Question 110: Motion in 2D & Vectors

Question

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{C}$. If the magnitude of $\vec{B}$ is doubled, the new resultant vector becomes perpendicular to $\vec{A}$, then the magnitude of $\overrightarrow{\mathrm{C}}$ is

Accepted Answer

$$ \begin{aligned} & \vec{A}+\vec{B}=\vec{C} \quad \vec{A} \ & C^2=A^2+B^2+2 A B \cos 	heta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $$
and $	an \alpha=\frac{\mathrm{B} \sin 	heta}{\mathrm{A}+\mathrm{B} \cos 	heta}$
When $B$ is doubled, resultant is perpendicular to $\vec{A}$
$$	herefore C_1^2=A^2+4 B^2+4 A B \cos 	heta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
From right angled triangle PSR
$$ \begin{aligned} & (2 B)^2=C_1^2+A^2 \ & C_1^2=4 B^2-A^2 \end{aligned} $$
Substituting in (ii) and solving,
$$ \mathrm{A}^2+2 \mathrm{AB} \cos 	heta=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$$
Substituting (iii) in (i), $\mathrm{C}=\mathrm{B}$

Answer

4 B

Answer

3 B

Answer

B

Answer

2 B

Motion in 2D & Vectors

The resultant of two vectors \vec{A} and \vec{B} is \vec{C}. If the magnitude of \vec{B} is doubled, the new resultant vector becomes perpendicular to \vec{A}, then the magnitude of \overrightarrow{\mathrm{C}} is

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