14-05-09 Collision

JEE Advanced 2015 Guidance › Question Of the Day questions › 14-05-09 Collision

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

Abhishek Priyam ·2009-05-14 03:43:17

Final answer?

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msp ·2009-05-14 05:37:58

(M1+M2)v=M1V

v=√g(H+D)

V=(1+M2/M1)√g(H+D)

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aposlil ·2009-12-03 09:31:16

What is the answer?

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eureka123 ·2009-12-03 09:49:20

This 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 v_relative=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

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aposlil ·2009-12-03 09:56:28

Kk thanx

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14-05-09 Collision

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