Chapter 3: Q52. (page 72)

Two vectors, ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ , add to a resultant ${\vec{v}}_{R} = {\vec{v}}_{1} + {\vec{v}}_{2}$ . Describe ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ if (a) $v_{R} = v_{1} + v_{2}$ , (b) $v_{R}^{2} = v_{1}^{2} + v_{2}^{2}$ , and (c) $v_{1} + v_{2} = v_{1} - v_{2}$ .

Short Answer

Expert verified

(a) ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ are in the same direction when $v_{R} = v_{1} + v_{2}$ .

(b) ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ are perpendicular when $v_{R}^{2} = v_{1}^{2} + v_{2}^{2}$ .

Step by step solution

Step 1. Given data

If two vectors are $\vec{a}$ and $\vec{b}$ ,and the angle between them is $θ$ , using the triangle rule of vector addition, the resultant vector has a magnitude $R = \sqrt{a^{2} + b^{2} + 2 a b \cos θ}$ .

Given data:

The first vector is ${\vec{v}}_{1}$ .

The second vector is ${\vec{v}}_{2}$ .

Assumptions:

Let $θ$ be the angle between the vectors for part (a), and $ϕ$ be the angle between the vectors for part (b).

Step 2. Calculation for part (a)

It is given that the magnitude of the resultant vector is $v_{R} = v_{1} + v_{2}$ .

You know that the resultant is

$\begin{matrix} v_{R} = \sqrt{v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos θ} \\ v_{1} + v_{2} = \sqrt{v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos θ} \\ {(v_{1} + v_{2})}^{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos θ \\ v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos θ \end{matrix}$

Comparing both sides of the above equation,

$\begin{matrix} \cos θ = 1 \\ θ = \cos^{- 1} (1) \\ θ = 0 ° \end{matrix}$

Hence, ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ are in the same direction when $v_{R} = v_{1} + v_{2}$ .

Step 3. Calculation for part (b)

The given condition is $v_{R}^{2} = v_{1}^{2} + v_{2}^{2}$ .

You know that the resultant is

$\begin{matrix} v_{R} = \sqrt{v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos ϕ} \\ {v_{R}}^{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos ϕ \\ v_{1}^{2} + v_{2}^{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} \cos ϕ \end{matrix}$

Comparing both sides of the above equation,

role="math" localid="1644912448369" $\begin{matrix} \cos ϕ = 0 \\ ϕ = \cos^{- 1} (0) \\ ϕ = 90 ° \end{matrix}$

Hence, ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ are perpendicular when $v_{R}^{2} = v_{1}^{2} + v_{2}^{2}$ .

Step 4. Calculation for part (c)

The given condition is $v_{1} + v_{2} = v_{1} - v_{2}$ .

Now, simplifying the given condition,

$\begin{matrix} v_{1} + v_{2} - (v_{1} - v_{2}) = 0 \\ v_{1} + v_{2} - v_{1} + v_{2} = 0 \\ 2 v_{2} = 0 \\ v_{2} = 0 \end{matrix}$

Hence, the magnitude of ${\vec{v}}_{2}$ is zero when $v_{1} + v_{2} = v_{1} - v_{2}$ .

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Two vectors, ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ , add to a resultant ${\vec{v}}_{R} = {\vec{v}}_{1} + {\vec{v}}_{2}$ . Describe ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ if (a) $v_{R} = v_{1} + v_{2}$ , (b) $v_{R}^{2} = v_{1}^{2} + v_{2}^{2}$ , and (c) $v_{1} + v_{2} = v_{1} - v_{2}$ .

Short Answer

Step by step solution

Step 1. Given data

Step 2. Calculation for part (a)

Step 3. Calculation for part (b)

Step 4. Calculation for part (c)

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Two vectors,v→1andv→2, add to a resultantv→R=v→1+v→2. Describev→1andv→2if (a)vR=v1+v2, (b)vR2=v12+v22, and (c)v1+v2=v1-v2.

Short Answer

Step by step solution

Step 1. Given data

Step 2. Calculation for part (a)

Step 3. Calculation for part (b)

Step 4. Calculation for part (c)

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Two vectors, ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ , add to a resultant ${\vec{v}}_{R} = {\vec{v}}_{1} + {\vec{v}}_{2}$ . Describe ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ if (a) $v_{R} = v_{1} + v_{2}$ , (b) $v_{R}^{2} = v_{1}^{2} + v_{2}^{2}$ , and (c) $v_{1} + v_{2} = v_{1} - v_{2}$ .