| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile with bounce or impact |
| Difficulty | Standard +0.3 This is a standard two-stage projectile motion problem with a bounce. Part (i) requires routine application of SUVAT equations to find time of flight and range. Part (ii) involves using the given horizontal distance after bounce to find the vertical component of velocity after impact, then calculating resultant speed. While it requires careful organization of multiple steps, all techniques are standard M2 material with no novel insight required. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = (25\sin 70)t - gt^2/2\) | M1 | Uses \(0 = ut - gt^2/2\) |
| \(t = 4.7(0)\,\mathrm{s}\) | A1 | |
| \(OP = (25\cos 70 \times 4.7) = 40.2\,\mathrm{m}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(OP = 25^2\sin(2 \times 70)/g\) | M1 | Uses \(R = v^2\sin 2\alpha/g\) |
| \(OP = 40.2\,\mathrm{m}\) | A1 | |
| \(t[= 40.2/(25\cos 70)] = 4.7\,\mathrm{s}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = 25\sin 70 - 10t\) | M1 | Find time to greatest height and double it |
| \(t = 2.349, 2t = 4.70\) | A1 | |
| \(OP = (25\cos 70 \times 4.7) = 40.2\,\mathrm{m}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = x\tan 70 - gx^2/(2 \times 25^2\cos^270)\) | M1 | Use trajectory equation |
| \(x = 40.2\,\mathrm{m}\) | A1 | |
| \(t = 4.70\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(t[= 17.1/(25\cos 70)] = 2\,\mathrm{s}\) | B1 | Finds time of flight |
| \(-v = v - g \times 2\) | B1ft | Finds vertical component of speed |
| \(V^2 = 10^2 + (25\cos 70)^2\) | M1 | For squaring components |
| \(V = 13.2\,\mathrm{ms}^{-1}\) | A1 | [4] |
**(i)**
$0 = (25\sin 70)t - gt^2/2$ | M1 | Uses $0 = ut - gt^2/2$
$t = 4.7(0)\,\mathrm{s}$ | A1 |
$OP = (25\cos 70 \times 4.7) = 40.2\,\mathrm{m}$ | A1 |
OR
$OP = 25^2\sin(2 \times 70)/g$ | M1 | Uses $R = v^2\sin 2\alpha/g$
$OP = 40.2\,\mathrm{m}$ | A1 |
$t[= 40.2/(25\cos 70)] = 4.7\,\mathrm{s}$ | A1 |
OR
$0 = 25\sin 70 - 10t$ | M1 | Find time to greatest height and double it
$t = 2.349, 2t = 4.70$ | A1 |
$OP = (25\cos 70 \times 4.7) = 40.2\,\mathrm{m}$ | A1 |
OR
$0 = x\tan 70 - gx^2/(2 \times 25^2\cos^270)$ | M1 | Use trajectory equation
$x = 40.2\,\mathrm{m}$ | A1 |
$t = 4.70$ | A1 | [3]
**(ii)**
$t[= 17.1/(25\cos 70)] = 2\,\mathrm{s}$ | B1 | Finds time of flight
$-v = v - g \times 2$ | B1ft | Finds vertical component of speed
$V^2 = 10^2 + (25\cos 70)^2$ | M1 | For squaring components
$V = 13.2\,\mathrm{ms}^{-1}$ | A1 | [4] | [7]5 A ball is projected with velocity $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of $70 ^ { \circ }$ above the horizontal from a point $O$ on horizontal ground. The ball subsequently bounces once on the ground at a point $P$ before landing at a point $Q$ where it remains at rest. The distance $P Q$ is 17.1 m .\\
(i) Calculate the time taken by the ball to travel from $O$ to $P$ and the distance $O P$.\\
(ii) Given that the horizontal component of the velocity of the ball does not change at $P$, calculate the speed of the ball when it leaves $P$.
\hfill \mbox{\textit{CAIE M2 2012 Q5 [7]}}