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(a) Explain perfect inelastic collision and derive an expression for loss of kinetic energy in perfect inelastic collision. |
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Answer» Solution :In a perfectly INELASTIC or completely inelastic collision, the objects stick together permanently aft er collision such that they move with common velocity. Let the two bodies with masses `m_(1) "and" m_(2)` move with initial velocities `u_(1) "and" u_(2)` respectively before collision. Aft er perfect inelastic collision both the objects move together with a common velocity v as shown in figure. SINCE, the linear momentum is conserved during collisions,  The common velocity can be computed by `v = (m_(1)u_(1) + m_(2)u_(2))/(m_(1) + m_(2))`...(1) Loss of kinetic ENERGY in perfect inelastic collision: In perfectly inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, heat, light etc. Let `KE_(i)` be the total kinetic energy before collision and `KE_(f)` be the total kinetic energy after collision. Total kinetic energy before collision, `KE_(e) = 1/2 m_(1)u_(1)^(2) + 1/2 m_(2)u_(2)^(2)`...(2) Total kinetic energy after collision, `KE_(f) = 1/2 (m_(1) + m_(2))v^(2)`...(3) Then the loss of kinetic energy is Loss of `KE, DeltaQ = EK_(f) - KE_(i)` = `1/2 (m_(1) + m_(2)) v^(2) - 1/2 m_(1)u_(1)^(2) - 1/2 m_(2)u_(2)^(2)`...(4) Substituting equation (1) in equation (4), and on simplifying (expand v by USING the algebra `((a + b)^(2) = a^(2) + b^(2) + 2AB)`, we get Loss of KE, `DeltaQ = 1/2 ((m_(1)m_(2))/(m_(1)+ m_(2))) (u_(1) – u_(2))^(2)`...(5) |
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