If kinetic energy increased by 300% then increasement of momentum
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It would be K.e= undertoot 2Xmomentum
Answered by
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We have,
p=√(2mK)……,where K is the Kinetic Energy.
Let K1 and K2 be the initial and final K.E.
Let p1 and p2 be the initial and final momentum.
According to ques.,
K2 = K1+K1*300%
K2 = K1+3*K1
K2 = 4*K1
p1 = √(2mK1)
p2 = √(2mK2)
p2 = √(2m*4*K1)
p2 = 2√(2mK1)
p2 = 2*p1
Now,
% increase in momentum=(p2-p1)/p1*100%
= (2*p1-p1)/p1*100%
= 100%
p=√(2mK)……,where K is the Kinetic Energy.
Let K1 and K2 be the initial and final K.E.
Let p1 and p2 be the initial and final momentum.
According to ques.,
K2 = K1+K1*300%
K2 = K1+3*K1
K2 = 4*K1
p1 = √(2mK1)
p2 = √(2mK2)
p2 = √(2m*4*K1)
p2 = 2√(2mK1)
p2 = 2*p1
Now,
% increase in momentum=(p2-p1)/p1*100%
= (2*p1-p1)/p1*100%
= 100%
pranjalsingh49:
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