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

 Let v and y be the initial and final velocities of the body. Then

$$\begin{gathered} \frac{\frac{1}{2}m{\left(v^{\text{'}}\right)}^2}{\frac{1}{2}{\mathrm{mv}}^2}=\frac{81}{100} \\ \therefore \frac{v^{\text{'}}}{v}=\frac{9}{10} \\ \mathrm{or} \frac{{\mathrm{mV}}^{\text{'}}}{\mathrm{mV}}=\frac{90}{100}=90\mathrm{\%} \\ \mathrm{So}, \mathrm{the} \mathrm{decrease} \mathrm{in} \mathrm{momentum} \mathrm{is} 10\%. \end{gathered}$$