PROBLEM AND COMPLETION OF PHYSICAL VECTOR
Problem 1 The quantities below which are not included as vector quantities are....
a. energy
b. speed
c. force
d. momentum
e. acceleration
Physics Solving: A
Vector magnitude is a quantity that has a value and direction. Examples are displacement, speed, acceleration, momentum force, impulses, magnetic fields, and others
Problem 2 The quantities below which are not included in the scalar scale are ...
a. mass
b. time
c. temperature
d. speed
e. Density
Physics Solving: D
Scalar Magnitude is a quantity that has only value and has no direction. Examples are all principal quantities, energy, speed, effort, area, power, volume, and so on
Problem 3 Consider the following picture.
a. 4 N
b. 6 N
c. 8 N
d. 10 N
e. 12 N
Physics Solving: D
\[\small \begin{align*} R&=\sqrt{F_1^2+F_2^2}\\R&=\sqrt{8^2+6^2}\\R&=\sqrt{100}\\R&=10N \end{align*}\]
Problem 4 Consider the following picture.
Three styles of F1, F2, and F3 have direction and magnitude like in the following picture. The correct relationship for all three styles is ....a. F1 + F2 = F3
b. F2 + F3 = F1
c. F1 + F3 = F2
d. F1 + F2 = F3 = 0
e. F1 = F3 = F2
Physics Solving: C
F3 + F1 = F2 or F1 + F3 = F2
Problem 5 Two vectors of magnitude of 6 units and 8 units respectively. The magnitude of the resultant vector that is not possible is ....
a. 14 units
b. 2 units
c. 10 units
d. 9 units
e. 1 unit
Physics Solving: E
resultant of two vectors is between the difference in the number of both vectors.
(8-6)≤R≤(8+6)2≤R≤14(8-6)≤R≤(8+6)2≤R≤14
Problem 6 A force whose magnitude is F has the same vector component on the x-axis and the y-axis. The angle between the two vectors on the horizontal axis is...
a. 30 °
b. 45 °
c. 60 °
d. 90 °
e. 0 °
Physics Solving: B
\[\begin{align*} F_x&=F\cos \theta\\F_y&=F\sin \theta\\ F_x&=F_y\\ F\cos \theta&=F\sin \theta\\\frac{\sin \theta}{\cos \theta}&=1\\\tan \theta &=1\\\theta&=45^o \end{align*}\]
Problem 7 If the resultant magnitude and the difference between the two vectors are the same, then the angle between the two vectors is ...
a. 0 °
b. 60 °
c. 90 °
d. 120 °
e. 180 °
Physics Solving: C
\[\small \begin{align*} A+B&=A-B\\\sqrt{A^2+B^2+2AB\cos \theta}&=\sqrt{A^2+B^2-2AB\cos \theta}\\\cos \theta &=-\cos \theta \\\cos \theta&=0\\\theta&=90^o \end{align*}\]
Problem 8 Two vectors have the same magnitude, that is F. If the resultant magnitude of the two vectors is equal to F, the size of the angle of the wedge is ...
a. 30 °
b. 45 °
c. 60 °
d. 90 °
e. 120 °
Physics Solving: E
\[\begin{align*} R&=\sqrt{A^2+B^2+2AB\cos \theta}|_{A=B}\\R&=\sqrt{A^2+A^2-2AA\cos \theta}|_{R=A}\\A^2&=A^2+A^2-2AA\cos \theta\\1&=-2\cos \theta \\\cos \theta&=-\frac{1}{2}\\\theta&=120^o \end{align*}\]
Problem 9 Two vectors each have 5 units and 12 units and are opposite each other. The difference between the two vectors is ....
a. 7 units
b. 12 units
c. 17 units
d. 30 units
e. 60 units
Physics Solving: C
\[\small \begin{align*} A-B&=A+(-B)\\A-B&=5+(-(-12))\\A-B&=17\textrm{ unit} \end{align*}\]
Problem 10 From the collection of vectors, a collection of vectors that do not ever produce a resultant equal to zero is ....
a. 5 N, 5 N, 5 N, 5 N
b. 5 N, 10 N, 15 N, 20 N
c. 5N, 5N, 40N, 40 N
d. 10 N, 15 N, 20 N, 45 N
e. 5 N, 5 N, 10N, 25 N
Physics Solving: E
Look at the largest vector for each choice, then add the other vectors. If the largest vector is still greater than the number of other vectors, the resultant vector may not be zero.
See choice E, the largest vector 25N. The number of other vectors = 5N + 5N + 10N = 20N. Because 25N> 20 N, it is not possible to produce a zero resultant.
Problem 11 Consider the following picture. ABCD square with 10 cm long sides. Point E divides BC into 2 equal parts. The length of the resultant vector AC with AE is ....
a. 10 √2 cm
b. 20 cm
c. 25 cm
d. 25 √2 cm
e. 15 √2 cm
Physics Solving: D
\[\begin{align*} \Sigma F_x &=C_x+E_x\\\Sigma F_x &=10+10=20N\\\\\Sigma F_y &=C_y+E_y\\\Sigma F_y &=5+10=15N\\\\F=&\sqrt{\Sigma F^2_x+\Sigma F^2_y}\\F=&\sqrt{20^2+15^2}=25N \end{align*}\]
Problem 12 Two vectors, A and B, are each 20 N and 40 N. Determine the resultant of the two vectors if they form an angle of 60 °.
Physics Solving:
\[\begin{align*} R&=\sqrt{A^2+B^2+2AB\cos \theta}\\R&=\sqrt{20^2+40^2+2(20)(40)\cos 60^o}\\R&=\sqrt{400+1600+1600(0,5)}\\R&=\sqrt{2800}\approx 53N \end{align*}\]
