JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
Blocks of masses \(m , 2 m , 4 m\) and \(8 m\) are arranged in a line on a frictionless floor. Another block of mass \(m ,\) moving with speed \(v\) along the same line (see figure) ollides with mass \(m\) in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass \(8 m\) starts moving the total energy loss is \(p \%\) of the original energy. Value of \('p'\) is close to
- A \(77\)
- B \(37\)
- C \(87\)
- D \(94\)
Answer & Solution
Correct Answer
(D) \(94\)
Step-by-step Solution
Detailed explanation
All collisions are perfectly inelastic, so affer the final collision, all blocks are moving together. So let the final velocity be \(v^{\prime},\) so on applying momentum conservation: \(mv =16 m v ^{\prime} \Rightarrow v ^{\prime}= v / 16\) Now initial energy…
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