
Inelastic collision
mv=16mv1
ΔKloss=21mv2−21(16M)(16v)2
=21mv2−21M16v2
=21mv2(1615)
Blocks of masses m,2m,4m and 8m are arranged in a line of a frictionless floor. Another block of mass m , moving with speed υ 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 \text{p%} of the original energy. Value of ‘p ’ is close to:

Held on 4 Sept 2020 · Verified 6 Jul 2026.
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