JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
Three blocks \(A, B\) and \(C\) are lying on a smooth horizontal surface, as shown in the figure. \(A\) and \(B\) have equal masses, \(m\) while \(C\) has mass \(M\). Block \(A\) is given an initial speed \(v\) towards \(B\) due to which it collides with \(B\) perfectly inelastically. The combined mass collides with \(C\), also perfectly inelastically. The combined mass collides with \(C\), also perfectly inelastically if \(\frac{5}{6}^{th}\) of the initial kinetic energy is lost in whole process. What is value of \(M/m\) ?
- A \(5\)
- B \(2\)
- C \(4\)
- D \(3\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
Apply \(LMC\) (Linear Momentum Conservation) \(mv = (2m + M) v'\) \(v' = \frac{{mv}}{{2m + M}}\) Initial energy \(\frac{1}{2}\) \(m{v^2}\) Final energy \(\frac{1}{2}\,\left( {2m + M} \right){\left( {\frac{{mv}}{{2m + M}}} \right)^2}\)…
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