| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile with bounce or impact |
| Difficulty | Standard +0.3 This is a standard projectile motion question with bounces, requiring routine application of SUVAT equations, coefficient of restitution, impulse-momentum theorem, and geometric series. All techniques are straightforward Further Maths mechanics content with clear signposting and no novel problem-solving required. The multi-part structure and 14 marks indicate moderate length, but each part follows predictable patterns (resolve components, apply e, sum series). |
| Spec | 3.02i Projectile motion: constant acceleration model6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (i) | P |
| Answer | Marks |
|---|---|
| NEL(cid:110)so 0.4(cid:117)39(cid:32)15.6(cid:110) | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (ii) | (cid:110)0.1(cid:11)15.6(cid:16)((cid:16)39)(cid:12)(cid:32)5.46 |
| Answer | Marks |
|---|---|
| [2] | 1.2 |
| 1.1 | Attempt at change in momentum. |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (iii) | (A) |
| Answer | Marks |
|---|---|
| 1 1 1 5 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | N |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (iii) | (B) |
| Answer | Marks |
|---|---|
| n 5 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 2.2a | E |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (iv) | 39 a |
Total time = 13 s
Total distance travelled while bouncing =
| Answer | Marks |
|---|---|
| 676m | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 3.2a | M |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (v) | Carries on after bouncing phase, with |
| horizontal velocity of 52ms–1 | E |
| Answer | Marks |
|---|---|
| [1] | 3.5b |
Question 10:
10 | (i) | P
At projection S
(cid:110)65sin(cid:68)(cid:32)39 and (cid:111)65cos(cid:68)(cid:32)52
After bounce, LM conserved horizontally, so
(cid:111)52
The vertical speed when it bounces is the same
as at projection
NEL(cid:110)so 0.4(cid:117)39(cid:32)15.6(cid:110) | B1
E1
B1
E1
[4] | 3.3
3.4
3.4
1.1 | Both correct
Accept "horizontal speed not
affected by impact"
Use of NEL
10 | (ii) | (cid:110)0.1(cid:11)15.6(cid:16)((cid:16)39)(cid:12)(cid:32)5.46 | M1
A1
[2] | 1.2
1.1 | Attempt at change in momentum.
Accept sign error
All signs correct
10 | (iii) | (A) | 39
0(cid:32)39T (cid:16)5T2 so T (cid:32) AG
1 1 1 5 | M1
A1
[2] | 3.4
1.1 | N
Use of appropriate suvat
10 | (iii) | (B) | 39
T (cid:32)(cid:11)0.4(cid:12)n(cid:16)1(cid:117)
n 5 | M1
A1
[2] | 3.1b
2.2a | E
Recognise geometric progression
Correct answer
10 | (iv) | 39 a
GP with a(cid:32) and r(cid:32)0.4: S (cid:32)
5 (cid:102) 1(cid:16)r
Total time = 13 s
Total distance travelled while bouncing =
676m | M1
A1
A1
[3] | 3.1b
I
1.1
C
3.2a | M
Use of formula
10 | (v) | Carries on after bouncing phase, with
horizontal velocity of 52ms–1 | E
B1
[1] | 3.5bIn this question take $g = 10$.
A smooth ball of mass 0.1 kg is projected from a point on smooth horizontal ground with speed 65 m s$^{-1}$ at an angle $\alpha$ to the horizontal, where $\tan\alpha = \frac{3}{4}$. While it is in the air the ball is modelled as a particle moving freely under gravity. The ball bounces on the ground repeatedly. The coefficient of restitution for the first bounce is 0.4.
\begin{enumerate}[label=(\roman*)]
\item Show that the ball leaves the ground after the first bounce with a horizontal speed of 52 m s$^{-1}$ and a vertical speed of 15.6 m s$^{-1}$. Explain your reasoning carefully. [4]
\item Calculate the magnitude of the impulse exerted on the ball by the ground at the first bounce. [2]
\end{enumerate}
Each subsequent bounce is modelled by assuming that the coefficient of restitution is 0.4 and that the bounce takes no time. The ball is in the air for $T_1$ seconds between projection and bouncing the first time, $T_2$ seconds between the first and second bounces, and $T_n$ seconds between the $(n-1)$th and $n$th bounces.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item \begin{enumerate}[label=(\Alph*)]
\item Show that $T_1 = \frac{39}{5}$. [2]
\item Find an expression for $T_n$ in terms of $n$. [2]
\end{enumerate}
\item According to the model, how far does the ball travel horizontally while it is still bouncing? [3]
\item According to the model, what is the motion of the ball after it has stopped bouncing? [1]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major Q10 [14]}}