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Grade: 12th pass
        If kinetic energy increased by 0.12% then what change happened in momentum...?
one year ago

Answers : (1)

Vikas TU
6869 Points
							
K.E is given as:
K.E. = P^2/2m...........(1)
After increased by 0.12% we get New K.E’ as:
K.E’ = 0.0012K.E = P’^2/2m..............(2)
Dividng the eqns. (1) and (2) we get,
(P/P’)^2 = 10000/12
P/P’ = 100/3.464 => 28.868
P’ = P/28.68 => 0.0346P
Change in Momentum =>
(P – 0.0346P)*100/P => 96.5%
one year ago
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