AQA M1 2013 January — Question 8 10 marks

Exam BoardAQA
ModuleM1 (Mechanics 1)
Year2013
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile passing through given point
DifficultyModerate -0.3 This is a standard M1 projectile question requiring straightforward application of kinematic equations with given time, horizontal distance, and vertical height. The three parts systematically guide students through finding horizontal velocity (distance/time), then using vertical motion with SUVAT to find the vertical component, and finally combining components to find V and α. No novel problem-solving or geometric insight required—just routine application of well-practiced techniques.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model

8 A golf ball is hit from a point on a horizontal surface, so that it has an initial velocity \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle \(\alpha\) above the horizontal. The ball travels through the air and after 2.4 seconds hits a vertical wall at a height of 3 metres. The wall is at a horizontal distance of 38.4 metres from the point where the ball was hit. The path of the ball is shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{ccc1db66-9700-4f22-905e-cc0bdf1fd3c1-18_300_1000_566_520} Assume that the weight of the ball is the only force that acts on it as it travels through the air.
  1. Find the horizontal component of the velocity of the ball.
  2. \(\quad\) Find \(V\).
  3. \(\quad\) Find \(\alpha\).

Question 8:
Part (a): Horizontal component of velocity
\(v_x = \frac{38.4}{2.4} = 16 \text{ ms}^{-1}\)
Part (b): Find \(V\)
Using vertical motion: \(s = u_y t - \frac{1}{2}gt^2\)
\(3 = u_y(2.4) - \frac{1}{2}(9.8)(2.4)^2\)
\(3 = 2.4u_y - 28.224\)
\(u_y = \frac{31.224}{2.4} = 13.01 \text{ ms}^{-1}\)
\(V = \sqrt{16^2 + 13.01^2} = \sqrt{256 + 169.26} \approx 20.6 \text{ ms}^{-1}\)
Part (c): Find \(\alpha\)
\(\tan\alpha = \frac{13.01}{16}\)
\(\alpha = \arctan(0.8131) \approx 39.2°\)
## Question 8:

**Part (a):** Horizontal component of velocity

$v_x = \frac{38.4}{2.4} = 16 \text{ ms}^{-1}$

**Part (b):** Find $V$

Using vertical motion: $s = u_y t - \frac{1}{2}gt^2$

$3 = u_y(2.4) - \frac{1}{2}(9.8)(2.4)^2$

$3 = 2.4u_y - 28.224$

$u_y = \frac{31.224}{2.4} = 13.01 \text{ ms}^{-1}$

$V = \sqrt{16^2 + 13.01^2} = \sqrt{256 + 169.26} \approx 20.6 \text{ ms}^{-1}$

**Part (c):** Find $\alpha$

$\tan\alpha = \frac{13.01}{16}$

$\alpha = \arctan(0.8131) \approx 39.2°$
8 A golf ball is hit from a point on a horizontal surface, so that it has an initial velocity $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\alpha$ above the horizontal. The ball travels through the air and after 2.4 seconds hits a vertical wall at a height of 3 metres. The wall is at a horizontal distance of 38.4 metres from the point where the ball was hit. The path of the ball is shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{ccc1db66-9700-4f22-905e-cc0bdf1fd3c1-18_300_1000_566_520}

Assume that the weight of the ball is the only force that acts on it as it travels through the air.
\begin{enumerate}[label=(\alph*)]
\item Find the horizontal component of the velocity of the ball.
\item $\quad$ Find $V$.
\item $\quad$ Find $\alpha$.
\end{enumerate}

\hfill \mbox{\textit{AQA M1 2013 Q8 [10]}}