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Discuss the completely inelastic collision in one dimension . |
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Answer» Solution :Here particle of mass `m_(1)` is moving speed `v_(1f)` in x direction . ` :. theta_(1)=theta_(2) = 0 ` From the law of CONSERVATION of momentum , momentum before collision and momentum after collision is same . ` :. M_(1)v_(1f)=(m_(1)+m_(2))v_(1f) ""...(1)` Energy is dissipated during collision , `DELTAK` = kinetic energy before collision- kinetic energy after collision ` 1/2 m_(1)v_(1)^(2)i - 1/2 (m_(1)+m_(2))v_(f)^(2) "" [:. v_(2i)=0] ` Substituting the value of EQUATION (1) , `DeltaK = 1/2m_(1)v_(1i)^(2)-1/2(m_(1)+m_(2))xx(m_(1)/(m_(1)+m_(2)))^(2)v_(1i)^(2)` `=1/2m_(1)v_(1i)^(2)[1-(m_(1))/(m_(1)+m_(2))] ` ` :. DeltaK= 1/2 m_(1)m_(2)v_(1i)^(2)` is positive . Note Ifparticles moves in the same direction and speed of particle of mass `m_(1) v_(1i)` and if `v_(1i) gt v_(2i)` them , after inelastic both having speed `v_(f)` `v_(f) = (m_(1)v_(1i)+m_(2)v_(2i))/(m_(1)+m_(2))` Kinetic energybefore collision , ` K_(i) = 1/2 m_(1)v_(1i)^(2) +1/2 m_(2)v_(2i)^(2)` Kinetic energy after collision , `K_(f) =1/2 (m_(1)+m_(2))v_(f)^(2)` ` :. K_(f) = 1/2(m_(1)+m_(2))((m_(1)v_(1i)+m_(2)v_(2i))^(2))/((m_(1)+m_(2))^(2))` `= 1/2((m_(1)v_(1i)+m_(2)v_(2i))^(2))/((m_(1)+m_(2)))` ` :. ` Energy dissipated after collision , `K_(i)-K_(f) =1/2 m_(1)v_(1i)^(2)+1/2m_(2)v_(2i)^(2)` `-1/2(-m_(1)v_(1i)+m_(2)v_(2i))^(2)/(m_(1)+m_(2))` `{m_(1)^(2)v_(1i)^(2)+m_(1)m_(2)v_(2i)^(2)+m_(1)m_(2)v_(1i)^(2)+m_(2)^(2)v_(2i)^(2)=1/2} (-(m_(1)^(2)v_(1i)^(2)+2m_(1)m_(2)v_(1i)v_(2i)+m_(2)^(2)v_(2i)^(2))/((m_(1)+m_(2))))` `m_(1)m_(2)v_(2i)^(2)+m_(1)m_(2)v_(1i)^(2)=1/2((-2m_(1)m_(2)v_(1i)v_(2i)+v_(2i)^(2)))/(m_(1)+m_(2))` `=(m_(1)m_(2))/(2(m_(1)+m_(2)))(v_(1i)^(2)-2v_(1i)^(2)-2v_(1i)v_(2i)+v_(2i)^(2))` `=(m_(1)m_(2))/(2(m_(1)+m_(2)))(v_(1i)^(2)-2v_(1i)v_(2i)+v_(2i)^(2))` `=(m_(1)m_(2))/(2(m_(1)+m_(2)))[v_(1i)-v_(2i)]^(2)`. Hence ,kineticenergy is conserved in inelastic collision . |
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