Question

chapter- FORCE , WORK ENERGY & POWER.

Calculate the ratio of K.E's of a moving body of its velocity is reduced by ⅓rd of the initial velocity.
$ke1 \div ke2 = m1 \div m2 \times v1 \times v1 \div v1\times v1 \div 3$

Answer 1

let the initial velocity be v

$ke1 = ( \frac{1}{2} )mv {}^{2}$
final velocity is one_third of initial velocity.

therefore.,


$ke2 = \frac{1}{2} m( \frac{v}{3}) {}^{2} = \frac{1}{2} m \frac{ {v}^{2} }{9}$
divide. ke1/ke2

$\frac{ke1}{ke2} = \frac{ \frac{1}{2} m {v}^{2} }{ \frac{1}{2} m \frac{ {v}^{2} }{9} } = \frac{ {v}^{2} }{ \frac{ {v}^{2} }{9} } = 9$
ratio = 9:1

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