| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| 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-stage projectile problem requiring resolution of velocities at multiple time points, use of energy/kinematics to find height, and analysis of a bounce with changed vertical component. It demands careful bookkeeping across three connected parts with non-trivial calculations, placing it moderately above average difficulty but within reach of well-prepared M2 students. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(U_H = 18\cos 30\) and \(U_V = 18\sin 30 + 2g (=29)\) | B1 | |
| \(U = \sqrt{[(18\cos 30)^2 + 29^2]}\) or \(\tan\theta = 29/(18\cos 30)\) | M1 | Uses Pythagoras's Theorem and trigonometry. |
| \(U = 32.9(24\ldots)\text{ m s}^{-1}\) | A1 | |
| \(\theta = 61.7^\circ\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 38^2 - (18\cos 30)^2 = (+/-29)^2 + 2gh\) | M1 | Uses 2 ways to find \(v\), the vertical velocity at the ground and equates. |
| \(h = 18\) | A1 | |
| OR \(mgh + m \times 32.924^2/2 = m \times 38^2/2\) | M1 | |
| \(h = 18\) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-\sqrt{[38^2-(18\cos 30)^2]} = 29-gt\) | M1 | Uses \(v = u + at\) for first part of flight. |
| \(t = 6.36(6)\) | A1 | |
| \(v = \sqrt{[20^2-(18\cos 30)^2]} = 12.5(3)\) | M1 | Uses \(v = u + at\) for second part of flight. |
| \(-12.5(3) = 12.5(3) - gt'\) | ||
| \(t' = 2.50(6)\) | A1 | |
| \(T\ (= 6.366 + 2.506) = 8.87\) | A1 | |
| Total | 5 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $U_H = 18\cos 30$ and $U_V = 18\sin 30 + 2g (=29)$ | B1 | |
| $U = \sqrt{[(18\cos 30)^2 + 29^2]}$ or $\tan\theta = 29/(18\cos 30)$ | M1 | Uses Pythagoras's Theorem and trigonometry. |
| $U = 32.9(24\ldots)\text{ m s}^{-1}$ | A1 | |
| $\theta = 61.7^\circ$ | A1 | |
| **Total** | **4** | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 38^2 - (18\cos 30)^2 = (+/-29)^2 + 2gh$ | M1 | Uses 2 ways to find $v$, the vertical velocity at the ground and equates. |
| $h = 18$ | A1 | |
| OR $mgh + m \times 32.924^2/2 = m \times 38^2/2$ | M1 | |
| $h = 18$ | A1 | |
| **Total** | **2** | |
## Question 7(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $-\sqrt{[38^2-(18\cos 30)^2]} = 29-gt$ | **M1** | Uses $v = u + at$ for first part of flight. |
| $t = 6.36(6)$ | **A1** | |
| $v = \sqrt{[20^2-(18\cos 30)^2]} = 12.5(3)$ | **M1** | Uses $v = u + at$ for second part of flight. |
| $-12.5(3) = 12.5(3) - gt'$ | | |
| $t' = 2.50(6)$ | **A1** | |
| $T\ (= 6.366 + 2.506) = 8.87$ | **A1** | |
| **Total** | **5** | |7 A small ball $B$ is projected from a point $O$ which is $h \mathrm {~m}$ above a horizontal plane. At time 2 s after projection $B$ has speed $18 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is moving in the direction $30 ^ { \circ }$ above the horizontal.\\
(i) Find the initial speed and the angle of projection of $B$.\\
$B$ has speed $38 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ immediately before it strikes the plane.\\
(ii) Calculate $h$.\\
$B$ bounces when it strikes the plane, and leaves the plane with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ but with its horizontal component of velocity unchanged.\\
(iii) Find the total time which elapses between the initial projection of $B$ and the instant when it strikes the plane for the second time.\\
\hfill \mbox{\textit{CAIE M2 2017 Q7 [11]}}