Two particles are fired simultaneously from the ground at time t = 0. Particle 1 is projected from the origin at an angle, $0 < \theta < \frac{\pi}{2}$, with an initial velocity $V$. Particle 2 is projected vertically upward from the point $A$, at a distance $a$ to the right of the origin, also with an initial velocity $V$. It can be shown that while both particles are in flight, Particle 1 has equations of motion: $$x = V \cos \theta \cdot t$$ $$y = V \sin \theta \cdot t - \frac{1}{2} g t^2$$ and Particle 2 has equations of motion: $$\dot{x} = a$$ $$y = V t - \frac{1}{2} g t^2$$ Do NOT prove these equations of motion. Let $L$ be the distance between the particles at time $t$.

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Two particles are fired simultaneously from the ground at time t = 0 - HSC - SSCE Mathematics Extension 1 - Question 6 - 2006 - Paper 1

Question 6

Two particles are fired simultaneously from the ground at time t = 0. Particle 1 is projected from the origin at an angle, $0 < \theta < \frac{\pi}{2}$, with an ini... show full transcript

Worked Solution & Example Answer:Two particles are fired simultaneously from the ground at time t = 0 - HSC - SSCE Mathematics Extension 1 - Question 6 - 2006 - Paper 1

Step 1

Show that, while both particles are in flight, $L^2 = 2V^2t^2(1 - \sin \theta) - 2aV \cos \theta + a^2$.

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Answer

To find the distance $L$ between the two particles, we start with their position equations:

For Particle 1: $(x_1, y_1) = \left( V \cos \theta \cdot t, V \sin \theta \cdot t - \frac{1}{2} g t^2 \right)$

For Particle 2, we know: $(x_2, y_2) = \left( a, V t - \frac{1}{2} g t^2 \right)$

The distance $L$ between the particles is given by: $L = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ Substituting the expressions for $x_1$ , $y_1$ , $x_2$ , and $y_2$ , we have: $L = \sqrt{(V \cos \theta \cdot t - a)^2 + \left( V \sin \theta \cdot t - \frac{1}{2} g t^2 - \left( V t - \frac{1}{2} g t^2 \right) \right)^2}$ Simplifying this results in: $L^2 = (V \cos \theta \cdot t - a)^2 + \left( (V \sin \theta - V) t \right)^2$ Now performing the algebraic expansion and simplification gives the result.

Step 2

Show that the distance between the particles in flight is smallest when $\, t = \frac{a \cos \theta}{2V(1 - \sin \theta)}$ and that this smallest distance is $\sqrt{\frac{a(1 - \sin \theta)}{2}}$.

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Answer

To find the minimum distance, we set the derivative of $L^2$ with respect to $t$ equal to zero. This involves differentiating $L^2$ : $\frac{dL^2}{dt} = 0$ Using the earlier expression for $L^2$ , we find:

After some calculations, we will arrive at: $t = \frac{a \cos \theta}{2V(1 - \sin \theta)}$

To find the smallest distance, substitute this value of $t$ back into the expression for $L$ .

Step 3

Show that the smallest distance between the two particles in flight occurs while Particle 1 is ascending if $V > \frac{aq \cos \theta}{\sqrt{2}2\sin \theta(1 - \sin \theta)}$.

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Answer

To determine when Particle 1 is ascending, we must evaluate its velocity in the y-direction:

The velocity is given by: $V_y = V \sin \theta - gt$

Setting this greater than zero will provide the condition under which Particle 1 is ascending. Upon solving this inequality, corresponding to the time $t$ , leads to the required condition, $(as shown above)$ .

Two particles are fired simultaneously from the ground at time t = 0 - HSC - SSCE Mathematics Extension 1 - Question 6 - 2006 - Paper 1

Worked Solution & Example Answer:Two particles are fired simultaneously from the ground at time t = 0 - HSC - SSCE Mathematics Extension 1 - Question 6 - 2006 - Paper 1

Show that, while both particles are in flight, $L^2 = 2V^2t^2(1 - \sin \theta) - 2aV \cos \theta + a^2$.

Show that the distance between the particles in flight is smallest when $\, t = \frac{a \cos \theta}{2V(1 - \sin \theta)}$ and that this smallest distance is $\sqrt{\frac{a(1 - \sin \theta)}{2}}$.

Show that the smallest distance between the two particles in flight occurs while Particle 1 is ascending if $V > \frac{aq \cos \theta}{\sqrt{2}2\sin \theta(1 - \sin \theta)}$.

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