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 :

Solution

Physics sir · 4 steps

Momentum stays mv the whole way, so the kinetic energy falls as 1/M while the lump grows.

1. Momentum is fixed. The floor is frictionless, so no outside horizontal force acts. Every collision keeps the total momentum at P = mv. Each collision only adds mass to the moving lump.
2. Energy of a lump. A lump of mass M carrying momentum P has kinetic energy P²/2M. With P fixed, the kinetic energy falls as 1/M. The original energy is K₀ = P²/2m.

   K_(final) = (mv)²/2(m+m+2m+4m+8m) = (mv)²/(2 × 16m) = K₀/16

3. When 8m moves. The block of mass 8m starts moving only when the moving lump hits it and sticks. The whole 16m then moves at v/16. Only 1/16 of K₀ is left as kinetic energy.
4. So. The loss is 15/16 = 93.75% of the original energy, so p ≈ 94, option (d).

The idea to keep. In a chain of sticking collisions, the momentum stays fixed and only the mass grows. Since K = P²/2M, the fraction of energy left is the initial mass divided by the final total mass.

Answer from the NTA final answer key. Solution by PrepWiser.