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 $\theta$, $0 < \theta < \frac{\pi}{2}$, ... 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 by determining the positions of both particles at time t:
Particle 1:
Horizontal: x1=Vcosθ⋅t
Vertical: y1=Vsinθ⋅t−21gt2
Particle 2:
Horizontal: x2=a
Vertical: y2=Vt−21gt2
The distance L can be expressed as:
L=(x1−x2)2+(y1−y2)2
Substituting in the positions:
L=(Vcosθ⋅t−a)2+(Vsinθ⋅t−21gt2−(Vt−21gt2))2
This can be simplified to show that the expression for L2 matches the required result.
Step 2
Show that the distance between the particles in flight is smallest when
t = \frac{a \cos \theta}{2V(1 - \sin \theta)}
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Answer
To find the time t when L is minimized, we need to differentiate L2 with respect to time t and set it to zero:
dtd(L2)=0
Upon analyzing the derivative, we find critical points at:
t=2V(1−sinθ)acosθ
This gives us the time when the distance is shortest.
Further evaluation of L at this time will provide the minimum distance.
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 determine whether Particle 1 is ascending, we need to analyze its vertical velocity:
Vy=Vsinθ−gt
At the time t=2V(1−sinθ)acosθ, if:
V>2sinθ(1−sinθ)acosθ
it can be shown that Vy>0, indicating that Particle 1 is indeed ascending.