinetic energy |
| Difficulty | Standard +0.8 This is a multi-part Further Mechanics collision question requiring conservation of momentum, Newton's restitution law, and kinetic energy analysis. While the individual techniques are standard for Further Maths, the algebraic manipulation is substantial (especially showing the given λ expression), and the final interpretation requires understanding of elastic/inelastic collisions. The 15-mark allocation and multi-step nature place it above average difficulty, but it follows a predictable structure for Further Mechanics collision problems. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | 3m(2u)(cid:14)m((cid:16)u)(cid:32)3mv(cid:14)mw |
| Answer | Marks |
|---|---|
| 4 | *M1 |
| Answer | Marks |
|---|---|
| [6] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Attempt at use of conservation of linear |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (ii) | (a) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | B1 | |
| [1] | 1.1 | m |
| 5 | (ii) | (b) |
| Answer | Marks |
|---|---|
| 2 (cid:169)4(cid:185) 2 (cid:169)4(cid:185) | B1FT |
| Answer | Marks |
|---|---|
| [2] | i |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | B1 for each term correct for their part (i) | |
| 5 | (iii) | p |
| Answer | Marks |
|---|---|
| 52 | B1FT |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | (cid:79)T (cid:32)T |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (iv) | (a) |
| (b) | (cid:79)(cid:32)1((cid:159)e(cid:32)1) so the collision is perfectly |
| Answer | Marks |
|---|---|
| inelastic | B1 |
| Answer | Marks |
|---|---|
| [3] | 2.4 |
| Answer | Marks |
|---|---|
| 2.4 | oe e.g. there is no loss in kinetic energy |
| Answer | Marks |
|---|---|
| oe e.g. the particles coalesce | language must be precise |
| Answer | Marks |
|---|---|
| (b) | a |
Question 5:
5 | (i) | 3m(2u)(cid:14)m((cid:16)u)(cid:32)3mv(cid:14)mw
v(cid:16)w(cid:32)(cid:16)e(cid:11)2u(cid:16)((cid:16)u)(cid:12)
1
v(cid:32) u(cid:11)5(cid:16)3e(cid:12)
4 | *M1
A1
*M1
A1
dep*M1
A1
[6] | 1.1a
1.1
1.1a
1.1
3.1b
1.1 | Attempt at use of conservation of linear
momentum
Attempt at use of restitution equation,
must be correct way round
Mustn be consistent with directions used
for conservation of linear momentum
eSolving simultaneous equations
5 | (ii) | (a) | 1 1
T (cid:32) (3m)(2u)2(cid:14) m((cid:16)u)2
before
2 2 | B1
[1] | 1.1 | m
5 | (ii) | (b) | 2 2
1 (cid:167)u(cid:183) 1 (cid:167)u(cid:183)
T (cid:32) (3m) (cid:11)5(cid:16)3e(cid:12)2 (cid:14) (m) (5(cid:14)9e)2
after (cid:168) (cid:184) (cid:168) (cid:184)
2 (cid:169)4(cid:185) 2 (cid:169)4(cid:185) | B1FT
e
B1FT
[2] | i
c
1.1
1.1 | B1 for each term correct for their part (i)
5 | (iii) | p
13 3 1
(cid:79)mu2 (cid:32) mu2(5(cid:16)3e)2 (cid:14) mu2(5(cid:14)9e)2
2 32 32S
27e2(cid:14)25
(cid:79)(cid:32)
52 | B1FT
M1
E1
[3] | 3.4
1.1
1.1 | (cid:79)T (cid:32)T
before after
Rearrange to make (cid:79) the subject
AG cancelling of mu2 factor must be seen
and expansion of brackets
5 | (iv) | (a)
(b) | (cid:79)(cid:32)1((cid:159)e(cid:32)1) so the collision is perfectly
elastic
25
(cid:79)(cid:32) (cid:159)e(cid:32)0 so the collision is perfectly
52
inelastic | B1
M1
A1
[3] | 2.4
1.1
2.4 | oe e.g. there is no loss in kinetic energy
Substitute (cid:79) and solve for e
oe e.g. the particles coalesce | language must be precise
language must be precise
--- 5(i) ---
5(i)
t
f
a
r
D
5(ii)(a)
5(ii)(b)
--- 5(iii) ---
5(iii)
t
f
5(iv)(a)
(b) | a
r
D\includegraphics{figure_5}
The masses of two spheres $A$ and $B$ are $3m$ kg and $m$ kg respectively. The spheres are moving towards each other with constant speeds $2u \, \text{m s}^{-1}$ and $u \, \text{m s}^{-1}$ respectively along the same straight line towards each other on a smooth horizontal surface (see diagram). The two spheres collide and the coefficient of restitution between the spheres is $e$. After colliding, $A$ and $B$ both move in the same direction with speeds $v \, \text{m s}^{-1}$ and $w \, \text{m s}^{-1}$, respectively.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $v$ in terms of $e$ and $u$. [6]
\item Write down unsimplified expressions in terms of $e$ and $u$ for
\begin{enumerate}[label=(\alph*)]
\item the total kinetic energy of the spheres before the collision, [1]
\item the total kinetic energy of the spheres after the collision. [2]
\end{enumerate}
\item Given that the total kinetic energy of the spheres after the collision is $\lambda$ times the total kinetic energy before the collision, show that
$$\lambda = \frac{27e^2 + 25}{52}.$$
[3]
\item Comment on the cases when
\begin{enumerate}[label=(\alph*)]
\item $\lambda = 1$,
\item $\lambda = \frac{25}{52}$. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS Q5 [15]}}