the momentum of a body is increased by 50% when is the % change in kinetic energy 3 marks
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Answered by
24
hello friend..!!
p = mv , if this increases by 50 % ( 0.50 p )
the new momentum p' = p + 0.50 p = 1.50 p
the intial kinetic energy (k) = 1/2 mv² = (mv)²/ 2m = p²/ 2m
the new kinetic energy (k') = (p')² / 2m .
therefore ,
k' / k = (p')² / p²
k'/k = ( 1.50 p) ² / p²
k'/k = 2.25
k' = 2.25 ( k)
k' = 2.25 k
therefore the percentage change is ,
(k' - k ) / k x 100 = ( 2.25 k - k ) / k x 100 = 125 % .
therefore the % change in the kinetic energy is 125 %
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hope it is useful...!!
p = mv , if this increases by 50 % ( 0.50 p )
the new momentum p' = p + 0.50 p = 1.50 p
the intial kinetic energy (k) = 1/2 mv² = (mv)²/ 2m = p²/ 2m
the new kinetic energy (k') = (p')² / 2m .
therefore ,
k' / k = (p')² / p²
k'/k = ( 1.50 p) ² / p²
k'/k = 2.25
k' = 2.25 ( k)
k' = 2.25 k
therefore the percentage change is ,
(k' - k ) / k x 100 = ( 2.25 k - k ) / k x 100 = 125 % .
therefore the % change in the kinetic energy is 125 %
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hope it is useful...!!
Answered by
0
Answer:
125%
Explanation:
pf = pi + 50/100 pi
pf = 150/100 pi
change in K.E = pf^2/2m - pi^2/2m
= 5pi^2/8m
% change = 5/4 *100 = 125%
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