Two masses m1 and m2(m2 > m1) are positioned as shown in figure, m1 being on the ground and m2 at a height h above the ground. When m2 is released, the speed at which it hits the ground will be
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a
2ghm1m2
b
2ghm1−m2m1+m2
c
2ghm1+m2m1−m2
d
2ghm2−m1m1+m2
answer is D.
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Detailed Solution
Since, a=m2−m1m2+m1g and v2−u2=2a s ⇒v2−02=2m2−m1m2+m1gh⇒v=2m2−m1m2+m1gh