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Derive an expression for kinetic energy in pure rolling. |
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Answer» Solution :In perfectly inclastic 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_(i)=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)` ...(1) Total kinetic energy after collision. `KE_(f)=(1)/(2)(m_(1)+m_(2))V^(2)` ...(2) Loss of KE, `DeltaQ=KE_(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)` ...(3) Substituting equation `v = (m_(1)u_(1)+m_(2)u_(2))/((m_(1)+m_(2)))` in equation (3), 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)` |
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