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 ... 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, $L^2 = 2V^2t^2(1 - \sin \theta) - 2aVt \cos \theta + a^2$.
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Answer
To find the distance L between the two particles, we can use the equations of motion. The position of Particle 1 at time t is given by:
x1=Vtcosθ y1=Vtsinθ−21gt2
For Particle 2, we have:
x2=a y2=Vt−21gt2.
The distance L between the two particles can be expressed as:
L=(x1−x2)2+(y1−y2)2.
Substituting the expressions for x1, x2, y1, and y2 into this equation, we can simplify and show that:
L2=2V2t2(1−sinθ)−2aVtcosθ+a2.
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 $\frac{a}{2} \sqrt{1 - \sin \theta}$.
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Answer
To minimize the distance L, we need to differentiate L2 with respect to t and set the derivative equal to zero. We find:
dtd(L2)=0.
Solving for t gives:
t=2V(1−sinθ)acosθ.
Next, substituting this value back into the expression for L2 yields:
Lmin2=4a2(1−sinθ),
therefore,
Lmin=2a1−sinθ.
Step 3
Show that the smallest distance between the two particles in flight occurs while Particle 1 is ascending if $V > \frac{a \cos \theta}{\sqrt{2}\sin \theta(1 - \sin \theta)}$.
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Answer
To establish that Particle 1 is ascending when the minimum distance occurs, we evaluate the velocity of Particle 1:
Vy=Vsinθ−gt.
Substituting the expression for t gives:
Vy=Vsinθ−g2V(1−sinθ)acosθ.
If we require Vy>0, we must show:
V>2sinθ(1−sinθ)acosθ,
which confirms that Particle 1 is still ascending.
