| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Range of coefficient of restitution |
| Difficulty | Standard +0.3 This is a standard Further Maths mechanics collision problem requiring conservation of momentum and Newton's restitution law. Part (a) involves routine algebraic manipulation of two simultaneous equations (5 marks for showing a given result). Part (b) tests understanding through three straightforward applications: coalescence (trivial, e=0), direction reversal (simple inequality), and energy loss (standard calculation). While it requires multiple techniques, all are textbook applications with no novel insight needed. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03i Coefficient of restitution: e |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | 2(2)+3(−1)=2v +3v and v −v =3e |
| Answer | Marks |
|---|---|
| 5 5 | M1 |
| Answer | Marks |
|---|---|
| [5] | 3.3 |
| Answer | Marks |
|---|---|
| 2.2a | Use of CLM – correct number of terms |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (b) | (i) |
| [1] | 3.4 | |
| (ii) | 1 |
| Answer | Marks |
|---|---|
| 9 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 2.5 | Setting their expression forv <0(condone 0 or =) |
| Answer | Marks |
|---|---|
| (iii) | 1 1 |
| Answer | Marks |
|---|---|
| 3 | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Correct expression for the KE after collision with |
Question 7:
7 | (a) | 2(2)+3(−1)=2v +3v and v −v =3e
A B B A
1 1
v = (1−9e) v = (1+6e)
A B
5 5 | M1
M1
A1
M1
A1
[5] | 3.3
3.3
1.1
1.1
2.2a | Use of CLM – correct number of terms
Use of NEL (e on correct side) – correct number of
terms
Use of NEL must be consistent with CLM
( 2−(−1))
oe NEL v −v =−e
A B
Solve simultaneously to find v or v
A B
AG for v - sufficient working must be shown
B
7 | (b) | (i) | 0.2 (ms- 1) | B1
[1] | 3.4
(ii) | 1
v <0⇒ (1−9e)<0
A
5
1
<e„ 1
9 | M1
A1
[2] | 3.1b
2.5 | Setting their expression forv <0(condone 0 or =)
A
1
– or for v >0 if v = (9e−1)from reversing the
A A
5
motion of A in (a)
www
(iii) | 1 1
KE before collision: (2)(2)2 + (3)(−1)2 (=5.5)
2 2
1 1−9e 2 1 1+6e 2
KE after collision: (2) + (3)
2 5 2 5
11 1 ( ) 3 ( )
− 1−18e+81e2 − 1+12e+36e2 =3
2 25 50
2
27−27e2 =15⇒e=
3 | B1
B1ft
M1
A1
[4] | 1.1
1.1
3.3
1.1 | Correct expression for the KE after collision with
correct v and their v
B A
Setting up an equation in e using KE before – KE
after = 3 with correct number of terms – dependent on
one previous B markTwo small uniform smooth spheres A and B, of masses 2 kg and 3 kg respectively, are moving in opposite directions along the same straight line towards each other on a smooth horizontal surface. Sphere A has speed 2 m s$^{-1}$ and B has speed 1 m s$^{-1}$ before they collide. The coefficient of restitution between A and B is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that the velocity of B after the collision, in the original direction of motion of A, is $\frac{1}{5}(1 + 6e)$ m s$^{-1}$ and find a similar expression for the velocity of A after the collision. [5]
\item The following three parts are independent of each other, and each considers a different scenario regarding the collision between A and B.
\begin{enumerate}[label=(\roman*)]
\item In the collision between A and B the spheres coalesce to form a combined body C. State the speed of C after the collision. [1]
\item In the collision between A and B the direction of motion of A is reversed. Find the range of possible values of $e$. [2]
\item The total loss in kinetic energy due to the collision is 3 J. Determine the value of $e$. [4]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q7 [12]}}