| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics B AS (Further Mechanics B AS) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| 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 an oblique collision problem requiring resolution of velocities, conservation of momentum in two perpendicular directions, Newton's experimental law, and energy calculations. While it involves multiple steps and careful geometric reasoning (particularly part (a) requiring physical insight about perpendicular motion), the techniques are standard for Further Mechanics. The given angle constraint (tan α = 0.75) simplifies calculations. This is harder than typical A-level mechanics but represents a well-structured Further Maths question rather than requiring exceptional insight. |
| 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 impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | Component of momentum of A parallel to line of |
| Answer | Marks | Guidance |
|---|---|---|
| stationary). | B1 | 2.4 |
| along loc | Allow |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | COM: 0.5×0.5×cos𝛼−0.4×0.6cos60° = | |
| 0.4𝑏 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| b = 0.2 | A1 | 1.1 |
| Comp of vel of B perp loc is 0.6sin60° | M1 | 1.1 |
| β = 68.9 | A1 | 1.1 |
| Answer | Marks |
|---|---|
| (c) | 1 1 |
| Answer | Marks |
|---|---|
| 2 | M1 |
| A1 | 1.2 |
| 1.1 | Could consider total Kinetic |
| Answer | Marks | Guidance |
|---|---|---|
| = 0.05 (J) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (d) | NEL:𝑏−𝑎 = −𝑒(−0.6×cos60°−0.5×cos𝛼) | M1 |
| 0.2 (−0) = −𝑒(−0.6×0.5−0.5×0.8) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | A1 | 1.1 |
Question 4:
4 | (a) | Component of momentum of A parallel to line of
centres is greater (and opposite) that of B, (so after
impact total mom must be to right, so B cannot be
stationary). | B1 | 2.4 | Some consideration of momentum
along loc | Allow
0.50.5cos0.40.6cos60
[1]
(b) | COM: 0.5×0.5×cos𝛼−0.4×0.6cos60° =
0.4𝑏 | M1 | 3.3 | Where b is vel of B along
loc after collision
b = 0.2 | A1 | 1.1
Comp of vel of B perp loc is 0.6sin60° | M1 | 1.1 | May see tan𝛽 = 0.6sin60°÷0.2 | SOI
β = 68.9 | A1 | 1.1 | 68.948…
[4]
(c) | 1 1
×0.5×(0.5cos𝛼)2+ ×0.4×(0.6cos60°)2−
2 2
1
×0.4×𝑏2
2 | M1
A1 | 1.2
1.1 | Could consider total Kinetic
Energy rather than just along
LoC
= 0.05 (J) | A1 | 1.1
[3]
(d) | NEL:𝑏−𝑎 = −𝑒(−0.6×cos60°−0.5×cos𝛼) | M1 | 3.3
0.2 (−0) = −𝑒(−0.6×0.5−0.5×0.8) | A1 | 1.1
2
e = or 0.286
7 | A1 | 1.1
[3]4 Two small smooth discs, A of mass 0.5 kg and B of mass 0.4 kg , collide while sliding on a smooth horizontal plane.
Immediately before the collision A and B are moving towards each other, A with speed $0.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $B$ with speed $0.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Before the collision the direction of motion of A makes an angle $\alpha$ with the line of centres, where $\tan \alpha = 0.75$, and the direction of motion of B makes an angle of $60 ^ { \circ }$ with the line of centres, as shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{37798594-8cb0-48aa-8401-090f09e25dff-4_506_938_687_244}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
After the collision, one of the discs moves in a direction perpendicular to the line of centres, and the other disc moves in a direction making an angle $\beta$ with the line of centres.
\begin{enumerate}[label=(\alph*)]
\item Explain why the disc which moves perpendicular to the line of centres must be A .
\item Determine the value of $\beta$.
\item Determine the kinetic energy lost in the collision.
\item Determine the value of the coefficient of restitution between A and B .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2021 Q4 [11]}}