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InterviewSolution
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What are collisions ? Explain the possible types of collision ? Develop the theory of one dimensional elastic collision . |
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Answer» Solution :A Process in which the motion of a system of particles changes but keeping the total momentum conserved is called collision . Collisions are two types : (1) ELASTIC (2) inelastic . To show relative velocity of approach before collision is EQUAL to relative velocity of separation after collision Let two bodies of masses`m_(1) ,m_(2)` are moving with velocities`u_(1) ,u_(2)` along the same LINE in same direction collied elastically . Let their velocities after collision are `v_(1) and v_(2)` .  According to the LAW of conservation of linear momentum According to the law of conservation of linear momentum `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)` or`m_(1)(u_(1)-v_(1))=m_(2)(v_(2)-u_(2)) "" ...(1)` According to the law of conservation of kinetic energy `(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)=(1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2)` `m_(1)(u_(1)^(2)-v_(1)^(2))=m_(2)(v_(1)^(2)-u_(1)^(2))""...(2)` Dividing eqn . (2) by (1) `(u_(1)^(2)-v_(1)^(2))/(u_(1)-v_(1))=(v_(2)^(2)-u_(2)^(2))/(v_(2)-u_(2))or` ` u_(1)+v_(1)=u_(2)+v_(2)rArru_(1)-u_(2)=v_(2)-v_(1) "" ...(3)` i.e , In elastic collisions relative velocity of approach of the two bodies before collision = relative velocity of separation of the two bodies collision . Velocities of two bodies after elastic collision : To Find `v_(1), " put "v_(2)=u_(1)-u_(2)+v_(1)" in eqn".(1)` `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)(u_(1)-u_(2)+v_(1))` `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)u_(1)-m_(2)u_(2)+m_(2)v_(1)` `m_(1)u_(1)+m_(2)u_(2)-m_(2)u_(1)+m_(2)u_(2)=v_(1)(m_(1)+m_(2))` `u_(1)(m_(1)-m_(2))+2m_(2)u_(2)=v_(1)(m_(1)+m_(2))` `thereforev_(1)=(m_(1)-m_(2))/(m_(1)+m_(2))" "u_(1)+(2m_(2))/(m_(1)+m_(2))" " u_(2)""....(4) ` Similarly , `v_(2)=(u_(2)(m_(1)-m_(2)))/(m_(1)+m_(2))+(2m_(2)u_(1))/(m_(1)+m_(2))"" ...(5) ` |
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