if the k.E of a body is increased by 69% then momentum is increased by
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we know that momentum (p) = mv and KE = mv²/2
=> KE = p²/2m
so let the initial KE be KE only
=> KE1 = p²/2m => p1 = √2mKE1
when the KE is increase by 69% the new KE = KE2
KE2 = 1.69 KE1
1.69 KE1 = 1.69 p²/2m => p2 = √2m(1.69)KE1
=> p2 = 1.3 p1
=> percentage increase in momentum = (p2 - p1)× 100
=> (1.3p1 - p1)100 => 0.3p1 × 100
=> 30%
=> KE = p²/2m
so let the initial KE be KE only
=> KE1 = p²/2m => p1 = √2mKE1
when the KE is increase by 69% the new KE = KE2
KE2 = 1.69 KE1
1.69 KE1 = 1.69 p²/2m => p2 = √2m(1.69)KE1
=> p2 = 1.3 p1
=> percentage increase in momentum = (p2 - p1)× 100
=> (1.3p1 - p1)100 => 0.3p1 × 100
=> 30%
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Answer:
The momentum is increased by 30%.
Explanation:
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