Question

the kinetic energy is increased by 21% what percentage increase in magnitude of linear momentum

no need to give ierelative answers ​

Answer 1

$\large\rm { E_{k2} = \frac{121}{100} E_{k1}}$

$\large\rm { \frac{1}{2} mv^{2}_{2} = \frac{121}{100} \cdot \frac{1}{2} mv^{2}_{1}}$

$\large\rm { \therefore v_{2} = \frac{11}{10} v_{1}}$

$\large\rm { mv_{2} = \frac{11}{10} mv_{1}}$ ( multiplying both sides by m)

so $\large\rm { \frac{p_{2}}{p_{1}} - 1 = \frac{1}{10} }$

so $\large\rm { \frac{p_{2} - p_{1}}{p_{1}} × 100 = \bf{10}}$

So, the percentage increase in the magnitude of linear momentum is 10 %.

Answer 2

Explanation:

adding: l²+m²,m²+n²,n²+l²