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Explain the special caes of elastic collision in one dimension . |
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Answer» Solution :Case 1 . If the TWO mases are equal `(m_(1)=m_(2)) v_(1f)=0, v_(2f)=v_(1i)` The first mass comes to rest and pushes off the second mass with its initial SPEED on collision . `:. ` Velocity of first mass after collision , `v_(1f)=((m_(1)-m_(2))/(m_(1)+m_(2)))v_(1i)` `= (0/(2m_(1)))v_(1i)` = 0 and velocity of second mass `v_(2f) =((2m_(1)v_(1i))/(m_(1)+m_(2)))` `=(2mv_(1i))/(m+m) ""[ :. m_(1)=m_(2)=m " supposed"] ` `v_(2f) =v_(1i)` ` :. ` The velocity of first ball after collision . Case 2: If `m_(2) gt gt m_(1)` then second ball is more HEAVIER then first one . ` :. ` The velocity of first after collision , `v_(1f) =((m_(1)-m_(2))/(m_(1)+m_(2)))v_(1i)` Neglecting the `m_(1)` compared to `m_(2)` `v_(1f) =((0-m_(2)))/((0+m_(2)))v_(1i)`. ` :. v_(1f) =-v_(1i)` `:. ` First ball having same velocity after collision and `v_(2f) =(2m_(1)v_(1i))/(m_(1)+m_(2)),` here `m_(1)=0` ` = (2xx0xxv_(1i))/(0+m_(2))=0` Hence, there is no effect in velocityof heavier ball and so it REMAIN static at its POSITION . Case 3 : If `m_(1) gt gt m_(2)` The velocity after collision of first ball , `v_(if) = ((m_(1)-m_(2))/(m_(1)+m_(2)))v_(1i)` here `m_(2) = 0 ` `v_(1f) =((m_(1)-0)/(m_(1)+0))v_(1i) = v_(1i)` velocity of second ball , `v_(2f) = (2m_(1)v_(1i))/(m_(1)+m_(2)) `,here `m_(2) = 0 ` `v_(2f) = 2v_(li)` |
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