Question 118: Motion in 2D & Vectors

Question

The resultant of two vector $$\mathbf{A}$$ and $$\mathbf{B}$$ is $$\mathbf{C}$$. If the magnitude of $$\mathbf{B}$$ is doubled, the new resultant vector becomes perpendicular to A. Then, the magnitude of $$\mathbf{C}$$ is

Accepted Answer

Given, $$\mathbf{A}+\mathbf{B}=\mathbf{C}$$
Resultant of $$A$$ and $$B$$ is
$$|C|=\sqrt{|\mathbf{A}|^2+|\mathbf{B}|^2+2|\mathbf{A}||\mathbf{B}| \cos 	heta} \quad 	ext{... (i)}$$
When magnitude of B is doubled, new resultant (R$$_1$$) is perpendicular to A.
$$\begin{aligned}
	herefore \quad \left|\mathbf{R}_1ight| \cdot \mathbf{A} & =0 \
|\mathbf{A}+2 \mathbf{B}| \cdot \mathbf{A} & =0 \
|\mathbf{A}|^2+2|\mathbf{A}||\mathbf{B}| \cos 	heta & =0 \
\cos 	heta & =\frac{-|\mathbf{A}|}{2|\mathbf{B}|}
\end{aligned}$$
Substituting this value in Eq. (i), we get
$$\begin{array}{cc} 
& |C|=\sqrt{|\mathbf{A}|^2+|B|^2+2|\mathbf{A}||\mathbf{B}|\left(\frac{-|\mathbf{A}|}{2|B|}ight)} \
\Rightarrow & |C|^2=|\mathbf{A}|^2-|\mathbf{A}|^2+|\mathbf{B}|^2 \
& |\mathbf{C}|^2=|\mathbf{B}|^2 \
	herefore & |C|=|B| \operatorname{or} C=B
\end{array}$$

Answer

2B

Answer

B

Answer

3B

Answer

4B

Motion in 2D & Vectors

The resultant of two vector $\mathbf{A} and \mathbf{B} is \mathbf{C}. If the magnitude of \mathbf{B} is doubled, the new resultant vector becomes perpendicular to A. Then, the magnitude of \mathbf{C}$ is

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