JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
\(3 {MV}_{0}=2 {MV}_{2}+{MV}_{1}\) \(3 {V}_{0}=2 {V}_{2}+{V}_{1}\) \(120=2 {V}_{2}+60 \Rightarrow {V}_{2}=30\, {m} / {s}\) \(\frac{\Delta {K} {E}}{{K} . {E} .}=\frac{\frac{1}{2} {MV}_{1}^{2}+\frac{1}{2} 2 {M} V_{2}^{2}-\frac{1}{2} 3 {M} V_{0}^{2}}{\frac{1}{2} 3 {M} V_{0}^{2}}\)…
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