| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Two possible trajectories through point |
| Difficulty | Challenging +1.2 This is a substantial multi-part projectile motion question requiring derivation of trajectory equation, discriminant analysis for quadratic in tan α, and interpretation of regions. While it involves several steps and connects algebra to physical meaning, the techniques are standard for Further Maths mechanics: eliminating t from parametric equations, applying discriminant conditions, and sketching parabolic envelopes. The conceptual demand is moderate—recognizing the envelope of trajectories—but execution follows well-established methods. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (i) | x(cid:32)28cos(cid:68)t y(cid:32)28sin(cid:68)t(cid:16)4.9t2 |
| Answer | Marks |
|---|---|
| AG | B1 |
| Answer | Marks |
|---|---|
| [5] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | Both |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (ii) | (A) |
| Answer | Marks |
|---|---|
| 160 | E |
| Answer | Marks |
|---|---|
| [3] | 2.2a |
| Answer | Marks |
|---|---|
| 1.1 | Finding the discriminant |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (ii) | (B) |
| [1] | 1.1 | |
| 12 | (iii) | Labels curve with S |
| Labels region above curve with T | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (iv) | Two real distinct roots means two distinct |
| Answer | Marks |
|---|---|
| trajectory through a point on the curve | B1 |
| Answer | Marks |
|---|---|
| [3] | 2.3 |
| Answer | Marks |
|---|---|
| 2.2a | N |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (v) | Graph gives x = 80 when y = 0, so |
| Answer | Marks |
|---|---|
| Yes, according to the model | B1 |
| [1] | I |
| Answer | Marks |
|---|---|
| C | M |
| Answer | Marks | Guidance |
|---|---|---|
| Question | AO1 | AO2 |
Question 12:
12 | (i) | x(cid:32)28cos(cid:68)t y(cid:32)28sin(cid:68)t(cid:16)4.9t2
x
Use t (cid:32) to eliminate t from y giving
28cos(cid:68)
x (cid:167) x (cid:183) 2
y(cid:32)28sin(cid:68)(cid:117) (cid:16)4.9(cid:117)(cid:168)
(cid:184)
28cos(cid:68) (cid:169)28cos(cid:68)(cid:185)
x2
y(cid:32)tan(cid:68)x(cid:16) sec2(cid:68)
160
x2
y(cid:32)tan(cid:68)x(cid:16) (cid:11) 1(cid:14)tan2(cid:68) (cid:12)
160
x2 x2
(cid:159) tan2(cid:68)(cid:16)xtan(cid:68)(cid:14) y(cid:14) (cid:32)0
160 160
160 160y
(cid:159)tan2(cid:68)(cid:16) tan(cid:68)(cid:14) (cid:14)1(cid:32)0
x x2
AG | B1
M1
A1
A1
A1
[5] | 3.4
1.1
1.1
2.1
I
C
1.1 | Both
Process mNust be completed
Simplification. May leave
E
1
cos2(cid:68)
M Using sec2(cid:68)(cid:32)1(cid:14)tan2(cid:68)
12 | (ii) | (A) | We require the discriminant to be positive so
(cid:167) 160(cid:183) 2 (cid:167)160y (cid:183) P
(cid:168)(cid:16) (cid:184) (cid:33)4 (cid:168) (cid:14)1 (cid:184)
(cid:169) x (cid:185) (cid:169) x2 (cid:185)
1602
so (cid:33)160y(cid:14)x2 S
4
x2
and y(cid:31)40(cid:16) AG
160 | E
M1
A1
E1
[3] | 2.2a
1.1
1.1 | Finding the discriminant
Obtain a ky term
Completely shown
12 | (ii) | (B) | Clearly indicates region below curve with R | A1
[1] | 1.1
12 | (iii) | Labels curve with S
Labels region above curve with T | B1
B1
[2] | 1.1
1.1
12 | (iv) | Two real distinct roots means two distinct
values of tanα (and so α) and so
R: two distinct trajectories through points in
this region
T: no real roots means no trajectories go
through those points
S: equal roots means there is a single
trajectory through a point on the curve | B1
B1
B1
[3] | 2.3
2.2a
2.2a | N
E
12 | (v) | Graph gives x = 80 when y = 0, so
No, because model does not take air resistance
into account which would slow it down; or
Yes, according to the model | B1
[1] | I
3.5a
C | M
Reason must be seen
Question | AO1 | AO2 | AO3(PS) | AO3(M) | TotalsFig. 12 shows $x$- and $y$- coordinate axes with origin O and the trajectory of a particle projected from O with speed 28 m s$^{-1}$ at an angle $\alpha$ to the horizontal. After $t$ seconds, the particle has horizontal and vertical displacements $x$ m and $y$ m.
Air resistance should be neglected.
\includegraphics{figure_12}
\begin{enumerate}[label=(\roman*)]
\item Show that the equation of the trajectory is given by
$$\tan^2\alpha - \frac{160}{x}\tan\alpha + \frac{160y}{x^2} + 1 = 0.$$ (*) [5]
\item \begin{enumerate}[label=(\Alph*)]
\item Show that if (*) is treated as an equation with $\tan\alpha$ as a variable and with $x$ and $y$ as constants, then (*) has two distinct real roots for $\tan\alpha$ when $y < 40 - \frac{x^2}{160}$. [3]
\item Show the inequality in part (ii)(A) as a locus on the graph of $y = 40 - \frac{x^2}{160}$ in the Printed Answer Booklet and label it R. [1]
\end{enumerate}
\end{enumerate}
S is the locus of points $(x, y)$ where (*) has one real root for $\tan\alpha$.
T is the locus of points $(x, y)$ where (*) has no real roots for $\tan\alpha$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Indicate S and T on the graph in the Printed Answer Booklet. [2]
\item State the significance of R, S and T for the possible trajectories of the particle. [3]
\end{enumerate}
A machine can fire a tennis ball from ground level with a maximum speed of 28 m s$^{-1}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{4}
\item State, with a reason, whether a tennis ball fired from the machine can achieve a range of 80 m. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major Q12 [15]}}