if kinetic energy of a particle is decreased by 19% . By what % its momentum is changed??
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what r the options I think it is 19%
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We know that
KE = 1/2 m v^2 and momentum = mv
{ the mass will remain constant }
Let initial KE = K = 1/2 m v^2
after 19 % decrease it will be
K - 19 % of K = K - 0.19K = 0.81 K
Or , 0.81 K = 1/2 m v'^2
so, v'^2 = 2×0.81 K / m
or, v' = √(2×0.81K / m ) is the new velocity
So, new momentum = m × √(2×0.81K / m )
= m × √(2×0.81 (1/2 m v^2) / m )
= m × √0.81v^2
= m × 0.09v = 0.09 mv
or, 0.09 mv = 9/100 mv
= 9% of initial momentum
KE = 1/2 m v^2 and momentum = mv
{ the mass will remain constant }
Let initial KE = K = 1/2 m v^2
after 19 % decrease it will be
K - 19 % of K = K - 0.19K = 0.81 K
Or , 0.81 K = 1/2 m v'^2
so, v'^2 = 2×0.81 K / m
or, v' = √(2×0.81K / m ) is the new velocity
So, new momentum = m × √(2×0.81K / m )
= m × √(2×0.81 (1/2 m v^2) / m )
= m × √0.81v^2
= m × 0.09v = 0.09 mv
or, 0.09 mv = 9/100 mv
= 9% of initial momentum
neetcommitted75:
in question it is not mention that mass is constant and correct answer is - 10%
mass always remains constant. it need not to be given. and sorry about the calculation error....it will be 0.9 mv in place of 0.09 mv and hence the decrease will be mv - 0.9mv = 0.1 mv which is 10% of mv and as it's decrease so it will come out to be -10% . Thanks.
thank you
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