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Particles A moves along the line y=4 \sqrt{3} with constant velocity \vec{v} of magnitude 2.0 \mathrm{~m} / \mathrm{s} and directed parallel to the positive x-axis. Particle B starts at the origin with zero speed and constant acceleration a (of magnitude 4.0 \mathrm{~m} / \mathrm{s}^2 ) at the same instant that the particle A passes The y-axis. The angle \theta between z and y-axis that would result in a collision between these two particles should have a value equal t.

Option: 1

30^{\circ}


Option: 2

45^{\circ}


Option: 3

50^{\circ}


Option: 4

80^{\circ}


Answers (1)

best_answer

If t is the time of collision, then for particle A
$$ x=v t -----\left ( 1 \right )
and for particle B,
\begin{aligned} & x=0+\frac{1}{2}(a \sin \theta) t^2 \\ & y=0+\frac{1}{2}(a \cos \theta) t^2 \end{aligned}
i.e..,
\begin{aligned} v t & =\frac{1}{2}(a \sin \theta) t^2 -----\left ( 2 \right )\\ 4 \sqrt{3} & =\frac{1}{2}(a \cos \theta) t^2-----\left ( 3 \right ) \end{aligned}
on substituting a=4 \mathrm{~m} / \mathrm{s}^2, v=2 \mathrm{~m} / \mathrm{s} and simplifying we get.
\begin{aligned} \frac{\sqrt{3}}{2} & =\cos \theta \\ & \Theta =\pi / 6=30^{\circ} \text { } \end{aligned}

Posted by

Devendra Khairwa

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