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5.Work, Energy, Power and Collision
hard
A block moving horizontally on a smooth surface with a speed of $40\, {m} / {s}$ splits into two parts with masses in the ratio of $1: 2$. If the smaller part moves at $60\, {m} / {s}$ in the same direction, then the fractional change in kinetic energy is :-
A
$\frac{1}{3}$
B
$\frac{2}{3}$
C
$\frac{1}{8}$
D
$\frac{1}{4}$
(JEE MAIN-2021)
Solution
$3 {MV}_{0}=2 {MV}_{2}+{MV}_{1}$
$3 {V}_{0}=2 {V}_{2}+{V}_{1}$
$120=2 {V}_{2}+60 \Rightarrow {V}_{2}=30\, {m} / {s}$
$\frac{\Delta {K} {E}}{{K} . {E} .}=\frac{\frac{1}{2} {MV}_{1}^{2}+\frac{1}{2} 2 {M} V_{2}^{2}-\frac{1}{2} 3 {M} V_{0}^{2}}{\frac{1}{2} 3 {M} V_{0}^{2}}$
$=\frac{{V}_{1}^{2}+2 {V}_{2}^{2}-3 {V}_{0}^{2}}{3 {V}_{0}^{2}}$
$=\frac{3600+1800-4800}{4800}$
$=\frac{1}{8}$
Standard 11
Physics
