| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile with bounce or impact |
| Difficulty | Standard +0.8 This is a multi-part projectile problem requiring understanding that 'moving horizontally' means vertical velocity is zero at maximum height, then applying energy/kinematics to find the angle, followed by analyzing the rebound motion with changed velocity. It requires more insight than standard projectile questions and involves multiple connected steps across three parts, but uses well-established M2 techniques without requiring novel problem-solving approaches. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For using \(\dot{y} = \dot{y}_0 - gt\) with \(\dot{y} = 0\) \((t = 2\sin\alpha)\) | M1 | |
| For using \(y = \dot{y}_0 t - \frac{1}{2}gt^2\) with \(t\) as found and \(y = 7.2\), or show \(t = 1.2\) as in (ii) | M1 | |
| Alternatively for using \(y_{max} = \frac{V^2\sin^2\alpha}{2g}\) with \(y_{max} = 7.2\) and \(V = 20\), or \(\dot{y}^2 = \dot{y}_0^2 - 2gy\) with \(\dot{y} = 0\) | M2 | |
| \(7.2 = \frac{400\sin^2\alpha}{20}\) | A1 | |
| Angle is 36.9° | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Speed on hitting the wall is \(20 \times 0.8\) | B1ft | use of ball rebounding at 10 ms\(^{-1}\) scores B0 |
| For using \(y = 0 - \frac{1}{2}gt^2\) \(\left(-7.2 = -\frac{1}{2}(10)t^2\right)\) or \(0 = \dot{y} - gt\) \((0 = 12 - 10t)\) | M1 | |
| \(t = 1.2\) | A1 | |
| Distance is 9.6 m | A1ft | No ft if rebound velocity \(= 10\) ms\(^{-1}\) |
| Alternative: speed on hitting wall \(= 20 \times 0.8\); use trajectory equation with \(\theta = 0°\) | B1ft, M1 | |
| \(-7.2 = x\tan 0° - \frac{gx^2}{2(8)^2\cos^2 0°}\) | A1ft | allow ft with halving attempt including 10 |
| \(x = 9.6\) m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dot{y} = \mp 10 \times 1.2\) | B1ft | |
| \(\theta = \tan^{-1}\left(\mp\frac{\dot{y}}{\dot{x}}\right)\) (\(\dot{x}\) must have halving attempt. Allow \(\dot{x} = 10\)) | M1 | |
| Required angle is 56.3° | A1 |
# Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| For using $\dot{y} = \dot{y}_0 - gt$ with $\dot{y} = 0$ $(t = 2\sin\alpha)$ | M1 | |
| For using $y = \dot{y}_0 t - \frac{1}{2}gt^2$ with $t$ as found and $y = 7.2$, or show $t = 1.2$ as in (ii) | M1 | |
| Alternatively for using $y_{max} = \frac{V^2\sin^2\alpha}{2g}$ with $y_{max} = 7.2$ and $V = 20$, or $\dot{y}^2 = \dot{y}_0^2 - 2gy$ with $\dot{y} = 0$ | M2 | |
| $7.2 = \frac{400\sin^2\alpha}{20}$ | A1 | |
| Angle is 36.9° | A1 | |
# Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Speed on hitting the wall is $20 \times 0.8$ | B1ft | use of ball rebounding at 10 ms$^{-1}$ scores B0 |
| For using $y = 0 - \frac{1}{2}gt^2$ $\left(-7.2 = -\frac{1}{2}(10)t^2\right)$ or $0 = \dot{y} - gt$ $(0 = 12 - 10t)$ | M1 | |
| $t = 1.2$ | A1 | |
| Distance is 9.6 m | A1ft | No ft if rebound velocity $= 10$ ms$^{-1}$ |
| Alternative: speed on hitting wall $= 20 \times 0.8$; use trajectory equation with $\theta = 0°$ | B1ft, M1 | |
| $-7.2 = x\tan 0° - \frac{gx^2}{2(8)^2\cos^2 0°}$ | A1ft | allow ft with halving attempt including 10 |
| $x = 9.6$ m | A1 | |
# Question 5(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dot{y} = \mp 10 \times 1.2$ | B1ft | |
| $\theta = \tan^{-1}\left(\mp\frac{\dot{y}}{\dot{x}}\right)$ ($\dot{x}$ must have halving attempt. Allow $\dot{x} = 10$) | M1 | |
| Required angle is 56.3° | A1 | |
---5 A stone is projected from a point on horizontal ground with a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $\alpha ^ { \circ }$ above the horizontal. The stone is moving horizontally when it hits a vertical wall at a point 7.2 m above the ground.\\
(i) Find the value of $\alpha$.
After rebounding at right angles from the wall the speed of the stone is halved. Find\\
(ii) the distance from the wall of the point at which the stone hits the ground,\\
(iii) the angle which the direction of motion of the stone makes with the horizontal, immediately before the stone hits the ground.
\hfill \mbox{\textit{CAIE M2 2003 Q5 [11]}}