Question

The kinetic energy of a body is increased by $21\%$. What is the percentage increase in the magnitude of linear momentum of the body?

• A. $100\%$
• B. $11\%$
• C. $10\%$
• D. $21\%$

Answer 1

The correct option is C $10\%$

Given $K_2=K_1+\frac{21}{100}{K}_1=$ $\frac{121}{100}{K}_1$
$\frac{K_2}{K_1}=\frac{121}{100}$
$\frac{\frac{1}{2}m{v}_2^2}{\frac{1}{2}m{v}_1^2}=\left(\frac{121}{100}\right)$

$v_2^2=\frac{121}{100}{v}_1^2$

$v_2=\frac{11}{10}{v}_1$

$m{v}_2=\frac{11}{10}m{v}_1$

$p_2=\frac{11}{10}{p}_1$
Change in momentum is $\Delta p=p_2-p_1=\frac{11}{10}{p}_1-p_1=\frac{1}{10}{p}_1$
Thus percentage increase in linear momentum is $\frac{\Delta p}{p_1}\times 100=\frac{\frac{1}{10}{p}_1}{p_1}\times 100=10\%$