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Q.39, find t he loss of energyduring the collision. |
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Answer» SOLUTION :`K_(1)` = Total KE before collision `=1/2mv^(2)+1/2(3m)v^(2).` `k_(2)=` Total KE after collision `=1/2(4m)V ^(2)` PUT the VALUE of V FOUND in Q. 39 `DeltaK=` loss of ` KE=K_(1)-K_(2)` `K_(1)` total KE before collision `=1/2mv^(2)+1/2(3m)v^(2)=2mv^(2)` `=2m5/8v^(2)asV=sqrt((5)/(8))v` `=5/4mv^(2)` `Deltak=` loss of KE during the collision `=K_(1)-K_(2)` `=2mv^(2)-5/4mv^(2)` `DeltaK=3/4mv^(2)` |
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