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NEET UG PHYSICS MCQ: LAWS OF MOTION

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A particle is projected with a velocity 200  ms−1200\mathrm{\;m}{\mathrm{s}^{ - 1}} at an angle of 600{60^0}. At the highest point, it explodes into three particles of equal masses. One goes vertically upwards with a velocity 100  ms−1100\mathrm{\;m}{\mathrm{s}^{ - 1}}, the second particle goes vertically downwards. What is the velocity of third particle?
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Correct option: C
300 ms−1300\,m{s^{ - 1}}
Option C (300,ms−1300\\,m{s^{ - 1}}) is correct because it is derived from applying the conservation of momentum at the highest point of the trajectory after the explosion.

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