3(M1+M2)v=M1V
v=√g(H+D)
V=(1+M2/M1)√g(H+D)
Q.1 A ball B1 with mass(very small) m1 sits on the top of another ball B2 of mass m2. The bottom of B2 is at a height H above the ground and the bottom of B1 is at ht H+D above ground. Both ball are dropped from rest. All collisions are elastic. Find the height to which B1 rises.
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5 Answers
24This is a famous ques
height=D+9H
soln..
before collison ,both move with √2gH=v
after B2 collides it satrts moving upwrd with v(becoz e=1)
but B1 stilll moves down with v
so vrelative=v-(-v)=2v
after the balls collide with each other,the basket balls speed remains same i.e v upward whereas relative velocity is 2v
so velocity of small ball after collison is in upward direction(becoz mass very small)=2v+v=3v
applying energy conservation
we get height rise=D+(9v2/2g)
but v2=2gH
=> h=D+9H