Question

A certain vector in the $x y$ plane has an $x$-component of $12\, m$ and a $y$-component of $8\, m$. It is then rotated in the $x-y$ plane so that its $x$-component is halved. Then ifs new $y$-component is approximately

• A. 14 m
• B. 13.11 m
• C. 10 m
• D. 2.0 m

Answer 1

Answer: (B)

13.11 m

Answer 2

Let A be vector in $x-y$ plane. Its $x$ and $y$ components are
$A_{x}=12 m$ and $A_{y}=8 m$
$A=\sqrt{A_{x}^{2}+A_{y}^{2}}$
$=\sqrt{(12)^{2}+(8)^{2}}$
$A=\sqrt{208} m$
When the vector is rotated in $x-y$ plane, then $x$ component become halved and its new y component
$A_{y}'=\sqrt{\left(\frac{A_{x}}{2}\right)^{2}+A_{y}^{'2}} \sqrt{208}=\sqrt{(6)^{2}+A_{y}^{\prime 2}}$
$A_{y}'=\sqrt{208-36}$
$A_{y}'=\sqrt{172}$
$=13.11 \,cm$