Blocks of masses m,2m 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:

Option 1 - <p>77</p>
Option 2 - <p>87</p>
Option 3 - <p>37</p>
Option 4 - <p>94</p>
2 Views|Posted 7 months ago
Asked by Shiksha User
1 Answer
A
7 months ago
Correct Option - 4
Detailed Solution:


? v

mv=16mv1

V1=V16

Δk loss =12mv2-12 (16m)V162

12mv21516

% loss =1516*100=93.75%

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Similar Questions for you

Let ‘h’ be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

h=ut+12at2

=0×1+12×10×1×1

= 5m

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Physics NCERT Exemplar Solutions Class 12th Chapter Five 2025

Physics NCERT Exemplar Solutions Class 12th Chapter Five 2025

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