2023HSCShort answer

Q13 (b) (iii)

3 marks

Question

Particle $A$ is projected from the origin with initial speed $v\ \mathrm{m\,s^{-1}}$ at an angle $\theta$ with the horizontal plane. At the same time, particle $B$ is projected horizontally with initial speed $u\ \mathrm{m\,s^{-1}}$ from a point that is $H$ metres above the origin, as shown in the diagram.

The position vector of particle $A$ , $t$ seconds after it is projected, is given by

\mathbf r_A(t)=\begin{pmatrix}vt\cos\theta\\ vt\sin\theta-\frac12gt^2\end{pmatrix}.

(Do NOT prove this.)

The position vector of particle $B$ , $t$ seconds after it is projected, is given by

\mathbf r_B(t)=\begin{pmatrix}ut\\ H-\frac12gt^2\end{pmatrix}.

(Do NOT prove this.)

The angle $\theta$ is chosen so that $\tan\theta=2$ .

The two particles collide.

(iii) Show that

H=\frac{2u^2}{g}.

Uncropped source page 13 for question 13.b.iii; requires human review and cropping — Reviewed visual · source page 13

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Evaluates the dot product of the velocity vectors and equates to 0, or equivalent merit	2
Observes that the dot product of the velocity vectors is 0, or equivalent merit	1

Sample answer:

At $t=T$ the vectors $\vec v_A$ and $\vec v_B$ are perpendicular, so $\vec v_A\cdot\vec v_B=0$ .

\begin{aligned} \begin{pmatrix}v\cos\theta\\v\sin\theta-gT\end{pmatrix}\cdot\begin{pmatrix}u\\-gT\end{pmatrix}&=0\\ uv\cos\theta-gT(v\sin\theta-gT)&=0\\ u^2-2ugT+g^2T^2&=0\\ (u-gT)^2&=0\\ u=gT=\frac{gH}{2u}\\ H&=\frac{2u^2}{g} \end{aligned}

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Syllabus alignment

Vectors

Introduction to VectorsRepresent and operate with two-dimensional vectors; use magnitude, direction, scalar products and vectors in geometric proofs and projectile motion.

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