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Q.

A particle of mass 200 g is dropped from a height of 50 m and another particle of mass 100 g is simultaneously projected up from the ground along the same line, with a speed of 100 m/s . The velocity of the centre of mass after 1 s will be:

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answer is D.

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Detailed Solution

After t = 1 sec, the velocity of : 200 g mass will be v1 = 0 + 10(1) = 10 m/s (down) and that of 100 g mass will be v2 = 100 - 10(1) = 90 m/s(up)Velocity v of centre of mass will be v = 200(10) + 100(-90)200 +100 = 703m/s up
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