Question

The kinetic energy of a body decreases by 19%.what is the percentage decrease in its linear momentum?

Answer 1

Please find below the answer..

Answer 2

Solution :

$\tt K = \frac{1}{2} m {v}^{2} = \frac{ {p}^{2} }{2m} \\ \\ \therefore \tt p = \sqrt{2mK}$

When KE decreases by 19 %. New KE is

$\tt K' = \frac{100 - 19}{100} \\ \\ \tt K' = \frac{81K}{100}$

New momentum

$\tt p' = \sqrt{2mK'} \\ \\ \tt \implies \sqrt{2m \frac{81}{100}K } \\ \\ \tt \implies \frac{9}{10 } \sqrt{2mK} \\ \\ \tt \implies \frac{9}{10} p$

Percentage decrease in linear momentum

$\tt \implies \frac{p - p'}{p} \times 100 \\ \\ \tt \implies \frac{(p - \frac{9p}{10} )100 }{10} \\ \\ \tt \implies10\%$

Linear momentum decreases by 10 %