Edexcel M2 — Question 8 15 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile clearing obstacle
DifficultyStandard +0.3 This is a standard M2 projectile motion question requiring routine application of SUVAT equations with resolved components. All parts follow textbook methods: resolving velocity, finding time from horizontal motion, calculating vertical displacement, finding maximum height using v=0, and optimizing range. The 15 marks reflect length rather than conceptual difficulty—no novel problem-solving or geometric insight required, making it slightly easier than average.
Spec3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model

In a fairground game, a contestant bowls a ball at a coconut 6 metres away on the same horizontal level. The ball is thrown with an initial speed of 8 ms\(^{-1}\) in a direction making an angle of 30° with the horizontal. \includegraphics{figure_8}
  1. Find the time taken by the ball to travel 6 m horizontally. [2 marks]
  2. Showing your method clearly, decide whether or not the ball will hit the coconut. [4 marks]
  3. Find the greatest height reached by the ball above the level from which it was thrown. [4 marks]
  4. Find the maximum horizontal distance from which it is possible to hit the coconut if the ball is thrown with the same initial speed of 8 m s\(^{-1}\). [3 marks]
  5. State two assumptions that you have made about the ball and the forces which act on it as it travels towards the coconut. [2 marks]

AnswerMarks Guidance
(a) \(x = 8 \cos 30° \cdot t\)M1 A1 When \(x = 6\), \(t = 0.866 \text{ s}\)
(b) Then \(y = 8 \sin 30° \cdot t - 4.9t^2 = -0.21 \text{ m}\)M1 A1 M1 A1 so does not hit coconut
(c) When \(v_y = 0\), \(8 \sin 30° - 9.8t = 0\)M1 A1 \(t = 0.408\)
Then \(y = 4(0.408) - 4.9(0.408)^2 = 0.816 \text{ m}\)M1 A1
(d) Max range when projected at \(45°\)B1 M1 \(y = 0\) when \(t = 1.154\)
Then \(x = 1.154(8 \cos 45°) = 6.53 \text{ m}\)A1
(e) Ball = particle; assumed gravity is only force acting on ballB1 B1
Total: 15 marks
**(a)** $x = 8 \cos 30° \cdot t$ | M1 A1 | When $x = 6$, $t = 0.866 \text{ s}$

**(b)** Then $y = 8 \sin 30° \cdot t - 4.9t^2 = -0.21 \text{ m}$ | M1 A1 M1 A1 | so does not hit coconut

**(c)** When $v_y = 0$, $8 \sin 30° - 9.8t = 0$ | M1 A1 | $t = 0.408$

Then $y = 4(0.408) - 4.9(0.408)^2 = 0.816 \text{ m}$ | M1 A1 |

**(d)** Max range when projected at $45°$ | B1 M1 | $y = 0$ when $t = 1.154$

Then $x = 1.154(8 \cos 45°) = 6.53 \text{ m}$ | A1 |

**(e)** Ball = particle; assumed gravity is only force acting on ball | B1 B1 |
**Total: 15 marks**
In a fairground game, a contestant bowls a ball at a coconut 6 metres away on the same horizontal level. The ball is thrown with an initial speed of 8 ms$^{-1}$ in a direction making an angle of 30° with the horizontal.

\includegraphics{figure_8}

\begin{enumerate}[label=(\alph*)]
\item Find the time taken by the ball to travel 6 m horizontally. [2 marks]
\item Showing your method clearly, decide whether or not the ball will hit the coconut. [4 marks]
\item Find the greatest height reached by the ball above the level from which it was thrown. [4 marks]
\item Find the maximum horizontal distance from which it is possible to hit the coconut if the ball is thrown with the same initial speed of 8 m s$^{-1}$. [3 marks]
\item State two assumptions that you have made about the ball and the forces which act on it as it travels towards the coconut. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q8 [15]}}