| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Basic trajectory calculations |
| Difficulty | Standard +0.3 This is a straightforward projectiles question requiring standard SUVAT equations and the speed formula. Part (a) uses the relationship between initial speed, time, and final speed to find the angle; part (b) applies standard range formulas. While it requires careful algebra and understanding of vector components, it follows a well-practiced method with no novel insight needed. Slightly above average difficulty due to the two-part structure and algebraic manipulation required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Components of velocity: \(\rightarrow 25\cos\theta\), \(\uparrow 25\sin\theta - 2g\) | B1 | |
| Speed \(= \sqrt{(25\cos\theta)^2 + (25\sin\theta - 2g)^2}\) | M1 A1 | Expression for speed or square of speed |
| \((25\cos\theta)^2 + (25\sin\theta - 2g)^2 = 15^2\); \(625 - 100g\sin\theta + 4g^2 = 225\) | M1 | Attempt to solve and find value for \(\sin\theta\) |
| \(\sin\theta = \frac{800}{1000} = \frac{4}{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Time of flight \(= \left(\frac{2 \times 25\sin\theta}{g}\right) = 4\) s | B1 | |
| Range \(= \frac{2 \times 25\sin\theta}{g} \times 25\cos\theta\) | M1 | Any equivalent method |
| Range \(= 60\) m | A1 | CWO |
| Alternative: \(y = \frac{4}{3}x - \frac{1}{45}x^2\) | B1 | Equation of trajectory |
| Substitute \(y = 0\) and solve | M1 | |
| \(60\) m | A1 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Components of velocity: $\rightarrow 25\cos\theta$, $\uparrow 25\sin\theta - 2g$ | B1 | |
| Speed $= \sqrt{(25\cos\theta)^2 + (25\sin\theta - 2g)^2}$ | M1 A1 | Expression for speed or square of speed |
| $(25\cos\theta)^2 + (25\sin\theta - 2g)^2 = 15^2$; $625 - 100g\sin\theta + 4g^2 = 225$ | M1 | Attempt to solve and find value for $\sin\theta$ |
| $\sin\theta = \frac{800}{1000} = \frac{4}{5}$ | A1 | |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Time of flight $= \left(\frac{2 \times 25\sin\theta}{g}\right) = 4$ s | B1 | |
| Range $= \frac{2 \times 25\sin\theta}{g} \times 25\cos\theta$ | M1 | Any equivalent method |
| Range $= 60$ m | A1 | CWO |
| **Alternative:** $y = \frac{4}{3}x - \frac{1}{45}x^2$ | B1 | Equation of trajectory |
| Substitute $y = 0$ and solve | M1 | |
| $60$ m | A1 | |
---3 A particle $P$ is projected with speed $25 \mathrm {~ms} ^ { - 1 }$ at an angle $\theta$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. After 2 s the speed of $P$ is $15 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\sin \theta$.
\item Find the range of the flight.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q3 [8]}}