OCR H240/03 2018 March — Question 11 12 marks

Exam BoardOCR
ModuleH240/03 (Pure Mathematics and Mechanics)
Year2018
SessionMarch
Marks12
TopicProjectiles
TypeTwo projectiles meeting - 2D flight
DifficultyChallenging +1.8 This is a challenging projectile motion problem requiring students to set up and solve simultaneous equations for two projectiles meeting at a point, involving double-angle formulas and algebraic manipulation across multiple steps. While the techniques are standard M1 content, the coordination of two projectiles with related angles (α and 2α) and the 7-mark allocation indicates substantial problem-solving beyond routine exercises.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02i Projectile motion: constant acceleration model

In this question you must show detailed reasoning. \includegraphics{figure_11} A football \(P\) is kicked with speed \(25\,\text{m}\,\text{s}^{-1}\) at an angle of elevation \(\alpha\) from a point \(A\) on horizontal ground. At the same instant a second football \(Q\) is kicked with speed \(15\,\text{m}\,\text{s}^{-1}\) at an angle of elevation \(2\alpha\) from a point \(B\) on the same horizontal ground, where \(AB = 72\) m. The footballs are modelled as particles moving freely under gravity in the same vertical plane and they collide with each other at the point \(C\) (see diagram).
  1. Calculate the height of \(C\) above the ground. [7]
  2. Find the direction of motion of \(P\) at the moment of impact. [4]
  3. Suggest one improvement that could be made to the model. [1]

Part (i):
AnswerMarks Guidance
\((25\sin\alpha)t - 4.9t^2 = (15\sin(2\alpha)t - 4.9t^2\)M1 Use \(s = ut + \frac{1}{4}at^2\) for both, and equate
\(25\sin\alpha = 15\sin(2\alpha)\)A1
\(25\sin\alpha = 30\sin\alpha\cos\alpha \Rightarrow \cos\alpha = \ldots\)M1 Correct use of double angle formula and attempt to solve for \(\cos\alpha\)
\(\cos\alpha = \frac{5}{6}\) (and \(\sin\alpha = \frac{1}{6}\sqrt{11}\))A1 ; \(\alpha = 33.557\ldots°\)
\((25\cos\alpha)t + (15\cos(2\alpha))t = 72 \Rightarrow t = \ldots\)M1 Use \(s = ut\) for both, equate total to 72 and attempt to solve for \(t\)
\(t = 2.7\)A1
Height of \(C\) is \((25\sin\alpha)t - 4.9t^2 = 1.59\) mA1 ; 1.591 028 8...
Part (ii):
AnswerMarks Guidance
\(v_b = 25\cos\alpha\)B1ft With their value of \(\cos\alpha\); \(v_b = 20.833\ldots\)
\(v_v = 25\sin\alpha - 9.8t\)B1ft With their values of \(\sin\alpha\) and \(t\); \(v_v = \pm 12.640\ldots\)
\(\tan\theta = \frac{v_v}{v_b}\)M1 \(\theta\) is angle with horizontal; condone sign error/ambiguity for this mark
Direction is 31.2° below the horizontalA1 ; 31.247 93...
Part (iii):
AnswerMarks Guidance
e.g. include the dimensions of the footballs in the model of the motion; e.g. use a more accurate value of \(g\) in the model of the motion; e.g. include air resistance in the model of the motionB1 DR
## Part (i):

$(25\sin\alpha)t - 4.9t^2 = (15\sin(2\alpha)t - 4.9t^2$ | M1 | Use $s = ut + \frac{1}{4}at^2$ for both, and equate

$25\sin\alpha = 15\sin(2\alpha)$ | A1 | 

$25\sin\alpha = 30\sin\alpha\cos\alpha \Rightarrow \cos\alpha = \ldots$ | M1 | Correct use of double angle formula and attempt to solve for $\cos\alpha$

$\cos\alpha = \frac{5}{6}$ (and $\sin\alpha = \frac{1}{6}\sqrt{11}$) | A1 | ; $\alpha = 33.557\ldots°$

$(25\cos\alpha)t + (15\cos(2\alpha))t = 72 \Rightarrow t = \ldots$ | M1 | Use $s = ut$ for both, equate total to 72 and attempt to solve for $t$

$t = 2.7$ | A1 | 

Height of $C$ is $(25\sin\alpha)t - 4.9t^2 = 1.59$ m | A1 | ; 1.591 028 8...

## Part (ii):

$v_b = 25\cos\alpha$ | B1ft | With their value of $\cos\alpha$; $v_b = 20.833\ldots$

$v_v = 25\sin\alpha - 9.8t$ | B1ft | With their values of $\sin\alpha$ and $t$; $v_v = \pm 12.640\ldots$

$\tan\theta = \frac{v_v}{v_b}$ | M1 | $\theta$ is angle with horizontal; condone sign error/ambiguity for this mark

Direction is 31.2° below the horizontal | A1 | ; 31.247 93...

## Part (iii):

e.g. include the dimensions of the footballs in the model of the motion; e.g. use a more accurate value of $g$ in the model of the motion; e.g. include air resistance in the model of the motion | B1 | DR
In this question you must show detailed reasoning.

\includegraphics{figure_11}

A football $P$ is kicked with speed $25\,\text{m}\,\text{s}^{-1}$ at an angle of elevation $\alpha$ from a point $A$ on horizontal ground. At the same instant a second football $Q$ is kicked with speed $15\,\text{m}\,\text{s}^{-1}$ at an angle of elevation $2\alpha$ from a point $B$ on the same horizontal ground, where $AB = 72$ m. The footballs are modelled as particles moving freely under gravity in the same vertical plane and they collide with each other at the point $C$ (see diagram).

\begin{enumerate}[label=(\roman*)]
\item Calculate the height of $C$ above the ground. [7]
\item Find the direction of motion of $P$ at the moment of impact. [4]
\item Suggest one improvement that could be made to the model. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/03 2018 Q11 [12]}}