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A bulletof mass m fired at 30^(@) to the horizontal leaves the barrel of the gun with a velocity v . The bullet hits a soft target at a height h above the ground whileit is moving downward andground while it is movingdownward andemerge outwith half the kinetic energyit had before hitting the target . Which of the following statements are correct in respect of bullet after it emergesout ofteh target ? |
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Answer» The velocity of the bullet will be reduced to half its initial value  9B) From law of conservation of energy , `U_(i)+K_(i) =U_(f)+K_(f)` ` 0+1/2 mv^(2) =U_(f) +K_(f)` `((v)^(2))/2 = (v^(2))/2 = - gh` `(v)^(2)=v^(2)-2gh rArr v = sqrt(v^(2)-2gh) " "...(i)` where v is speed of the bullet just before hittingthe target . Let speed after emerging form the target is v . then , `1/2 (mv)^(2) = 1/2 [1/2m(v)^(2)]` (given) ` :. 1/2 m(v)^(2) = 1/4 m(v)^(2) ` ` :. 1/2 m(v)^(2) = 1/4 m[v^(2)-2gh]` ` (v")^(2) =1/4 m[v^(2)-2gh]` `(v")^(2) =(v^(2)-2gh)/2` `= (v^(2))/2 - gh` `v.. =sqrt((v^(2))/2-gh)" " ...(ii)` By taking ratio (1) to (2) `(v.)/(v..) =(sqrt(v^(2)-2gh))/((sqrt(v^(2)-2gh))/(sqrt(2)))=sqrt(2)` `v .. = (v.)/(sqrt(2))` ` = v^(2) (v/2)` `(v..)/(v/2)=sqrt(2)` ` = 1.414 GT 1 ` ` v.. gt (v.)/2 ` (D) As the velocity of the bullet CHANGES to v.. which is less than v HENCE , path followed will CHANGE andthe bullet reaches at point B insteadof A . as shown . (D) As the bulletis passing throughthe targetthe loss in energy of the bullet is transferred to particles of the target . Therefore , thier internal energy increases . |
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