| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, direction deflected given angle |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring resolution of velocities along/perpendicular to line of centres, application of conservation of momentum in both directions, use of Newton's experimental law, and algebraic manipulation to derive an inequality involving mass ratios. Part (b) requires calculating energy loss. The multi-step nature, oblique collision geometry, and proof element make this significantly harder than standard A-level mechanics, though it follows a recognizable framework for Further Maths collision problems. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | u =3, u =−2 |
| Answer | Marks |
|---|---|
| A A | B1 |
| Answer | Marks |
|---|---|
| [6] | 3.3 |
| Answer | Marks |
|---|---|
| 2.1 | Resolving horizontal |
| Answer | Marks |
|---|---|
| AG | Signs may be reversed throughout |
| Answer | Marks | Guidance |
|---|---|---|
| ‘ | (b) | 1 1 |
Total initial KE = ×2×52+ ×6×42 =73
| Answer | Marks |
|---|---|
| 2 2 | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Perpendicular components found |
| Answer | Marks |
|---|---|
| from (a). | NB If method mark for |
Question 7:
7 | (a) | u =3, u =−2
Ax Bx
m ×3+m ×−2=m v +m ×0
A B A Ax B
2m
v =3− B
Ax
m
A
0−v
e= Ax or v =−5e
3−−2 Ax
2m 3 m 3
e≥0⇒ B − ≥0⇒ B ≥
5m 5 m 2
A A
2m 3 m
e≤1⇒ B − ≤1⇒ B ≤4
5m 5 m
A A | B1
M1
A1
M1
A1
A1
[6] | 3.3
3.4
1.1
3.4
2.1
2.1 | Resolving horizontal
components of u and u . Accept
A B
and
but must have opposite signs or 𝜋𝜋
𝑢𝑢di 𝐴𝐴 re=ct5iocnoss in𝛼𝛼dicate𝑢𝑢d 𝐵𝐵 on= d−ia4grcaoms 3 .
Conservation of momentum
Restitution
AG
AG | Signs may be reversed throughout
May be seen in (b)
( 2m )
0− 3− B
m 2m 3
e= A = B −
3−−2 5m 5
A
7
‘ | (b) | 1 1
Total initial KE = ×2×52+ ×6×42 =73
2 2
v =u , v =u =2 3
Ay Ay By By
v =−3
Ax
KE Loss =
73− (1 ×2×(32+42)+ 1 ×6×( 2 3 )2 ) =12 J
2 2 | B1
M1
M1
A1
[4] | 1.1
3.4
3.4
1.1 | Perpendicular components found
and unchanged
Using their formula for v
Ax
from (a). | NB If method mark for
conservation of momentum not seen
in (a) then award M1 in (a) if either
m ×3+m ×−2=m v or
A B A Ax
2×3+6×−2=2v seen here
Ax
If method mark for restitution not
seen in (a) then award M1 in (a) if
seen here.7 Two smooth circular discs $A$ and $B$ of masses $m _ { A } \mathrm {~kg}$ and $m _ { B } \mathrm {~kg}$ respectively are moving on a horizontal plane. At the instant before they collide the velocities of $A$ and $B$ are as follows, as shown in the diagram below.
\begin{itemize}
\item The velocity of $A$ is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\alpha$ to the line of centres, where $\tan \alpha = \frac { 4 } { 3 }$.
\item The velocity of $B$ is $4 \mathrm {~ms} ^ { - 1 }$ at an angle of $\frac { 1 } { 3 } \pi$ radians to the line of centres.\\
\includegraphics[max width=\textwidth, alt={}, center]{c6445493-9802-46ca-b7eb-7738a831d9ee-5_469_873_548_274}
\end{itemize}
The direction of motion of $B$ after the collision is perpendicular to the line of centres.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { 3 } { 2 } \leqslant \frac { m _ { B } } { m _ { A } } \leqslant 4$.
\item Given that $\mathrm { m } _ { \mathrm { A } } = 2$ and $\mathrm { m } _ { \mathrm { B } } = 6$, find the total loss of kinetic energy as a result of the collision.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2021 Q7 [10]}}