JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(50\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
\({P}_{{i}}={P}_{{f}}\) \({m} \times 40=\frac{{m}}{2} \times {v}+\frac{{m}}{2} \times 60\) \(40=\frac{{v}}{2}+3\) \(\Rightarrow {v}=20\) \(({K} . {E} .)_{{I}}=\frac{1}{2} {m} \times(40)^{2}=800\, {m}\)…
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