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A particle moves in the X-Y plane under the influence of a force such that its linear momentum isvecp(t) = A[(wedge)icos (kt) -(wedge) sin (kt)] where A and k are constants. The angle between the force and the momentum is: |
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Answer» `0^(@)` `vecf=(dvecp)/(DT)=(d)/(dt)[A(hati coskt-hatj sin kt)]` Thus `vecF.vecp=A k[-veci sin kt-hatj COS kt]` `[A(hati cos kt-hatjsin kt)]=0` The angle between `vecF` and `vecp is 90^(@)` . Hence (d) is the correct choice. |
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