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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

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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 in... 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

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

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Answer

To find the distance LL between the two particles, we can use the equation derived from their positions:

  1. The positions of the particles are given as:

    • Particle 1:
      x1=Vcos⁡θ⋅t,y1=Vsin⁡θ⋅t−12gt2x_1 = V\cos\theta \cdot t, \\ y_1 = V\sin\theta \cdot t - \frac{1}{2}gt^2
    • Particle 2:
      x2=a,y2=Vt−12gt2x_2 = a, \\ y_2 = Vt - \frac{1}{2}gt^2

  2. The distance LL is given by: L=(x1−x2)2+(y1−y2)2L = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} Substituting the equations: L=(Vcos⁡θ⋅t−a)2+(Vsin⁡θ⋅t−12gt2−Vt+12gt2)2L = \sqrt{(V\cos\theta \cdot t - a)^2 + \left(V\sin\theta \cdot t - \frac{1}{2}gt^2 - Vt + \frac{1}{2}gt^2\right)^2}

  3. Simplifying:

    • The yy-component simplifies to: y1−y2=(Vsin⁡θ−V)ty_1 - y_2 = (V\sin\theta - V)t
    • Plug this into LL: L=(Vcos⁡θ⋅t−a)2+(Vsin⁡θ−V)2t2L = \sqrt{(V\cos\theta \cdot t - a)^2 + (V\sin\theta - V)^2 t^2}

  4. Finally, to derive L2L^2: L2=(Vcos⁡θ⋅t−a)2+(V(sin⁡θ−1)t)2L^2 = (V\cos\theta \cdot t - a)^2 + (V(\sin\theta - 1)t)^2 This leads to the equation as required.

Step 2

(ii) 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}{2}(1 - \sin \theta)}.

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Answer

To find the minimum distance, we need to differentiate LL with respect to tt and set it to zero.

  1. Initially, we find L2L^2 from previous steps and differentiate: d(L2)dt=0\frac{d(L^2)}{dt} = 0

  2. Using the result of the differentiation:

    • We set tt using the formula derived: t=acos⁡θ2V(1−sin⁡θ)t = \frac{a \cos \theta}{2V(1 - \sin \theta)}

  3. Substitute back into the distance formula to find the smallest distance: Lmin=a2(1−sin⁡θ)L_{min} = \sqrt{\frac{a}{2}(1 - \sin \theta)}.

Step 3

(iii) 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 show the condition for Particle 1 to be ascending:

  1. Particle 1 ascends as long as its vertical velocity is positive:

    • The vertical component is given by Vsin⁡θ−gt>0V\sin\theta - gt > 0.

  2. Substitute tt at its minimum distance into this condition: Vsin⁡θ>gacos⁡θ2V(1−sin⁡θ)V\sin\theta > g \frac{a \cos \theta}{2V(1 - \sin \theta)}

  3. Rearranging gives: V>acos⁡θ2sin⁡θ(1−sin⁡θ)V > \frac{a \cos \theta}{\sqrt{2} \sin \theta(1 - \sin \theta)} This proves the necessary condition for Particle 1's ascent.

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