An oil drop of mass 3.25X10^-15 kg falls vertically with uniform velocity, through the air between vertical parallel plates whic
h are 2 cm apart. When a p.d. of 1000V is applied to the plates ,the drop moves to the positively charged plate, being inclined at 45 degree to the vertical . Calculate the charge on the drop.
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Answer:
the third piece landed meters in a direction 45 degrees from the observer's right towards the front
Explanation:
Due to conservation of momentum, the momentum of the third piece must add to the other two momenta in vector from to render zero (initial momentum of the firework on a plane parallel to ground. Therefore the momentum components of the third piece must have a component to the right equal in magnitude to the momentum of the piece that traveled to the left, and a momentum component pointing to the front of the observer of equal magnitude to that of the piece that traveled behind the observer. Then the direction of the third momentum must be at 45 degrees from the observer's right towards the observer's front.
Also, the magnitude of a momentum vector of such components, is given by the Pythagorean theorem. Recall also that the magnitudes of the momentum vectors of the first two pieces must be equal since they traveled equal distances.
Then the distance traveled by the third piece must also keep this proportionality: