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If a particle covers half the circle of radius R with constant speed thenA. momentum change is mvrB. chanfe in K.E is `1//2mv^(2)`C. change in K.E. is `mv^(2)`D. change in K.E. is zero |
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Answer» Correct Answer - D As momentum is a vector quantity `therefore` Change in momentum `Delta P=2mv sin (theta //2)` `=2mv sin (90)=2mv` But kinetic energy remains always constant so change in kinetic energy is zero. |
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