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High School Physics : Understanding Scalar and Vector Quantities

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Example Questions

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Example Question #51 : High School Physics

$\vec{A}=-i-2j$

What are the magnitude and angle, CCW from the x-axis, of $2\vec{A}$ ?

Possible Answers:

$m=4.47,\ \theta=243^o$

$m=4.47,\ \theta=127^o$

$m=-4.47,\ \theta=117^o$

$m=2.24,\ \theta=-63.4^o$

$m=4.47,\ \theta=63.4^o$

Correct answer:

$m=4.47,\ \theta=243^o$

Explanation:

When multiplying a vector by a constant (called scalar multiplication), we multiply each component by the constant.

$2\vec{A}=2(-1-2j)=-2i-4j$

The magnitude of this new vector is found with these new components:

$\sqrt{x^2+y^2}=\sqrt{(-2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=4.47$

To calculate the angle we must first find the inverse tangent of $\frac{y}{x}$ :

$\arctan(\frac{y}{x})=\arctan(\frac{-4}{-2})=\arctan(2)=63.4^o$

This is the principal arctan, but it is in the first quadrant while our vector is in the third. We to add the angle 180° to this value to arrive at our final answer.

$\theta=180^o+63.4^o=243^o$

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Example Question #31 : Understanding Scalar And Vector Quantities

Vector $\vec{A}$ has a magnitude of 2.24 and is at an angle of 63.4° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 3.16 at an angle of 342° CCW from the x-axis.

Find $\vec{A}+\vec{B}$ by using the nose-to-tail graphical method.

Possible Answers:

$m=5.00,\ \theta=14.0^o$

$m=4.12,\ \theta=76.0^o$

$m=2.24,\ \theta=14.0^o$

$m=4.21,\ \theta=14.0^o$

$m=2.24,\ \theta=-14.0^o$

Correct answer:

$m=4.21,\ \theta=14.0^o$

Explanation:

First, construct the two vectors using ruler and protractor:

Place the tail of $\vec{B}$ at the nose of $\vec{A}$ :

Construct the resultant $\vec{R}$ from the tail of $\vec{A}$ to the nose of $\vec{B}$ :

With our ruler and protractor, we find that $\vec{R}$ is 4.12 at an angle of 14.0° CCW from the x-axis.

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Example Question #31 : Understanding Scalar And Vector Quantities

$\vec{A}=-2.00i+3.00j$

$\vec{B}=-1.00i-2.00j$

Find the magnitude and angle CCW from the x-axis of $\vec{A}-\vec{B}$ using the nose-to-tail graphical method.

Possible Answers:

$m=-5.10,\ \theta=101^o$

$m=5.10,\ \theta=258^o$

$m=4.90,\ \theta=78.7^o$

$m=4.00,\ \theta=101^o$

$m=5.10,\ \theta=101^o$

Correct answer:

$m=5.10,\ \theta=101^o$

Explanation:

Construct $\vec{A}$ and $\vec{B}$ from their x- and y-components:

Since we are subtracting, reverse the direction of $\vec{B}$ :

Form $\vec{A}-\vec{B}=\vec{A}+(-\vec{B})$ by placing the tail of $-\vec{B}$ at the nose of $\vec{A}$ :

Construct and measure the resultant, $\vec{R}$ , from the tail of $\vec{A}$ to the nose of $-\vec{B}$ using a ruler and protractor:

$m_{\vec{R}}=5.10,\ \theta_{\vec{R}}=101^o$

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Example Question #172 : Matrices And Vectors

Express a vector with magnitude 2.24 directed 63.4° CCW from the x-axis in unit vector form.

Possible Answers:

$-2.00\text{i} - 1.00\text{j}$

$-1.00\text{i} - 2.00\text{j}$

$\text{1.00i + 2.00j}$

$\text{1.00i + 1.00j}$

$\text{2.00i + 1.00j}$

Correct answer:

$\text{1.00i + 2.00j}$

Explanation:

The x-coordinate is the magnitude times the cosine of the angle, while the y-coordinate is the magnitude times the sine of the angle.

$\text{magnitude}=2.24,\ \theta=63.4^o$

$x = 2.24 \cos(63.4^o) = 1.00$

$y = 2.24 \sin(63.4^o) = 2.00$

The resultant vector is: $\text{1.00i + 2.00j}$ .

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Example Question #171 : Matrices And Vectors

Vector $\vec{A}$ has a magnitude of 2.24 and is at an angle of 63.4° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 3.61 and is at an angle of 124° CCW from the x-axis.

Find $\vec{A}-2\vec{B}$ by using the nose-to-tail graphical method.

Possible Answers:

$m=6.40,\ \theta=321^o$

$m=3.00,\ \theta=-38.7^o$

$m=6.40,\ \theta=38.7^o$

$m=9.00,\ \theta=321^o$

$m=9.00,\ \theta=38.7^o$

Correct answer:

$m=6.40,\ \theta=321^o$

Explanation:

First, construct the two vectors using ruler and protractor:

$2\vec{B}$ is twice the length of $\vec{B}$ , but in the same direction:

Since we are subtracting, reverse the direction of $2\vec{B}$ :

Form $\vec{A}-2\vec{B}=\vec{A}+(-2\vec{B})$ by placing the tail of $-2\vec{B}$ at the nose of $\vec{A}$ :

Construct and measure the resultant $\vec{R}$ from the tail of $\vec{A}$ to the nose of $-2\vec{B}$ with a ruler and protractor.

$m_{\vec{R}}=6.40,\ \theta_{\vec{R}}=321^o$

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Example Question #51 : High School Physics

Vector $\vec{A}$ has a magnitude of 2.24 and is at an angle of 63.4° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 3.16 at an anlge of 342° CCW from the x-axis.

Find $\vec{A}+\vec{B}$ by using the parallelogram graphical method.

Possible Answers:

$m=4.12,\ \theta=14.0^o$

$m=2.24,\ \theta=-14.0^o$

$m=4.12,\ \theta=76.0^o$

$m=5.00,\ \theta=14.0^o$

$m=2.24,\ \theta=14.0^o$

Correct answer:

$m=4.12,\ \theta=14.0^o$

Explanation:

First, construct the two vectors using ruler and protractor:

Place the tails of both vectors at the same point:

Construct a parallelogram:

Construct and measure the resultant using ruler and protractor:

$m_{\vec{R}}=4.12,\ \theta_{\vec{R}}=14.0^o$

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Example Question #175 : Matrices And Vectors

$\vec{A}=-1.00i-2.00j$

$\vec{B}=3.00i-1.00j$

Find $\vec{A}-\vec{B}$ using the parallelogram graphical method.

Possible Answers:

$m=-5.00,\ \theta=14.0^o$

$m=4.90,\ \theta=78.7^o$

$m=5.10,\ \theta=258^o$

$m=4.12,\ \theta=194^o$

$m=5.10,\ \theta=101^o$

Correct answer:

$m=4.12,\ \theta=194^o$

Explanation:

Construct $\vec{A}$ and $\vec{B}$ from their x- and y-components:

Since we are subtracting, reverse the direction of $\vec{B}$ :

Place the tails of $\vec{A}$ and $-\vec{B}$ at the same point:

Construct a parallelogram:

Construct and measure the resultant using a ruler and protractor.

$m_{\vec{R}}=4.12,\ \theta_{\vec{R}}=194.^o$

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Example Question #176 : Matrices And Vectors

Vector $\vec{A}$ has a magnitude of 3.61 and is at an angle of 124° CCW from the x-axis. Vector $\vec{B}$ has a magnitude of 2.24 at an anlge of 63.4° CCW from the x-axis.

Find $2\vec{A}-\vec{B}$ using the parallelogram graphical method.

Possible Answers:

$m=9.00,\ \theta=38.7^o$

$m=6.40,\ \theta=141^o$

$m=3.00,\ \theta=-38.7^o$

$m=6.40,\ \theta=33.7^o$

$m=9.00,\ \theta=321^o$

Correct answer:

$m=6.40,\ \theta=141^o$

Explanation:

First, construct the two vectors using ruler and protractor:

$2\vec{A}$ is twice the length of $\vec{A}$ , but in the same direction:

Since we are subtracting, reverse the direction of $\vec{B}$ :

Place the tails of $2\vec{A}$ and $-\vec{B}$ at the same point:

Construct a parallelogram:

Construct and measure the resultant using ruler and protractor:

$m_{\vec{R}}=6.40,\ \theta_{\vec{R}}=141^o$

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Example Question #177 : Matrices And Vectors

$\vec{A}=1.00i+2.00j$

$\vec{B}=3.00i-1.00j$

Find $\vec{A}+\vec{B}$ .

Possible Answers:

$\vec{R}=3.16i+2.23j$

$\vec{R}=4.00i+1.00j$

$\vec{R}=3.00i+2.00j$

$\vec{R}=4.00i+2.00j$

$\vec{R}=1.00i+4.00j$

Correct answer:

$\vec{R}=4.00i+1.00j$

Explanation:

Finding the resultant requires us to add like components:

$\vec{R}=\vec{A}+\vec{B}$

$\vec{R}=(1.00i+2.00j)+(3.00i-1.00j)$

$\vec{R}=(1.00+3.00)i+(2.00-1.00)j$

$\vec{R}=4.00i+1.00j$

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Example Question #11 : Evaluate Geometric Vectors

$\vec{A}=3.00i-1.00j$

$\vec{B}=-2.00i+3.00j$

Find $2\vec{A}+\vec{B}$ .

Possible Answers:

$\vec{R}=3.16i+2.23j$

$\vec{R}=4.00i+1.00j$

$\vec{R}=3.00i+2.00j$

$\vec{R}=1.00i+4.00j$

$\vec{R}=4.00i+2.00j$

Correct answer:

$\vec{R}=4.00i+1.00j$

Explanation:

Finding the resultant requires us to add like components:

$\vec{R}=2\vec{A}+\vec{B}$

$\vec{R}=2(3.00i-1.00j)+(-2.00i+3.00j)$

$\vec{R}=(6.00i-2.00j)+(-2.00i+3.00j)$

$\vec{R}=(6.00-2.00)i+(-2.00+3.00)j$

$\vec{R}=4.00i+1.00j$

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High School Physics : Understanding Scalar and Vector Quantities

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #51 : High School Physics

Example Question #31 : Understanding Scalar And Vector Quantities

Example Question #31 : Understanding Scalar And Vector Quantities

Example Question #172 : Matrices And Vectors

Example Question #171 : Matrices And Vectors

Example Question #51 : High School Physics

Example Question #175 : Matrices And Vectors

Example Question #176 : Matrices And Vectors

Example Question #177 : Matrices And Vectors

Example Question #11 : Evaluate Geometric Vectors

All High School Physics Resources

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