CBSE-XI-Physics
02: Physics and Mathematics
- #2Let
A→andB→be the two vectors of magnitude 10 unit each. If they are inclined to the X-axis at angle 30° and 60° respectively, find the resultant.Ans : Angle between `` \stackrel{\to }{\mathrm{A}}\mathrm{and}\stackrel{\to }{\mathrm{B}}``, θ = 60° - 30° = 30°
`` \left|\stackrel{\to }{\mathrm{A}}\right|=\left|\stackrel{\to }{\mathrm{B}}\right|=10\mathrm{units}
\mathrm{The}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{resultant}\mathrm{vector}\mathrm{is}\mathrm{given}\mathrm{by}
\mathrm{R}=\sqrt{{A}^{2}+{A}^{2}+2AA\mathrm{cos}\theta }
=\sqrt{{10}^{2}+{10}^{2}+2\times 10\times 10\times \mathrm{cos}30°}
=\sqrt{200+200\mathrm{cos}30°}
=\sqrt{200+200\times \frac{\sqrt{3}}{2}}
=19.3\mathrm{units}``
Let β be the angle between `` \stackrel{\to }{\mathrm{R}}\mathrm{and}\stackrel{\to }{\mathrm{A}}``.
`` \therefore \mathrm{\beta }={\mathrm{tan}}^{-1}\left(\frac{A\mathrm{sin}30°}{A+A\mathrm{cos}30°}\right)
\Rightarrow \mathrm{\beta }={\mathrm{tan}}^{-1}\left(\frac{10\mathrm{sin}30°}{10+10\mathrm{cos}30°}\right)
\Rightarrow \mathrm{\beta }={\mathrm{tan}}^{-1}\left(\frac{1}{2+\sqrt{3}}\right)={\mathrm{tan}}^{-1}\left(\frac{1}{3.372}\right)
\Rightarrow \mathrm{\beta }={\mathrm{tan}}^{-1}\left(0.26795\right)=15°``
Angle made by the resultant vector with the X-axis = 15° + 30° = 45°
∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the X-axis.
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