# Vector Physics Problems and Solutions

Zubair Sensei
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## 13 Problems with Their Each Solution to Practice Vector in Physics

We present 13 problems with solutions to help you learn vector in physics. We also provide images that help you understand the problems and the solutions better. Have a good study.

### 1. Displacement vector

Slide the image above
Sabrina walked 75 meters to the east. Then she turned 30 degrees to the left and walked 25 meters. Determine the magnitude of Sabrina’s displacement vector.
Solution:
Slide the image above to see the resultant vector. The resultant vector magnitude calculation is as follows.
\begin{aligned} R=&\sqrt{A^2+B^2+2AB\cos{\theta}}\\ =&\sqrt{75^2+25^2+2(75)(25)\cos{30^{\circ}}}\\ =&97.46 \;\text{meter} \end{aligned}
Based on the above calculations, the displacement of the Sabrina is 97.46 meters.

### 2. Find the resultant magnitude of $\overrightarrow{A}-\overrightarrow{B}$

Slide the image above
“Satuan” is a word in Indonesian language. It translates as “unit” in english.
Look at the image above. $\overrightarrow{A}$ has magnitude of 45 units. At the same time, $\overrightarrow{B}$ is 30 units in magnitude. Determine the magnitude of $\overrightarrow{A}-\overrightarrow{B}$.
Solution:
First we have to realize that the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\theta=30^{\circ}+67^{\circ}=97^{\circ}$. Then we can calculate the magnitude.
\begin{aligned} R=&\sqrt{A^2+B^2-2AB\cos{\theta}}\\ =&\sqrt{45^2+30^2-2(45)(30)\cos{97^{\circ}}}\\ \approx&\;57\;\text{unit} \end{aligned}
So, the magnitude of $\overrightarrow{A}-\overrightarrow{B}$ is about 57 units.
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### 3. Find angle and the magnitude of a resultant vector.

Look at the image above. Find the magnitude of $\overrightarrow{R}$ and find $\alpha$.
Solution:
Take a deep look to the picture, we have to find $\alpha$ first. The method is below.
\begin{aligned} \frac{A}{\sin{\beta}}=&\frac{B}{\sin{\alpha}}\\ \frac{40}{\sin{66.53^{\circ}}}=&\frac{30}{\sin{\alpha}}\\ \sin{\alpha}=&0.687\\ \alpha=&\sin^{-1}{0.687}\\ \approx&\;43.47^{\circ} \end{aligned}
Once we know $\alpha$, we can calculate $\theta$.
\begin{aligned} \theta=&\alpha+\beta\\ =&43.47^{\circ}+66.53^{\circ}\\ =&110^{\circ} \end{aligned}
Since we already know $\theta$ we can finally calculate the magnitude of $\overrightarrow{R}$.
\begin{aligned} R=&\sqrt{A^2+B^2+2AB\cos{\theta}}\\ =&\sqrt{40^2+30^2+2(40)(30)\cos{110^{\circ}}}\\ \approx&40.98 \;\text{units} \end{aligned}
Based on the above calculation, we get $\alpha=43.47^{\circ}$ and $\left | R \right |= 40.98\;\text{units}$

### 4. Find the x-component and y-component of a vector and write it down using $\hat{i}$ and $\hat{j}$ notation

Look at the previous problem and find the x-component and y-component of $\overrightarrow{R}$. Then write $\overrightarrow{R}$ in $\hat{i}$ and $\hat{j}$ notation.
Solution:
We could find the y-component
\begin{aligned} R_y=&40.98 \times \cos(66.53)\\ =&16.32 \;\text{units} \end{aligned}
Then find the x-component
\begin{aligned} R_x=&40.98 \times \sin(66.53)\\ =&37.59 \;\text{units}\\ \end{aligned}
Note that $\overrightarrow{R}$ is pointing left and down which means the unit vectors will be $(-\hat{i})$ and $(-\hat{j})$. We can write the vector $\overrightarrow{R}$ as follows.
\begin{aligned} \overrightarrow{R}=&R_x(-\hat{i})+R_x(-\hat{j})\\ =&-37.59\;\hat{i}+-16.32\;\hat{j} \end{aligned}
So, we have $R_x=37.59$ units, $R_y=16.32$ units, and we write $\overrightarrow{R}=-37.59 \;\hat{i}+R_x-16.32\;\hat{j}$

### 5. Resultant of 3 vectors

• $\overrightarrow{A}=4\hat{i}+5\hat{j}$
• $\overrightarrow{B}=9\hat{i}-7\hat{j}$
• $\overrightarrow{C}=-3\hat{i}+2\hat{j}$
Find the resultant of $\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}$. Determine the resultant magnitude.
Solution:
First, we operate using $\hat{i}$ and $\hat{j}$ notation.
\begin{aligned} \overrightarrow{R}=&\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}\\ =&(4\;\hat{i}+5\;\hat{j})-(9\;\hat{i}-7\;\hat{j})+(-3\;\hat{i}+2\;\hat{j})\\ =&(4-9-3)\;\hat{i}+(5+7+2)\;\hat{j}\\ =&-8\;\hat{i}+14\;\hat{j} \end{aligned}
Then we can calculate the resultant magnitude as follows.
\begin{aligned} R=&\sqrt{-8^2+14^2}\\ =&16.12 \; \text{units} \end{aligned}
Based on our calculation, the resultant length of the $\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}$ is 16.12 units.
The image below shows the whole vector we operate.

### 6. Finding the angles of a vector

Find the angle of the resultant vector from the previous problem.
Solution:
First, we calculate the angle between the negative x-axis and $\overrightarrow{R}$
\begin{aligned} \theta=&\tan^{-1}\frac{14}{-8}\\ =&60.26^{\circ} \end{aligned}
Next, we calculate the angle between the positive x-axis and $\overrightarrow{R}$
\begin{aligned} \alpha=&180^{\circ}-\theta\\ =&180^{\circ}-60.26^{\circ}\\ =&119.74^{\circ} \end{aligned}
So, $\overrightarrow{R}$ is $119.74^{\circ}$ from the positive x-axis.

### 7. Finding angles between 2 vectors which have same magnitude

It is known that the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ have the same magnitude, and that is L. Let $R_1$ be the resultant of $\overrightarrow{A}+\overrightarrow{B}$ and $R_2$ is the resultant of $\overrightarrow{A}-\overrightarrow{B}$. If $\frac{R_1}{R_2}=\sqrt{3}$, determine the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$.
Solution:
We operate $\frac{R_1}{R_2}=\sqrt{3}$ further as below.
\begin{aligned} R_1=&R_2\sqrt{3}\\ \sqrt{A^2+B^2+2AB\cos{\theta}}=&\sqrt{A^2+B^2-2AB\cos{\theta}}\times \sqrt{3}\\ A^2+B^2+2AB\cos{\theta}=&(A^2+B^2-2AB\cos{\theta})\times 3\\ L^2+L^2+2L^2\cos{\theta}=&(L^2+L^2-2L^2\cos{\theta})\times 3\\ 2L^2+2L^2\cos{\theta}=&6L^2-6L^2\cos{\theta}\\ 8L^2\cos{\theta}=&4L^2\\ \cos{\theta}=&\frac{4}{8}\\ =&\frac{1}{2}\\ \end{aligned}
It turns out that we get the value $\cos{\theta}$. Then we can calculate $\theta$.
\begin{aligned} \theta=&\cos^{-1}\frac{1}{2}\\ =&60^{\circ} \end{aligned}
Based on the calculation above, we get the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $60^{\circ}$
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### 8. Resultant of 2 vectors in a square

Look at the picture above. Determine the resultant vector of $\overrightarrow{A}+\overrightarrow{B}$.
Solution:
Get another look to the picture again. The head of $\overrightarrow{A}$ is in the center of the right side of the square. This means that $\overrightarrow{A}$ has a y-component of length 6 units and an x-component of 12 units. We can write vector A using the notation $\hat{i}$ and $\hat{j}$.
\begin{aligned} \overrightarrow{A}=12\;\hat{i}-6\;\hat{j}\end{aligned}
Look at $\overrightarrow{B}$ and realize that we can write $\overrightarrow{B}$ as follows.
\begin{aligned} \overrightarrow{B}=6\;\hat{i}-12\;\hat{j}\end{aligned}
Next, we calculate the resultant vector.
\begin{aligned} \overrightarrow{R}=18\;\hat{i}-18\;\hat{j}\end{aligned}
\begin{aligned} R=&\sqrt{18^2+{(-18)}^2}\\ \approx&\;25.46 \; \text{units}\end{aligned}
So the resultant vector has a length of 25.46 units.

### 9. Calculate the resultant of 3 vectors which 2 of them are symmetric

Slide the image above
Look at the picture above. There are 3 vectors that have the same length, which is 30 units. Determine the resultant magnitude of the addition of these three vectors.
Solution:
Take a more deep look at $\overrightarrow{A}$ and $\overrightarrow{B}$. The y-components of the two vectors have the same magnitude but opposite directions, so they cancel each other out. So the resultant of $\overrightarrow{A}+\overrightarrow{B}$ is simply an addition of the x-component of the two vectors and points to the left.
$\left | A+B \right |$ is the resultant magnitude of the $\overrightarrow{A}+\overrightarrow{B}$ which we can calculate as follows.
\begin{aligned} \left | A+B \right |=& A_x+B_x\\ =&A \sin{20}+B \sin{20}\\ =&30 \sin{20}+30 \sin{20}\\ \approx&\; 20.521 \; \text{units} \end{aligned}
Next, we can calculate the magnitude of the resultant vector.
\begin{aligned} R=&\sqrt{{\left | A+B \right |}^2 + C^2}\\ =&\sqrt{20.521^2+30^2}\\ =&36,.47 \; \text{units} \end{aligned}
Based on the calculation we have done, the resultant magnitude of the addition of the three vectors is 36.35 units.

### 10. Resultant of Coulomb Forces

A point charge experiences 3 different forces, namely $\overrightarrow{F_1}$, $\overrightarrow{F_2}$, and $\overrightarrow{F_3}$ as shown in the figure above. Determine the resultant force experienced by the point charge.
Solution:
We describe all the vectors into x-component and y-component. You can see the calculation in the table below.

Force

x-component

Y-Component

$\overrightarrow{F_1}$

\begin{aligned}F_{1x}=&F_1 \cos{30}\\=&8 \times \frac{1}{2}\sqrt{3}\\=&4\sqrt{3}\;\mu\text{N to the right} \end{aligned}

\begin{aligned}F_{1x}=&F_1 \cos{30}\\=&8 \times \frac{1}{2}\sqrt{3}\\=&4\sqrt{3}\;\mu\text{N to top} \end{aligned}

$\overrightarrow{F_2}$

\begin{aligned}F_{2x}=&F_2 \sin{45}\\=&6 \times \frac{1}{2}\sqrt{2}\\=&3\sqrt{2} \;\mu \text{N to the left} \end{aligned}

\begin{aligned}F_{2y}=&F_2 \cos{45}\\=&6 \times \frac{1}{2}\sqrt{2}\\=&3\sqrt{2} \; \mu\text{N to top} \end{aligned}

$\overrightarrow{F_3}$

\begin{aligned}F_{3x}=&F_3 \sin{30}\\=&10 \times \frac{1}{2}\\=&5 \; \mu\text{N to the left} \end{aligned}

\begin{aligned}F_{3y}=&F_3 \cos{30}\\=&10 \times \frac{1}{2}\sqrt{3}\\=&5\sqrt{2} \;\mu \text{N to bottom} \end{aligned}

$\overrightarrow{R}$

\begin{aligned} R_x=&4\sqrt{3}-3\sqrt{2}-5\\=&2.314 \;\mu\text{N to the left}\end{aligned}

\begin{aligned}R_y=&4+3\sqrt{2}-5\sqrt{3}\\=&0.418\; \mu\text{N to bottom}\end{aligned}

After we describe $F_1$, $F_2$, and $F_3$, then we add up these components to get the x and y component of the resultant vector. The calculations are in the last row of the table above. When adding components, be careful with the (-)(+) sign. Remember negative means the force is pointing to the left/bottom and positive means the force is pointing to the right/top.

Based on the table above, we can write $\overrightarrow{R}$ in $\hat{i}$ and $\hat{j}$ notation.
$\overrightarrow{R}=-2.314\;\hat{i}-0.418\;\hat{j}$
And finally, we can calculate the resultant magnitude.
\begin{aligned} R=&\sqrt{(-2,314)^2+(-0,418)^2}\\ =&2,35\; \mu N \end{aligned}
So, the point charge experiences a coulomb force of 2.35 $\mu N$

### 11. Parabolic motion distance

Dilon(red shirt) and Coach Saeed are practicing table tennis. Dilon hits the ball at point A and the ball then moves to B, to C, and to D. It is known that the velocity of the ball at point B is 2.5 m/s and the elevation is $\alpha=40^{\circ}$. The ball then undergoes parabolic motion and reaches point C in 0.328 seconds. Find the distance(s) between points B and C.
Solution:
Parabolic motion is a combination of 2 motions at once, namely constant velocity motion and constant accelaration motion.
In this case, as the ball moves, the light at the top will cause the ball to cast a shadow on the table. Well, even though the ball is moving parabolic, its shadow is actually moving in a straight line from point B to point C with velocity $v_x=v \cos{40}$.
To find the distance(s) we do the calculation below.
\begin{aligned} S=&v_x \times t\\ =&v \cos{40} \times 0.328\\ =&2.5 \times \cos{40} \times 0.328\\ =&0.628 \; \text{meter}\\ \approx&63 \; \text{cm} \end{aligned}
We have done a calculation which states that the distance(d) between B and C is about 63 cm.

### 12. Finding the acceleration vector of an object

An object experiences 2 forces as shown above. Determine the direction($\theta$) and the magnitude of the object’s acceleration.
Solution:
We can calculate the resultant force first.
\begin{aligned} F_R=&\sqrt{{F_1}^2+{F_2}^2}\\ =&\sqrt{{20}^2+{5}^2}\\ =&\sqrt{425}\\ =&5\sqrt{17} \;\text{N} \end{aligned}
Then using Newton’s law formula $F_R=ma$ we can find the acceleration of the object.
\begin{aligned} a=&\frac{F}{m}\\ =&\frac{5\sqrt{17}}{5}\\ =&\sqrt{17} \; \text{m/s}^2 \end{aligned}
The direction of the object’s acceleration is the same as the direction of the resultant force and can be calculated as follows.
\begin{aligned} \theta=&\tan^{-1}\frac{\text{opposite}}{\text{adjacent}}\\ =&\tan^{-1}\frac{5}{20}\\ \approx&\;14^{\circ} \end{aligned}
So the object’s acceleration is $\sqrt{17}\;\text{m/s}^2$ with direction about $-14^{\circ}$ from the negative y-axis.

### 13. Analyzing Newton's law using vectors

Slide the image above
An object is under the influence of many forces as shown above. If the object remains at rest, determine the magnitude of $F_2$ and N(normal force).
Solution:
An object is said to be at rest if it satisfies two conditions.
The first condition is that the force to the right is equal to the force to the left. Look at the second slide of the image above. In this case, the force to the right is $F_1$ while the force to the left is $F_{2x}$. So to fulfill the first condition we must write as follows.
\begin{aligned} F_{2x}=F_1=20 \; \text{Newton}\\ \end{aligned}
Since we already know $F_{2x}$, we can find $F_2$. The method is below.
\begin{aligned} F_{2x}=&F_2 \cos{30^{\circ}}\\ F_2=&\frac{F_{2x}}{\cos{30^{\circ}}}\\ =&\frac{20}{\cos{30^{\circ}}}\\ =&\frac{40}{3}\sqrt{3}\;\text{Newton} \end{aligned}
The second condition is the upward force must be equal to the downward force. Slide the image above and see. In this case, there are 2 upward forces, namely $F_{2y}$ and N(normal force). While there is one downward force, namely w(weight). To fulfill the condition that the upward force equals the downward force, we must write the following mathematical equation.
$N \;+\;F_{2y}=\;W$
We can operate equation above further
\begin{aligned} N \;+\;F_{2y}=&\;W\\ N\;+\; F_2 \;\sin{30}=&\;50\\ N\;+\; \frac{40}{3}\sqrt{3}\times \frac{1}{2}=&\;50\\ N=&50 – \frac{40}{6}\sqrt{3}\\ =&38.45 \; \text{Newton} \end{aligned}
So $F_2=\frac{40}{3}\sqrt{3}$ Newtons and $N=38,45$ Newton.
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