CAIE M2 2019 June — Question 1 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeDeriving trajectory equation
DifficultyModerate -0.8 This is a straightforward projectiles question requiring only standard recall and application of SUVAT equations. Part (i) asks for g=9.8 or 10, part (ii) uses sin(30°)=0.5 to find initial speed from the vertical component, and part (iii) applies horizontal motion with cos(30°). All steps are routine with no problem-solving or novel insight required.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model

1 A small ball is projected from a point \(O\) on horizontal ground at an angle of \(30 ^ { \circ }\) above the horizontal. At time \(t \mathrm {~s}\) after projection the vertically upwards displacement of the ball from \(O\) is \(\left( 14 t - k t ^ { 2 } \right) \mathrm { m }\), where \(k\) is a constant.
  1. State the value of \(k\). \includegraphics[max width=\textwidth, alt={}, center]{f3a35846-075d-4e03-ba6b-82774ef0e4f8-03_56_1563_495_331}
  2. Show that the initial speed of the ball is \(28 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  3. Find the horizontal displacement of the ball from \(O\) when \(t = 3\).

Question 1:
Part (i):
AnswerMarks Guidance
\(k = \frac{g}{2} = 5\)B1 Use the trajectory equation from the formula sheet

Total: 1 mark

Part (ii):
AnswerMarks Guidance
\(V\sin30 = 14\)M1 Use the trajectory equation from the formula sheet
\(V = 28 \text{ ms}^{-1}\)A1 (AG)

Total: 2 marks

Part (iii):
AnswerMarks Guidance
\(x = 28\cos30 \times 3\)M1 Use horizontal motion. Allow *their* V for M1
\(x = 72.7 \text{ m}\)A1

Total: 2 marks

**Question 1:**

**Part (i):**
$k = \frac{g}{2} = 5$ | B1 | Use the trajectory equation from the formula sheet

Total: 1 mark

---

**Part (ii):**
$V\sin30 = 14$ | M1 | Use the trajectory equation from the formula sheet
$V = 28 \text{ ms}^{-1}$ | A1 (AG) |

Total: 2 marks

---

**Part (iii):**
$x = 28\cos30 \times 3$ | M1 | Use horizontal motion. Allow *their* V for M1
$x = 72.7 \text{ m}$ | A1 |

Total: 2 marks

---
1 A small ball is projected from a point $O$ on horizontal ground at an angle of $30 ^ { \circ }$ above the horizontal. At time $t \mathrm {~s}$ after projection the vertically upwards displacement of the ball from $O$ is $\left( 14 t - k t ^ { 2 } \right) \mathrm { m }$, where $k$ is a constant.\\
(i) State the value of $k$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f3a35846-075d-4e03-ba6b-82774ef0e4f8-03_56_1563_495_331}\\

(ii) Show that the initial speed of the ball is $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\

(iii) Find the horizontal displacement of the ball from $O$ when $t = 3$.\\

\hfill \mbox{\textit{CAIE M2 2019 Q1 [5]}}