If a momentum of a particle is increased by 100% then percentage increase in kinetic energy of the particle is
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Kinetic energy = p^2/2m
If the momentum increases by 100% , then the p increases to 2p.
Substitute p =1
K before increase in momentum = p^2/2m = 1/2m - (1)
K' after increase in momentum by 100% = 2p^2/2m= 4/2m - (2)
Let p = 1
Divide (2) by (1) and substitute p=1
K'/K = (4/2m) / (1/2m)
= 4 ( 2m gets divided by 2m to give 1)
K'= 4K ( cross-multiplication)
Percentage increase = (K'-K / K) * 100
= (4K-K/K) *100
= (3K /K) *100
= 3*100
= 300%
When momentum of a body increases by 100% then its kinetic energy will increase by 300%.
hope it helps to you.
If the momentum increases by 100% , then the p increases to 2p.
Substitute p =1
K before increase in momentum = p^2/2m = 1/2m - (1)
K' after increase in momentum by 100% = 2p^2/2m= 4/2m - (2)
Let p = 1
Divide (2) by (1) and substitute p=1
K'/K = (4/2m) / (1/2m)
= 4 ( 2m gets divided by 2m to give 1)
K'= 4K ( cross-multiplication)
Percentage increase = (K'-K / K) * 100
= (4K-K/K) *100
= (3K /K) *100
= 3*100
= 300%
When momentum of a body increases by 100% then its kinetic energy will increase by 300%.
hope it helps to you.
Answered by
6
Kinetic energy = p^2/2m
If the momentum increases by 100% , then the p increases to 2p.
Substitute p =1
K before increase in momentum = p^2/2m = 1/2m - (1)
K' after increase in momentum by 100% = 2p^2/2m= 4/2m - (2)
Let p = 1
Divide (2) by (1) and substitute p=1
K'/K = (4/2m) / (1/2m)
= 4 ( 2m gets divided by 2m to give 1)
K'= 4K ( cross-multiplication)
Percentage increase = (K'-K / K) * 100
= (4K-K/K) *100
= (3K /K) *100
= 3*100
= 300%
When momentum of a body increases by 100% then its kinetic energy will increase by 300%.
hope it helps to you.
If the momentum increases by 100% , then the p increases to 2p.
Substitute p =1
K before increase in momentum = p^2/2m = 1/2m - (1)
K' after increase in momentum by 100% = 2p^2/2m= 4/2m - (2)
Let p = 1
Divide (2) by (1) and substitute p=1
K'/K = (4/2m) / (1/2m)
= 4 ( 2m gets divided by 2m to give 1)
K'= 4K ( cross-multiplication)
Percentage increase = (K'-K / K) * 100
= (4K-K/K) *100
= (3K /K) *100
= 3*100
= 300%
When momentum of a body increases by 100% then its kinetic energy will increase by 300%.
hope it helps to you.
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