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Obtain expression for velocities of the two bodies after elastic collision in one dimension . |
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Answer» SOLUTION :Consider elastic collision in one dimension as shown in figure . Suppose a ball of mass `m_(1)` moving with velocity`v_(1)` along X - axis undergoesan elasticcollision with a ball of mass `m_(2)` moving with velocity `v_(2i) and v_(1i) gt v_(2i)` Their final velocities are `v_(1F) and v_(2F)` respectively . Momentum is conserved in collision . `:. m_(1)v_(1i)+m_(2)v_(2i)=m_(1)v_(1f)+m_(2)v_(2f) ""...(1)` ` :. m_(1)(v_(1i)-v_(1f)) =m_(2)(v_(2f)-v_(2i))` Collision is elastic collision so kinetic energy isalso conserved . `1/2 m_(1)v_(1i)^(2)+m_(2)v_(2i)^(2) =m_(1)v_(1f)^(2)+m_(2)v_(2f)^(2) ....""(2)` ` :. m_(1)(v_(1i)-v_(1f))(v_(1i) +v_(1f))=m_(2)(v_(2f)-v_(2i)) (v_(2f)+v_(2i))` From equation(2) , `v_(1i)+v_(1f)=v_(2f)+v_(2i)` ` :. v_(1f)=v_(2f)+v_(2i)-v_(1i) ""....(3)` From equation(1) , `m_(1)v_(1i)+m_(2)v_(2i)=m_(1)(v_(2f)+v_(2i)-v_(1i))+m_(2)v_(2f)` `:. m_(1)v_(1i)+m_(2)v_(2i)=m_(1)v_(2f)=m_(1)v_(2f)+m_(1)v_(2f)+m_(1)v_(2i)-m_(1)v_(1)+m_(2)v_(2f)` `2m_(1)v_(1i)=m_(1)v_(2i)-m_(2)v_(2i)-m_(2)v_(2i)+m_(1)v_(2f)+m_(2)v_(2f)` `2m_(1)v_(1i)=(m_(1)-m_(2))v_(2i)+(m_(1)+m_(2))v_(2f)` If ball of mass `m_(2)`having intial velocity `v_(2i)=0` then `2m_(1)v_(1i) =(m_(1)+m_(2))v_(2f)` ` :. v_(2f) =(2m_(1)v_(1i))/(m_(1)+m_(2))` `v_(1f)=v_(2f)+v_(2i)-v_(1i)` ` = (2m_(1)v_(1i))/(m_(1)+m_(2))-v_(1i) "" [:. v_(2i)=0 " supposed"] ` `=(2m_(1)v_(1i)-m_(1)v_(1i)-m_(2)v_(1i))/(m_(1)+m_(2))` ` :. v_(1f) = (m_(1)v_(1i)-m_(2)v_(1i))/(m_(1)+m_(2))` ` :. v_(1f) = ((m_(1)-m_(2))v_(1i))/((m_(1)+m_(2)))` In theseequations if `m_(1),m_(2) and v_(1i) ` is giventhen thevalueof `v_(1f) and v_(2f)` can be calculated . |
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