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A constant force F acts for 1 sec, on a body of mass 1kg moving with perpendicular to its initial velocity, thenA. distance covered by the body is `(u+(F)/(2))`B. displacement of the body is `sqrt(u^(2)+((F)/(2))^(2))`C. change in kinetic energy of the body is `sqrt((1)/(2)(u^(2)+F^(2)))`D. momentum of the body is increased by F/2. |
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Answer» m=1kg `s_(y)=0+(1)/(2)at^(2)` `s_(y)=(1)/(2)((F)/(1))(I)^(2)=(F)/(2)` t=1sec, `s_(x)=u` `s=sqrt((s_(x))^(2)+(s_(y))^(2))=sqrt(u^(2)+((F)/(2))^(2))` `W_(F)=DeltaK` `Fxx(F)/(2)=(1)/(2)mV^(2)-(1)/(2)mu^(2)` `(F^(2))/(2)+(1)/(2)u^(2)=(V^(2))/(2)` `V=sqrt(F^(2)+u^(2))` J=`DeltaP` `FxxI=DeltaP=F` As constant force acts on the body and angle between u `&` F is neither `0^(@)` nor `180^(@)`, path is parabolic. |
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