| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2021 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Collision followed by wall impact |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring collision analysis with conservation of momentum, coefficient of restitution, friction work-energy calculations, and geometric series summation. While it involves several connected concepts and extended reasoning across parts (b)-(d), each individual step uses standard A-level Further Mechanics techniques without requiring novel insight. The geometric progression relationship in part (c) and the series sum in part (d) are moderately challenging but follow predictable patterns for this topic. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | Let the block move x metres before coming to rest. |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| max | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2µg | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| max | M1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| m | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0 =𝑣𝑣 +2 (−𝜇𝜇𝑚𝑚) 𝑥𝑥⟹𝑥𝑥 =2𝜇𝜇𝜇𝜇 | [3] | |
| (b) | mu=mv +8mv | |
| S B | M1* | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| B S | M1* | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑢𝑢 =𝑣𝑣𝑠𝑠+8 (0.8𝑢𝑢+𝑣𝑣𝑠𝑠) =9𝑣𝑣𝑠𝑠+6.4𝑢𝑢 | M1dep* | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| S B | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| B has speed 0.2u away from the wall | A1 | 2.4 |
| Answer | Marks |
|---|---|
| (c) | Each time S returns for impact it has 3 the speed it had |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | M1 | 3.5a |
| Answer | Marks | Guidance |
|---|---|---|
| distance moved after the previous impact. | A1 | 2.2a |
| Answer | Marks |
|---|---|
| squaring comes from) | May consider |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | After first impact, speed of block is 11.2×0.2=2.24 | B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| n=1 25 | A1 | 2.2a |
Question 6:
6 | (a) | Let the block move x metres before coming to rest.
1mv2 −Fx=0
2 | M1 | 1.1 | Work-energy principle – correct
number of terms
Since block is sliding, F =F =µmg
max | M1 | 3.4 | Use of F =µR
v2
1mv2 −µmgx=0⇒x=
2 2µg | A1 | 1.1 | N.B. answer given.
Alternative method:
Since block is sliding, F =F =µmg
max | M1 | M1 | 3.4 | 3.4
µmg
a=− =−µg
m | M1 | 1.1
So
2 | A1 | 1.1
2 2 𝑣𝑣
0 =𝑣𝑣 +2 (−𝜇𝜇𝑚𝑚) 𝑥𝑥⟹𝑥𝑥 =2𝜇𝜇𝜇𝜇 | [3]
(b) | mu=mv +8mv
S B | M1* | 3.3 | Conservation of linear momentum
–correct number of terms (allow
sign errors)
v −v =0.8u
B S | M1* | 3.3 | Newton’s experimental law – must
be consistent with CoLM (so signs
of in the two equations must be
different)
𝑢𝑢 =𝑣𝑣𝑠𝑠+8 (0.8𝑢𝑢+𝑣𝑣𝑠𝑠) =9𝑣𝑣𝑠𝑠+6.4𝑢𝑢 | M1dep* | 3.4 | Att𝑣𝑣e 𝑆𝑆 mpt at eliminating either
variable – dependent on both
previous M marks
⇒v =−0.6u and v =0.2u
S B | A1 | 1.1 | Ignore incorrect signs.
S has speed 0.6u towards the wall
B has speed 0.2u away from the wall | A1 | 2.4 | Both correct. Accept other
appropriate descriptions of
direction (e.g. ‘opposite to original
direction of travel’, etc.)
[5]
(c) | Each time S returns for impact it has 3 the speed it had
5
previously; therefore after impact, the block will have also
have 3 the speed it had just after the previous impact …
5 | M1 | 3.5a | Argument using their value of v
S
from (b) – must relate this value to
B
(3)2
… so by part (a), the block will move only = 9 of the
5 25
distance moved after the previous impact. | A1 | 2.2a | Must reference result in part (a) (or
convincingly explain where the
squaring comes from) | May consider
that the ratio of
successive
distances
travelled by B
is v2
[2]
(d) | After first impact, speed of block is 11.2×0.2=2.24 | B1ft | 3.4 | Follow through their value of v
B
from (b)
2.242
So x = =1.792
1 2⋅1⋅(9.8)
7 | M1 | 1.1 | Using given result in (a) to find
distance travelled after first
collision
∞ 1.792
∑x = =2.8(m)
n 1− 9
n=1 25 | A1 | 2.2a | AG Use of infinite sum of a GP to
derive required result
[3]
PMT
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recorded or monitored6 A block rests on a horizontal surface. The coefficient of friction between the block and the surface is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item Show that if the block is given an initial speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it will move a distance of $\frac { \mathrm { v } ^ { 2 } } { 2 \mu \mathrm {~g} }$ before coming to rest.
Block B rests on the same horizontal surface as a sphere S . On the other side of S is a vertical wall, as shown below. The mass of $B$ is 8 times the mass of $S$.\\
\includegraphics[max width=\textwidth, alt={}, center]{b3e369f4-13f7-457b-9a43-04ed2e2a2bba-8_211_1013_662_244}
S is projected directly towards B with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and hits B . It is given that
\begin{itemize}
\item the coefficient of restitution between S and B is 0.8 ,
\item collisions between S and the wall are perfectly elastic,
\item the wall is perpendicular to the direction of motion of S and B .
\end{itemize}
Furthermore, you should model the contact between B and the surface as rough and model the contact between S and the surface as smooth.
\item Determine, in terms of $u$, expressions for
\begin{itemize}
\item the speed of S
\item the speed of B\\
immediately after the first collision between S and B . In each case stating the corresponding direction of motion.
\end{itemize}
It is given that B has sufficient time to come to rest before each subsequent collision with S .\\
Let $\mathrm { X } _ { \mathrm { n } }$ be the distance B moves after the $n$th impact between S and B .
\item Explain why $\mathrm { x } _ { \mathrm { n } + 1 } = \frac { 9 } { 25 } \mathrm { x } _ { \mathrm { n } }$.
\item Given that $u = 11.2$ and the coefficient of friction between B and the surface is $\frac { 1 } { 7 }$, show that B will travel a total distance that cannot exceed 2.8 m .
\section*{END OF QUESTION PAPER}
\section*{OCR \\
Oxford Cambridge and RSA}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2021 Q6 [13]}}