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5.Work, Energy, Power and Collision
hard
A block moving horizontally on a smooth surface with a speed of $40\, {ms}^{-1}$ splits into two equal parts. If one of the parts moves at $60\, {ms}^{-1}$ in the same direction, then the fractional change in the kinetic energy will be $x: 4$ where $x=..... .$
A
$4$
B
$10$
C
$1$
D
$50$
(JEE MAIN-2021)
Solution
${P}_{{i}}={P}_{{f}}$
${m} \times 40=\frac{{m}}{2} \times {v}+\frac{{m}}{2} \times 60$
$40=\frac{{v}}{2}+3$
$\Rightarrow {v}=20$
$({K} . {E} .)_{{I}}=\frac{1}{2} {m} \times(40)^{2}=800\, {m}$
$({K} . {E} .)_{{f}}=\frac{1}{2} \frac{{m}}{2} \cdot(20)^{2}+\frac{1}{2} \cdot \frac{{m}}{2}(60)^{2}=1000\, {m}$
$|\Delta {K} . {E} .|=|1000\, {m}-800 \,{m}|=200\, {m}$
$\frac{\Delta {K} . {E}}{({K} . {E} .)_{{i}}}=\frac{200 \,{m}}{800\, {m}}=\frac{1}{4}=\frac{{x}}{4}$
${x}=1$
Standard 11
Physics
