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: $$\dot{x} = a$$ $$y = V \cdot t - \frac{1}{2} g t^2$$ 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 i... 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 determine the distance $L$ between the two particles at time $t$ , we first express their coordinates in terms of $t$ .

For Particle 1:

$x_1 = V \cos \theta \cdot t$
$y_1 = V \sin \theta \cdot t - \frac{1}{2} g t^2$

For Particle 2:

$x_2 = a$
$y_2 = V \cdot t - \frac{1}{2} g t^2$

Thus, the distance $L$ can be expressed using the distance formula: $L = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$ Substituting the coordinates: $L = \sqrt{(V \cos \theta \cdot t - a)^2 + \left(V \sin \theta \cdot t - \frac{1}{2} g t^2 - V \cdot t + \frac{1}{2} g t^2\right)^2}$ Simplifying the second term gives: $L = \sqrt{(V \cos \theta \cdot t - a)^2 + (V \sin \theta \cdot t - V \cdot t)^2} = \sqrt{(V \cos \theta \cdot t - a)^2 + (V (\sin \theta - 1) \cdot t)^2}$

After further expansion and simplification, we find: $L^2 = 2V^2t^2(1 - \sin \theta) - 2aV \cos \theta + a^2$ .

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 value of $t$ that minimizes $L$ , we differentiate $L^2$ with respect to $t$ and set the derivative to zero:

Differentiate: $\frac{d(L^2)}{dt} = 4V^2t - 2aV \cos \theta = 0$
Solving for $t$ gives: $t = \frac{a \cos \theta}{2V(1 - \sin \theta)}$

Next, substitute this value of $t$ back into the expression for $L^2$ to find the smallest distance: $L_{min}^2 = \frac{a(1 - \sin \theta)}{2}$ Therefore, the smallest distance is: $L_{min} = \sqrt{ \frac{a(1 - \sin \theta)}{2}}$ .

Step 3

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

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Answer

The vertical motion of Particle 1 is determined by: $y = V \sin \theta \cdot t - \frac{1}{2} g t^2$

Particle 1 is ascending when $V \sin \theta - gt > 0$ . Thus, for Particle 1 to be ascending: $t < \frac{V \sin \theta}{g}$ Substituting the expression for $t$ will show that the condition follows if $V > \frac{aq \cos \theta}{\sqrt{2(1 - \sin \theta)}}$ . Consequently, this inequality will determine the relation of conditions under which Particle 1 is still in ascent when the distance is at its minimum.

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 $V > \frac{aq \cos \theta}{\sqrt{2(1 - \sin \theta)}}$.

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