Soal Vektor
1. Dua vektor di gambar di bawah ini saling tegak lurus satu sama lain dan besarnya sama. Jika C dan D adalah dua vektor, C adalah jumlahan A+B dan D adalah selisih A-B, berapakah sudut yang dibentuk antara vektor C dan D?
a) 0o
b) 30o
c) 45o
d) 60o
e) 90o
Penyelesaian: e)
Vectors A and B have equal magnitudes, so the angle between A and A+B is 45o. Similarly, the angle between A-B and A is also 45o. In conclusion, the angle between A-B and A+B is 90o and e) is the correct answer.
2. If the x-component of the vector A in the figure below is equal to 3cm and the y-component of A is equal to 4cm, what is the magnitude of A?
a) 3cm
b) 4cm
c) 5cm
d) 6cm
e) 7cm
Solution: c)
According to the Pythagorean theorem the magnitude of vector A is equal to 5cm, as the hypotenuse of a right triangle with the other two sides equal to 3cm and 4cm.
3. Which of the following is the angle between two vectors A and B, if the scalar product of A and B is equal to the magnitude of the vector product of A and B?
a) 0o
b) 30o
c) 45o
d) 60o
e) 90o
Solution: c)
The scalar product is ABcosα and the magnitude of the vector product is ABsinα. The only choice given that satisfies the equation cosα = sinα is α = 45o.
Soal Kinematika
1. A graduation hat is thrown vertically with a speed of 20m/s. How long does it take the hat to reach maximum height?a) t = .2s
b) t = .5s
c) t = .8s
d) t = 1s
e) t = 2s
Solution: e)
The initial speed of the hat is v1 = 20m/s and the final speed is 0m/s.
0 - v1 = -gt
t = v1/g = 2s.
2. A soccer ball is kicked with an horizontal speed v = 10m/s from the height h = 20m, as shown in the figure below. Calculate the time passed from the moment the ball has an horizontal speed until the moment it touches the ground.
a) t = .2s
b) t = .5s
c) t = .8s
d) t = 1s
e) t = 2s
Solution: e)
The y-axis component of the initial speed is 0m/s. When the ball reaches the ground, the y-axis component of the ball speed is vf.
vf2 - 0 = 2gh.
vf2 = 2gh
vf = 20m/s
vf - 0 = gt
t = vf/g = 2s.
3. Calculate distance d from the problem above.
a) d = 10m
b) d = 15m
c) d = 20m
d) d = 25m
e) d = 30m
Solution: c)
The x-axis component of the ball speed is constant, v = 10m/s. We also know from the previous problem that the time passed from the moment the ball has an horizontal speed until the moment it touches the ground is t = 2s.
Distance d = v·t = 20m.
4. The movement of a particle along the axis x is characterized by the graph below. What is the speed of the particle at t = 3s?
a) v = 0m/s
b) v = 1m/s
c) v = -1m/s
d) v = 2m/s
e) v = -2m/s
Solution: e)
The speed at t = 3s is:
v3 = Δx/Δt
v3 = (x3 - x2)/(t3 - t2)
v3 = (0 - 2)/(3 - 2)
v3 = -2m/s.
5. A car travels a distance d with an average speed v1. The velocity of the car during this trip is v2 and the maximum instantaneous speed of the trip is v3. Which of the following statements must be true?
a) v2 < v1 ≤ v3
b) v2 ≤ v1 ≤ v3
c) v1 ≤ v2 ≤ v3
d) v3 ≤ v1 ≤ v2
Solution: b)
Speed is the rate of change of distance with time and velocity is the rate of change of displacement with time. As the displacement can be only less or equal to the distance, the velocity v2 can be only less or equal to the speed v1: v2 ≤ v1
The maximum instantaneous speed v3 should be greater or equal to the speed v1: v3 ≥ v1.
In conclusion v2 ≤ v1 ≤ v3 and b) is the correct answer.
6. Calculate the displacement between t = 2s and t = 5s of an object that moves along an axis and has the speed characterized below.
a) d = .5m
b) d = 1m
c) d = 1.5m
d) d = 2m
e) d = 2.5m
Solution: c)
The displacement between t = 2s and t = 5s of the object is equal to the area between the graph and the x axis as seen below:
The area between t = 2s and t = 3s is equal to the area between t = 3s and t = 4s and of opposite sign, so at t = 4s the object is at the same position it was at t = 2s and the displacement between these two points is 0m.
The area between t = 4s and t = 5s is equal to 1(1 + 2)/2 = 1.5m.
In conclusion, c) is the correct answer.
Soal Hukum Newton
1. A particle is moving with a constant velocity, v. One or two forces are applied to the particle, as shown in the figures below. In which of the cases below, the forces applied do not change the speed of the particle?
a)
b)
c)
d)
Solution: d)
Answer d) is the only case when the two forces cancel each other so they will not change the speed of the particle. The forces in both a) and b) answers create forces and accelerations towards the top of the page and the force in answer c) deccelerates the particle.
2. An object of mass m = 20kg is pushed by four forces: East by a 10N force, West by a 30N force, South by a 20N force and North by a 20N force, What is the magnitude of the acceleration of the object?
a) .5m/s2
b) 1m/s2
c) 2m/s2
d) 10m/s2
e) 15m/s2
Solution: b)
The forces that push the object South and North cancel each other. The vectorial sum of the forces is a 20N force that pushes the object West.
20N = 20kg·a2
a = 1m/s2
3. Two forces F1 and F2 act on a block of mass m = 5kg, as shown in the figure below. What is the acceleration of the block if the friction between the surface and the block is negligible?
a) .5m/s2
b) 1m/s2
c) 2m/s2
d) 5m/s2
e) 10m/s2
Solution: d)
The horizontal component of force F1 is 20N·cos(60o) = 10N.
The horizontal component of force F2 is 30N·cos(60o) = 15N.
A horizontal force of 10N + 15N = 25N creates an acceleration of 25N/5kg = 5m/s2
4. Two forces F1 and F2 act on a particle as shown in the figure below. What is the orientation of the acceleration of the particle?
a) same orientation as F1
b) same orientation as F2
c) same orientation as the x axis
d) same orientation as the y axis
e) same orientation as the negative -x axis
Solution: d)
According to Newton's second law, the orientation of the acceleration is the same as the orientation of the vectorial sum of the forces that act on the object, so d) is the correct answer.
Soal Listrik Magnet
1. Sebuah partikel bermuatan positif q bergerak dengan kecepatan v dan melewati sebuah medan magnet B yang sejajar dengan arah kecepatan v. Berapakah besar gaya magnet yang dialami partikel?
a) F = qvB
b) F = -qvB
c) F = 0
d) F = qvB/2
e) F = -qvB/2
Jawab: c)
Gaya F, pada muatan q bergerak dengan kecepatan v dalam medan magnet B adalah F = q(v x B).
Besar gaya Fadalah F = q·v·B·sinθ di mana θ adalah sudut antara v dan B.
Pada soal v dan B vektor-vektor sejajar, sehingga sinθ = 0. Dapat disimpulkan F = 0.
2. In the figure below, a magnetic field of .01 T is applied locally to a wire carrying a current of intensity I = 10A. What is the magnitude of the magnetic force applied to the wire?
a) F = .3N
b) F = .4N
c) F = .5N
d) F = 1N
e) F = 3N
Solution: b)
The magnitude of a magnetic force applied to a current-carrying wire situated in a magnetic field is F = I·B·l·sinθ, where:
In out case sinθ = 4/l so l·sinθ = 4m.
F = 10A·.01T·4m = .4N
3. Three parallel long straight wires carry currents as shown in the picture below. If the currents in the wires #1 and #2 are I = 5A, and the magnetic force on wire #1 is equal to 0N, what is the current in wire #3?
a) I3 = 5A
b) I3 = 10A
c) I3 = 15A
d) I3 = 20A
e) I3 = 25A
Solution: d)
Wire #2 pulls wire #1 toward the bottom of the page while wire #3 pulls wire #1 toward the top of the page. The two forces on wire #1 must cancel.
I1·I2·μ·l/(2·π·d) = I1·I3·μ·l/(2·π·4d)
I2/d = I3/4d
I2 = I3/4
I3 = 4·5 = 20A.
4. Two metallic bars slide along metallic rails at speeds v1 = .1m/s and v2 = .2m/s as shown below. The bar and the rails are situated in a magnetic field of 4T. What is the induced voltage in the bar and rails?
a) 50mV
b) 60mV
c) 100mV
d) 120mV
e) 150mV
Solution: d)
e = -dΦ/dt = -B·dA/dt
dA/dt = l(dx/dt) = l(v1 + v2)
The induced voltage in the bar is B·l·(v1 + v2) = 120mV.
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