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QJEEM20Ph2c95: JEE Main 2020 Physics - Class 11 JEEM20Ph2c95 Solution: Blocks of masses m, 2m, 4m and 8m are arranged in a line on a…

Blocks of masses m, 2m, 4m and 8m are arranged in a line on a frictionless floor. Another block of mass m, moving with speed v along the same line (see figure) collides with mass m in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass 8m starts moving the total energy loss is p% of the original energy. Value of 'p' is close to : JEE Main 2020 (Online) 4th September Morning Slot Physics - Center of Mass and Collision Question 68 English

Multiple Choice Options

  • A 37
  • B 77
  • C 87
  • D 94

Step-by-Step Verified Solution

All collisions are perfectly inelastic, so after the final collision, all blocks are moving together. So let the final velocity be v', so on applying momentum conservation :

mv = 16mv'

⇒\Rightarrow v' = v16{v \over {16}}

Now initial energy Ei = 12mv2{1 \over 2}m{v^2}

Final energy : Ef = 12×16m(v16)2{1 \over 2} \times 16m{\left( {{v \over {16}}} \right)^2}

Energy loss : Ei – Ef

= 12mv2{1 \over 2}m{v^2} - 12×m(v16)2{1 \over 2} \times m{\left( {{v \over {16}}} \right)^2}

= 12mv2[1−116]{1 \over 2}m{v^2}\left[ {1 - {1 \over {16}}} \right]

= 12mv2[1516]{1 \over 2}m{v^2}\left[ {{{15} \over {16}}} \right]

Total energy loss is p% =
Energy loss
Original energy
 ×\times 100

= 12mv2[1516]12mv2×100{{{1 \over 2}m{v^2}\left[ {{{15} \over {16}}} \right]} \over {{1 \over 2}m{v^2}}} \times 100

= 93.75% ≃\simeq 94 %

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