Perfectly Inelastic Collision

Question

A 2 kg lump of clay moving at 6 m/s collides with and sticks to a stationary 4 kg block on a frictionless surface. What fraction of the initial kinetic energy is lost during this collision?

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A perfectly inelastic collision conserves momentum but not kinetic energy, since the bodies move together afterward and some energy converts to heat and deformation. First find the common velocity from momentum conservation: $v = (2 	imes 6)/(2 + 4) = 12/6 = 2	ext{ m/s}$. The initial kinetic energy is $	frac{1}{2}(2)(6)^2 = 36	ext{ J}$, while the final kinetic energy is $	frac{1}{2}(6)(2)^2 = 12	ext{ J}$. The energy lost is $36 - 12 = 24	ext{ J}$, so the fraction lost is $24/36 = 2/3$. The option 1/3 reports the fraction retained rather than lost. The option 1/2 ignores the mass ratio in the energy accounting. The option 1/6 confuses the velocity ratio with an energy ratio. It is important to appreciate that momentum is conserved in every collision, but kinetic energy is conserved only in the elastic case; here the missing energy reappears as heat, sound, and permanent deformation of the clay. The heavier the struck object is relative to the incoming projectile, the larger the fraction of the original kinetic energy that ends up being lost. This is the NCERT analysis of energy loss in completely inelastic collisions. As a check, the lost fraction equals $m_2/(m_1+m_2) = 4/6 = 2/3$, the known result when the struck body is initially at rest.

Perfectly Inelastic Collision

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