Question

The momentum of a body increased by $50$%. If $v_1$ is the initial velocity of the body the change in kinetic energy is

• A. $2{mv}_1^2$
• B. $\frac{5}{4}{mv}_1^2$
• C. $\frac{5}{8}{mv}_1^2$
• D. $\frac{3}{2}{mv}_1^2$

Answer 1

The correct option is C $\frac{5}{8}{mv}_1^2$

K.E = $\frac{1}{2}$mv${}^2$ = $\frac{1}{2}$pv = $\frac{1}{2}$ $\frac{p^2}{2m}$

p1 = p1p2 = p + $\frac{p}{2}$ = $\frac{3p}{2}$ = 1.5p

K.E${}_2$ = $\frac{1}{2}$ $\frac{p_2^2}{2m}$ = $\frac{1}{2}$ $\ast$ $\frac{(1.5p{)}^2}{m}$ = $\frac{9p^2}{8m}$

K.E${}_2$ = $\frac{9}{4}$ K.E${}_1$

Therefore change in kinetic energy is K.E = K.E${}_2$ - K.E${}_1$

= $\frac{9}{4}$ K.E${}_1$ - K.E${}_1$

= $\frac{5}{4}$ $\times$ $\frac{1}{2}mv{1}^2$ = $\frac{5}{8}m$ $v_1^2$