| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| 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 a bounce condition. While it has multiple parts and requires understanding of projectile trajectory and impact mechanics, all techniques are routine: using standard projectile equations, finding maximum height conditions, and applying given velocity changes after impact. The given sin θ = 3/5 simplifies calculations significantly. More straightforward than average due to the structured parts and standard methods. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | For using \(\frac{1}{2}R = l^2\sin\theta\cos\theta/g\) or \(0 = V\sin\theta - gt\) and \(19.2 = l\cos\theta\) |
| \(19.2 = l^2 \times (3/5) \times (4/5)/10\) | A1 | |
| \(l^2 = 20\) | A1 | 3 |
| (ii) \(5l^2 - 12t + 4 = 0\) | M1 | For substituting for \(V\) and \(\theta\) and solving \(V\sin\theta - \frac{1}{2}gt^2 = 4\); \(t = 0.4\) must be discarded (which may be implied by subsequent use of \(t = 2\) only) |
| \(t = 2\) | A1 | |
| \(x = 20 \times 2 \times (4/5)\) | M1 | For substituting for \(V\), \(\theta\) and \(t\) into \(x = l/t\cos\theta\) |
| Horizontal distance is 32 m | A1 | 4 |
| (iii) Immediately before impact \((x, y) = (20 \times 0.8, 20 \times 0.6 - 10 \times 2)\) | B1 fl | |
| Immediately after impact \((x, y) = (-8, -8)\) | B1 fl | |
| At ground \(-8t - \frac{1}{2}10^2 = -4\) \(\rightarrow 5t^2 + 8t - 4 = 0\) | M1 | For obtaining appropriate quadratic equation; or using \(4 = -t(V\sin\theta + 8) + 2\) or using the idea that \(t\) takes the discarded value in (ii) [if the M mark is scored in this way the \(y\) component is not required for either of the B marks] |
| \(t = 4 \times 2 \div (12 + 8)\) | ||
| Distance is 3.2m | A1 | 4 |
**(i)** | M1 | For using $\frac{1}{2}R = l^2\sin\theta\cos\theta/g$ or $0 = V\sin\theta - gt$ and $19.2 = l\cos\theta$ |
| $19.2 = l^2 \times (3/5) \times (4/5)/10$ | A1 | |
| $l^2 = 20$ | A1 | 3 |
**(ii)** $5l^2 - 12t + 4 = 0$ | M1 | For substituting for $V$ and $\theta$ and solving $V\sin\theta - \frac{1}{2}gt^2 = 4$; $t = 0.4$ must be discarded (which may be implied by subsequent use of $t = 2$ only) |
| $t = 2$ | A1 | |
| $x = 20 \times 2 \times (4/5)$ | M1 | For substituting for $V$, $\theta$ and $t$ into $x = l/t\cos\theta$ |
| Horizontal distance is 32 m | A1 | 4 |
**(iii)** Immediately before impact $(x, y) = (20 \times 0.8, 20 \times 0.6 - 10 \times 2)$ | B1 fl | |
| Immediately after impact $(x, y) = (-8, -8)$ | B1 fl | |
| At ground $-8t - \frac{1}{2}10^2 = -4$ $\rightarrow 5t^2 + 8t - 4 = 0$ | M1 | For obtaining appropriate quadratic equation; or using $4 = -t(V\sin\theta + 8) + 2$ or using the idea that $t$ takes the discarded value in (ii) [if the M mark is scored in this way the $y$ component is not required for either of the B marks] |
| $t = 4 \times 2 \div (12 + 8)$ | | |
| Distance is 3.2m | A1 | 4 |7 A stone is projected from a point $O$ on horizontal ground with speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ above the horizontal, where $\sin \theta = \frac { 3 } { 5 }$. The stone is at its highest point when it has travelled a horizontal distance of 19.2 m .\\
(i) Find the value of $V$.
After passing through its highest point the stone strikes a vertical wall at a point 4 m above the ground.\\
(ii) Find the horizontal distance between $O$ and the wall.
At the instant when the stone hits the wall the horizontal component of the stone's velocity is halved in magnitude and reversed in direction. The vertical component of the stone's velocity does not change as a result of the stone hitting the wall.\\
(iii) Find the distance from the wall of the point where the stone reaches the ground.
\hfill \mbox{\textit{CAIE M2 2006 Q7 [11]}}