Answer:
Explanation:
Given
charge of first body
charge of second body
Particle 1 is at origin and particle 2 is at
third Particle which charge +q must be placed left of because it will repel the q charge while
will attract it
suppose it is placed at a distance of x m
Answer:
The target's velocity is about 1320 m/s in the direction 265.7°.
Explanation:
In order for there to be a collision between missile and target, we must have ...
(target starting position) + (target movement) = (missile movement)
assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...
(10.2 s)(1350 m/s) = 13,770 m
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We are interested in the target movement, so we can solve for that:
(target movement) = (missile movement) - (target starting position)
In terms of meters, this is ...
(target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°
The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.
As a positive angle, the target's direction is ...
-94.3° +360° = 265.7°
The direction of the target's velocity is 265.7°.
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If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.
Using rectangular coordinates, we have ...
13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)
23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)
Then the difference is ...
(12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)
and the (3rd-quadrant) angle is ...
target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°
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The target's speed is found by dividing the distance it covers by the time it takes.
√(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s
