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Question

# Two particles A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s respectively (g=10m/s2). Find separation between them after one second.

A

0 m

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B

5 m

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C

10 m

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D

15 m

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Solution

## The correct option is B 5 m Method-I Position of A after 1 sec SA=ut−12gt2=5t−12times10×t2=5×1−12×10×12=5−5=0 i.e., the particle will return to ground at t=1 sec The position of B after 1 sec; SB=ut−12gt2=10×12=10−5=5m Hence separation between A and B, SB−SA=5m. Method-II Relative velocity method: Accleration of B w.r.t A −−→aBA=−→aB−−→aA=(−10)−(−10)=0 Initial relative velocity Also −−→vBA=−→vB−−→vA=10−5=5m/s Hence relative separation between particles ∴−−→SBA(in 1 sec)=−−→vBA×t=5×‘=5m ∴ Distance between A and B after 1 sec = 5 m.

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