| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile with bounce or impact |
| Difficulty | Challenging +1.2 Part (a) is a standard projectile time-of-flight derivation. Part (b) requires understanding that vertical motion is unchanged by horizontal collision. Part (c) involves setting up range equations with coefficient of restitution, requiring multiple steps but using familiar projectile and collision mechanics without novel insight. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model6.03b Conservation of momentum: 1D two particles6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\uparrow\, 0 = V\sin 75t - \frac{1}{2}gt^2\) | M1 | |
| \(t = \frac{2V}{g}\sin 75°\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical component of velocity is unchanged. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Horizontally to wall: \(15 = V\cos 75°t\); \(\left(t = \dfrac{15}{V\cos 75°}\right)\) | B1 | |
| \(eV\cos 75°\) | B1 | Speed after rebound. |
| \(T = \dfrac{15}{eV\cos 75°}\) | M1 | Time back to \(O\); \(\left(t = \dfrac{3}{5}T\right)\) |
| Vertically for whole flight: \(t + T = \dfrac{2V}{g}\sin 75°\) | M1 | |
| \(\dfrac{15}{V\cos 75°} + \dfrac{15}{eV\cos 75°} = \dfrac{2V}{g}\sin 75°\) | A1 | |
| \(V^2\cos 75°\sin 75° = 20g\) | M1 | |
| Multiply by 2: \(V^2\sin 150° = 40g\), \(\quad V^2 = 80g\) \(V = 4\sqrt{5g}\left(= 8.94\sqrt{g}\right)\) | A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\uparrow\, 0 = V\sin 75t - \frac{1}{2}gt^2$ | M1 | |
| $t = \frac{2V}{g}\sin 75°$ | A1 | AG |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical component of velocity is unchanged. | B1 | |
## Question 7(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Horizontally to wall: $15 = V\cos 75°t$; $\left(t = \dfrac{15}{V\cos 75°}\right)$ | B1 | |
| $eV\cos 75°$ | B1 | Speed after rebound. |
| $T = \dfrac{15}{eV\cos 75°}$ | M1 | Time back to $O$; $\left(t = \dfrac{3}{5}T\right)$ |
| Vertically for whole flight: $t + T = \dfrac{2V}{g}\sin 75°$ | M1 | |
| $\dfrac{15}{V\cos 75°} + \dfrac{15}{eV\cos 75°} = \dfrac{2V}{g}\sin 75°$ | A1 | |
| $V^2\cos 75°\sin 75° = 20g$ | M1 | |
| Multiply by 2: $V^2\sin 150° = 40g$, $\quad V^2 = 80g$ $V = 4\sqrt{5g}\left(= 8.94\sqrt{g}\right)$ | A1 | |
**Total: 7 marks**7 A particle $P$ is projected with speed $\mathrm { Vms } ^ { - 1 }$ at an angle $75 ^ { \circ }$ above the horizontal from a point $O$ on a horizontal plane. It then moves freely under gravity.
\begin{enumerate}[label=(\alph*)]
\item Show that the total time of flight, in seconds, is $\frac { 2 \mathrm {~V} } { \mathrm {~g} } \sin 75 ^ { \circ }$.\\
A smooth vertical barrier is now inserted with its lower end on the plane at a distance 15 m from $O$. The particle is projected as before but now strikes the barrier, rebounds and returns to $O$. The coefficient of restitution between the barrier and the particle is $\frac { 3 } { 5 }$.
\item Explain why the total time of flight is unchanged.
\item Find an expression for $V$ in terms of $g$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q7 [10]}}