If the kinetic energy is increased by 100%% what is the percentage change in its momentum
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Since, increase in KE by % means that KE is being doubled.
CASE 1: KE=mv²/2 = m²v²/2m =(mv)²/2m
KE= p²/2m {p is momentum}
p=√2(KE)m........(i)
CASE 2: 2KE= mv²/2 =(mv)²/2{doub. KE
2KE=p²/2m
p'=2√(KE)m........(ii)
Therefore, percentage change in 'p'(momentum);
%=p/p' =2{√(KE)m}× / √2(KE)m
=200 / √2
=10√2 / √2
=10% answerrrrrrrrrrrrrrrrrrr......
HOW THE ANSWER IS CORRECT AS I'M NOT CONFIRM SO PLEASE CHECK IN THE RESULT & IF CORRECT PLEASE THANK ME & MARK BRAINLIEST.
AND IF INCORRECT INFORM ME THROUGH COMMENT, I'LL TRY CORRECT IT.
THANK YOU.
CASE 1: KE=mv²/2 = m²v²/2m =(mv)²/2m
KE= p²/2m {p is momentum}
p=√2(KE)m........(i)
CASE 2: 2KE= mv²/2 =(mv)²/2{doub. KE
2KE=p²/2m
p'=2√(KE)m........(ii)
Therefore, percentage change in 'p'(momentum);
%=p/p' =2{√(KE)m}× / √2(KE)m
=200 / √2
=10√2 / √2
=10% answerrrrrrrrrrrrrrrrrrr......
HOW THE ANSWER IS CORRECT AS I'M NOT CONFIRM SO PLEASE CHECK IN THE RESULT & IF CORRECT PLEASE THANK ME & MARK BRAINLIEST.
AND IF INCORRECT INFORM ME THROUGH COMMENT, I'LL TRY CORRECT IT.
THANK YOU.
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