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the kinetic energy of a body is increased by 300%.calculate the percentage change in its momentum  
Asked by nabendumandal | 11 Sep, 2018, 09:09: PM
answered-by-expert Expert Answer
Kineteic Energy E = (1/2)m×v2 ..............(1)
 
where m is mass and v is speed.
 
let us take log on both side of eqn.(1) and differntiate after taking log.
It is assumed that mass do not change and it remains constant
 
we get begin mathsize 12px style fraction numerator partial differential E over denominator E end fraction space equals space 2 fraction numerator partial differential v over denominator v space space end fraction space space space space space o r space space space space space space open parentheses fraction numerator begin display style partial differential E end style over denominator begin display style E end style end fraction cross times 100 close parentheses space equals space 2 open parentheses fraction numerator begin display style partial differential v end style over denominator begin display style v space space end style end fraction cross times 100 close parentheses end style ..................(2)
From eqn.(2), we say that percentage increase of energy is twice the percentage increase of speed.
since it is given that increase in energy is 300%, increase in speed  is 150%
 
Momentum p = m×v  ..................(3)
 
Since momentum is directly proportional to speed v for the given mass m,
momentum also increased by 150% for the 300% increase of energy. 


Answered by Thiyagarajan K | 12 Sep, 2018, 10:52: AM

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