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)

One App. One Place for Learning.

Most popular questions from this chapter

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Circular Motion and Gravitation

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Study anywhere. Anytime. Across all devices.

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) $v_{1} + v_{2} = v_{1} - v_{2}$ .