an asteroid is approaching Earth 20 km/s. fortunately it will pass right in fron

Question

an asteroid is approaching Earth 20 km/s. fortunately it will pass right in front of the earth and not hit it. for the sake of this problem you can assume the Earth is moving in a straight line along the negative x axis and the asteroid is moving in a straight line along the negative Y axis with a speed "Vi" . after the "near encounter" the trajectory of the asteroid has deviated by 20 degrees. what is the final speed "Vf" of the asteroid? the speed of Earth in its orbit is 30 kilometers per second

Explanation / Answer

By momentum conservation in Y axis:

mvi = mvf*cos 20 degree

vf = vi/ cos 20 degree

= 20/ cos 20 degree

= 21.28 km/s Answer

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