NEET Class neet Answered

please answer this

Asked by Prashant DIGHE | 29 Jan, 2020, 10:20: PM

Expert Answer

Let t seconds be the time of collision after the particles started moving from top and bottom.

Vertical distance h₁ travelled by particle which is dropped from the height h is given by,

h₁ = (1/2)gt² .....................(1)

Vertical distance h₂ travelled by the partcle which is projected from ground with speed

begin mathsize 14px style square root of 2 space g space h end root end style

is given by,

h₂ = (

begin mathsize 14px style square root of 2 space g space h end root end style

) t - [ (1/2) g t² ] ..................(2)

from eqn.(1) and (2), h₁ + h₂ = h = (

begin mathsize 14px style square root of 2 space g space h end root end style

) t .....................(3)

hence time of collision from starting time is given by,

begin mathsize 14px style t space equals space square root of fraction numerator h over denominator 2 g end fraction end root end style

...................(4)

hence from eqn.(1) and eqn.(4), we get h₁ = (1/2)g (h/2g) = (1/4)h

Hence collision has taken place at a height (3/4)h above ground level

Since collision is completely inelastic, all the kinetic energy is lost and the particles stick together.

Hence after collision, the combined partcles start from rest to reach ground .

If time taken is τ for the combined particle to reach ground , then we have , (3/4) h = (1/2) g τ²

hence time

begin mathsize 14px style tau space equals space square root of 3 over 2 end root space open parentheses square root of h over g end root close parentheses end style

; hence time τ in units of

begin mathsize 14px style square root of h over g end root end style

begin mathsize 14px style square root of 3 over 2 end root space end style

Answered by Thiyagarajan K | 30 Jan, 2020, 08:56: PM

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