Question

KE of a body is increased by 44%.What is the percent increase in the momentum?

Answer 1

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Answer 2

Dear Student,

◆ Answer -

% increase in momentum = 20 %

● Explanation -

Initial kinetic energy of the body is -

KEi = 1/2 m.v^2

Final kinetic energy of the body is -

KEf = 1/2 m.v'^2

Percent increase in kinetic energy is calculated by -

% increase in KE = (KEf - KEi) / KEi

44/100 = (1/2 m.v'^2 - 1/2 m.v^2) / 1/2 m.v^2

44/100 = (v'^2 - v^2) / v^2

44/100 = (v'/v)^2 - 1

(v'/v)^2 = 1 + 44/100

(v'/v)^2 = 144/100

v'/v = 6/5

Percent increase in momentum is given by -

% increase in momentum = (p'-p)×100/p

% increase in momentum = (mv'-mv)×100/mv

% increase in momentum = 100v'/v - 100

% increase in momentum = 100×6/5 - 100

% increase in momentum = 120 - 100

% increase in momentum = 20 %

Hence, % increase in momentum is 20 %.

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