| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring multiple conservation principles (momentum in two directions, restitution), geometric reasoning about collision angles, and careful tracking through multiple collision events. However, it's highly structured with guided parts leading to specific results, making it more accessible than open-ended proof questions. The oblique collision setup and symmetry conditions require solid understanding but follow standard FM techniques. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | (a) | Because if the collision where inelastic then the |
| Answer | Marks |
|---|---|
| with B again (which we know it does). | B1 |
| [1] | 2.4 |
| (b) | Recollide at same point => v = 0 |
| Answer | Marks |
|---|---|
| u s i n m | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | Seen in solution |
| Answer | Marks |
|---|---|
| B | If v is used must be consistent in |
| Answer | Marks |
|---|---|
| (c) | Velocity of A in y direction unchanged |
| Answer | Marks |
|---|---|
| 5 e 3 | B1 |
| Answer | Marks |
|---|---|
| [7] | 3.4 |
| Answer | Marks |
|---|---|
| 3.2a | v = ucos |
| Answer | Marks |
|---|---|
| root | FT if sin and cos confusion in part b |
| Answer | Marks |
|---|---|
| (d) | In practice the plane cannot be smooth |
| Answer | Marks | Guidance |
|---|---|---|
| or NEL is only an approximation | B1 | |
| [1] | 3.5b | Any sensible limitation |
Question 8:
8 | (a) | Because if the collision where inelastic then the
velocity of A after the collision would be 0 (ie A
would ‘stick to the wall’) and so it would not collide
with B again (which we know it does). | B1
[1] | 2.4
(b) | Recollide at same point => v = 0
Ax
Com: usin = mv
B
v
NEL: e B =
u s i n
u s i n
m 1
=> e = =
u s i n m | B1
M1
M1
A1
[4] | 3.1b
3.3
3.3
1.1 | Seen in solution
or v + mv on RHS
Ax B
v v −
or e Bu A x =
s i n
AG
(v = eusin)
B | If v is used must be consistent in
Ax
both equations
(c) | Velocity of A in y direction unchanged
Either A’s velocity multiplied by (–)e or B’s
velocity multiplied by (–)(5/9)e after collision
with wall.
d d d d
+ or +
u c o s e u c o s e u s i n 5 e 2 u s i n
9
d d d d
+ = +
u c o s e u c o s e u s i n 5 e 2 u s i n
9
5 t a n 5 92
5 t a n + = +
e e e
2 5 2 5
2 e e 5 e 9 0 + − − =
2 2
2 5 2 e 1 e 5 1 8 0 + − =
6 3
(5e + 6)(5e – 3) = 0 => e = − o r
5 5
3 1 5
0 e 1 e = m = =
5 e 3 | B1
B1
M1
M1
M1
A1
A1
[7] | 3.4
3.4
3.1b
2.1
1.1
1.1
3.2a | v = ucos
Ay
Ignore inconsistency between
velocity and speed here.
Calculating time of travel for
either A or B from point of
collision to wall and back
Equating expressions that are in
terms of theta, e or m and a single
velocity
Reducing equation to 3 term
quadratic in e or m
Could be BC. –1.2 or 0.6
AG. Correctly rejecting negative
root | FT if sin and cos confusion in part b
Must be the correct component
Could see eg d – r
− 1 5 2 0 2 5
5 0
(d) | In practice the plane cannot be smooth
or the discs may not be the same size/radius
or air resistance may be significant
or NEL is only an approximation | B1
[1] | 3.5b | Any sensible limitation
PMT
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8 Two smooth circular discs, $A$ and $B$, have equal radii and are free to move on a smooth horizontal plane. The masses of $A$ and $B$ are 1 kg and $m \mathrm {~kg}$ respectively. $B$ is initially placed at rest with its centre at the origin, $O$. $A$ is projected towards $B$ with a velocity of $u \mathrm {~ms} ^ { - 1 }$ at an angle of $\theta$ to the negative $y$-axis where $\tan \theta = \frac { 5 } { 2 }$. At the instant of collision the line joining their centres lies on the $x$-axis.
There are two straight vertical walls on the plane. One is perpendicular to the $x$-axis and the other is perpendicular to the $y$-axis. The walls are an equal distance from $O$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{857eca7f-c42d-49a9-ac39-a2fb5bcb9cd5-7_944_1241_694_242}
After $A$ and $B$ have collided with each other, each of them goes on to collide with a wall. Each then rebounds and they collide again at the same place as their first collision, with disc $B$ again at $O$.
The coefficient of restitution between $A$ and $B$ is denoted by $e$. The coefficient of restitution between $A$ and the wall that it collides with is also $e$ while the coefficient of restitution between $B$ and the wall that it collides with is $\frac { 5 } { 9 } e$.
It is assumed that any resistance to the motion of $A$ and $B$ may be ignored.
\begin{enumerate}[label=(\alph*)]
\item Explain why it must be the case that the collision between $A$ and the wall that it collides with is not inelastic.
\item Show that $\mathrm { e } = \frac { 1 } { \mathrm {~m} }$.
\item Show that $m = \frac { 5 } { 3 }$.
\item State one limitation of the model used.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2022 Q8 [13]}}