Question

The K E of a body decreases by 19%. What is the
percentage decrease in momentum?​

Answer 1

Answer:

here is the right answer

Answer 2

Answer:

$ke \: = \: p {}^{2} \div 2m$

so,

$\sqrt[]{2mk} = \: p$

if KE decreases by 19% then, it is 81% of initial KE means,

K' = 0.81K

keeping value in equation 2

$\sqrt{2m(0.81k)} = p$

So, P' becomes 0.9 times of √2mk

P'= 0.9√2mk

and √2mk = P

Hence,

P' = 0.9 P

Means P' becomes 90% of P

Hence, P decreases by 10 %

Explanation:

Done