Question
At an archery tournament, a particular competition sees a ball launched into the air while an
archer attempts to hit it with an arrow.
The path of the ball is modelled by the equation
\(\binom{x}{y}=\binom{5}{0}+t\binom{u_{x}}{u_{y}-5t}\)
where x is the horizontal displacement from the archer and y is the vertical displacement
from the ground, both measured in metres, and t is the time, in seconds, since the ball
was launched.
- ux is the horizontal component of the initial velocity
- uy is the vertical component of the initial velocity
In this question both the ball and the arrow are modelled as single points. The ball is launched
with an initial velocity such that ux = 8 and uy = 10.
(a) (i) Find the initial speed of the ball.
(ii) Find the angle of elevation of the ball as it is launched. [4]
(b) Find the maximum height reached by the ball. [3]
(c) Assuming that the ground is horizontal and the ball is not hit by the arrow, find the x
coordinate of the point where the ball lands. [3]
(d) For the path of the ball, find an expression for y in terms of x . [3]
An archer releases an arrow from the point (0 , 2). The arrow is modelled as travelling in a
straight line, in the same plane as the ball, with speed 60 ms−1and an angle of elevation of 10∘.
(e) Determine the two positions where the path of the arrow intersects the path of the ball. [4]
(f) Determine the time when the arrow should be released to hit the ball before the ball
reaches its maximum height. [4]
▶️Answer/Explanation
Ans
6. (a) (i) \(\sqrt{10^{2}+8^{2}}\) (M1)
\(=12.8(12.8062…,\sqrt{164})(ms^{-1})\) (A1)
(ii) \(\tan ^{-1}\left ( \frac{10}{8} \right )\) (M1)
= 0.896 OR 51.3 (0.896055… OR 51.3401 ) …° A1
Note: Accept 0.897 or 51.4 from use of 10 arcsin \(\left ( \frac{10}{12.8} \right )\)
[4 marks]
(b) y= t (10 − 5t ) (M1)
Note: The M1 might be implied by a correct graph or use of the correct equation.
METHOD 1 – graphical Method
sketch graph (M1)
Note: The M1 might be implied by correct graph or correct maximum (eg t =1).
max occurs when y = 5 m A1
METHOD 2 – calculus
differentiating and equating to zero (M1)
\(\frac{dy}{dt}=10-10t=0\)
t =1
y (= 1(10 – 5)) = 5 m A1
METHOD 3 – symmetry
line of symmetry is t =1 (M1)
y (= 1(10 – 5)) = 5 m
[3 marks]
(c) attempt to solve t (10 − 5t ) = 0 (M1)
t = 2 (or t = 0 ) (A1)
x (= 5 + 8 × 2) = 21 m A1
Note: Do not award the final A1 if x = 5 is also seen.
[3 marks]
(d) METHOD 1
\(t=\frac{x-5}{8}\)
\(y=\left ( \frac{x-5}{8} \right )\left ( 10-5\times \frac{x-5}{8} \right )\)
METHOD 2
y = k (x − 5)( x − 21) A1
when x = 13, y = 5 so \(k=\frac{5}{(13-5)(13-21)}=-\frac{5}{64}\) M1A1
\(\left ( y=-\frac{5}{64}(x-5)(x-21) \right )\)
METHOD 3
if y = ax2 + bx + c
0 = 25a +5b + c
5 = 169a + 13b + c
0 = 441a + 21b + c M1A1
solving simultaneously \(a=-\frac{5}{64},b=\frac{130}{64},c=-\frac{525}{64}\) A1
\((y=-\frac{5}{64}x^{2}+\frac{130}{64}x-\frac{525}{64})\)
METHOD 4
use quadratic regression on (5, 0), (13, 5), (21, 0) M1A1
\(y=-\frac{5}{64}x^{2}+\frac{130}{64}x-\frac{525}{64}\) A1
Note: Question asks for expression; condone omission of “ y = ”.
[3 marks]
(e) trajectory of arrow is y = x tan 10 + 2 (A1)
intersecting y = x tan 10 +2 and their answer to (d) (M1)
(8.66, 3.53) ((8.65705…, 3.52647 ) … ) A1
(15.1, 4.66) ((15.0859 , 4.66006 ) … … ) A1
[4 marks]
(f) when \(x_{target}=8.65705…,t_{target}=\frac{8.65705…-5}{8}=0.457132…s\) (A1)
attempt to find the distance from point of release to intersection (M1)
\(\sqrt{8.65705…^{2}+(3.52647…-2)^{2}}(=8.79060…m)\)
time for arrow to get there is \(\frac{8.79060}{60}=0.146510…s\) (A1)
so the arrow should be released when
t = 0.311 (s) (0.310622… (s)) A1
[4 marks]
Total [21 marks]
Question
Ed walks in a straight line from point \({\text{P}}( – 1,{\text{ }}4)\) to point \({\text{Q}}(4,{\text{ }}16)\) with constant speed.
Ed starts from point \(P\) at time \(t = 0\) and arrives at point \(Q\) at time \(t = 3\), where \(t\) is measured in hours.
Given that, at time \(t\), Ed’s position vector, relative to the origin, can be given in the form, \({{r}} = {{a}} + t{{b}}\),
find the vectors \({{a}}\) and \({{b}}\).
Roderick is at a point \({\text{C}}(11,{\text{ }}9)\). During Ed’s walk from \(P\) to \(Q\) Roderick wishes to signal to Ed. He decides to signal when Ed is at the closest point to \(C\).
Find the time when Roderick signals to Ed.
▶️Answer/Explanation
Markscheme
\({{a}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right)\) A1
\({{b}} = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 4 \\ {16} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} {\frac{5}{3}} \\ 4 \end{array}} \right)\) (M1)A1
[3 marks]
METHOD 1
Roderick must signal in a direction vector perpendicular to Ed’s path. (M1)
the equation of the signal is \({\mathbf{s}} = \left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 12} \\ 5 \end{array}} \right)\;\;\;\)(or equivalent) A1
\(\left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right) + \frac{t}{3}\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 12} \\ 5 \end{array}} \right)\) M1
\(\frac{5}{3}t + 12\lambda = 12\) and \(4t – 5\lambda = 5\) M1
\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) A1
METHOD 2
\(\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right) \bullet \left( {\left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} { – 1 + \frac{5}{3}t} \\ {4 + 4t} \end{array}} \right)} \right) = 0\;\;\;\)(or equivalent) M1A1A1
Note: Award the M1 for an attempt at a scalar product equated to zero, A1 for the first factor and A1 for the complete second factor.
attempting to solve for \(t\) (M1)
\(t = 2.13\;\;\;\left( {\frac{{360}}{{169}}} \right)\) A1
METHOD 3
\(x = \sqrt {{{\left( {12 – \frac{{5t}}{3}} \right)}^2} + {{(5 – 4t)}^2}} \;\;\;\)(or equivalent)\(\;\;\;\left( {{x^2} = {{\left( {12 – \frac{{5t}}{3}} \right)}^2} + {{(5 – 4t)}^2}} \right)\) M1A1A1
Note: Award M1 for use of Pythagoras’ theorem, A1 for \({\left( {12 – \frac{{5t}}{3}} \right)^2}\) and A1 for \({(5 – 4t)^2}\).
attempting (graphically or analytically) to find \(t\) such that \(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0\left( {\frac{{{\text{d}}({x^2})}}{{{\text{d}}t}} = 0} \right)\) (M1)
\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) A1
METHOD 4
\(\cos \theta = \frac{{\left( {\begin{array}{*{20}{c}} {12} \\ 5 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)}}{{\left| {\left( {\begin{array}{*{20}{c}} {12} \\ 5 \end{array}} \right)} \right|\left| {\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)} \right|}} = \frac{{120}}{{169}}\) M1A1
Note: Award M1 for attempting to calculate the scalar product.
\(\frac{{120}}{{13}} = \frac{t}{3}\left| {\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)} \right|\;\;\;\)(or equivalent) (A1)
attempting to solve for \(t\) (M1)
\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) A1
[5 marks]
Total [8 marks]
Question
OACB is a parallelogram with \(\overrightarrow {{\text{OA}}} = \) a and \(\overrightarrow {{\text{OB}}} = \) b, where a and b are non-zero vectors.
a.Show that
(i) \({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2} + 2\)a \( \bullet \) b \( + |\)b\({|^2}\);
(ii) \({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).[4]
b.Given that \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right|\), prove that OACB is a rectangle.[4]
▶️Answer/Explanation
Markscheme
METHOD 1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = \overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{OC}}} \)
= (a + b) \( \bullet \) (a + b) A1
= a \( \bullet \) a + a \( \bullet \) b + b \( \bullet \) a + b \( \bullet \) b A1
= \(|\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{OA}}} } \right|^2} + {\left| {\overrightarrow {{\text{OB}}} } \right|^2} – 2\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}})\) A1
\(\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}}) = – \)(a \( \bullet \) b) A1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
(ii) METHOD 1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = \overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AB}}} \)
= (b − a) \( \bullet \) (b − a) A1
= b \( \bullet \) b − b \( \bullet \) a − a \( \bullet \) b + a \( \bullet \) a A1
= \(|\)a\({|^2}\) – 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = {\left| {\overrightarrow {{\text{AC}}} } \right|^2} + {\left| {\overrightarrow {{\text{BC}}} } \right|^2} – 2\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}})\) A1
\(\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}}) = \) a \( \bullet \) b A1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a \({|^2} – \) 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
[4 marks]
\(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2} \Rightarrow |\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\) R1(M1)
Note: Award R1 for \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2}\) and (M1) for \(|\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
a \( \bullet \) b \( = 0\) A1
hence OACB is a rectangle (a and b both non-zero)
with adjacent sides at right angles R1
Note: Award R1(M1)A0R1 if the dot product has not been used.
[4 marks]
Examiners report
In part (a), a significant number of candidates either did not use correct vector notation or simply did not use vector notation at all. A large number of candidates who appeared to adopt a scalar product approach, did not use the scalar product ‘dot’ and represented a \( \bullet \) b as ab. A few candidates successfully used the cosine rule with correct vector notation. A small number of candidates expressed a and b in general component form. In part (a) (ii), quite a number of candidates expressed \({\overrightarrow {{\text{AB}}} }\) as a – b rather than as b – a.
In part (b), some very well structured proofs were offered by a small number of candidates.