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Kinematics
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Modern Physics
Light & Wave
100
In 11.2 s an object displaced 100m west. What is the average velocity of the object?
8.93 m/s west
100
A net force of 15.0 N north is used to pull an object. If the acceleration of this object is 8.0 m/s2, what is the mass of the object?
F=ma
=(16.0kg)(2.0m/s2)
=32N
100
A 15.0 kg object is lifted at constant velocity from the floor to height of 1.50 m. How much work is done on the object?
W= mgh = (15.0 kg)(9.80 m/s2)(1.50 m) = 211 J
100
Time Dilation: An astronaut is travelling at a constant speed of 2.95 × 108 m/s relative to earth through space. According to the timing devices on the space vehicle, her trip lasted 0.500 years. How long did the trip last if measured relative to the earth?
First establish which time is the proper time.
t = to/ {√[1-(v2/c2)]} = 0.500 years / {√[1- (2.95 × 108 m/s / 3.00 × 108 m/s)2]}
= 2.75 years
100
A wave has a frequency of 5.0 × 10-1 Hz and a speed of 3.3 × 10-1 m/s. What is the wavelength of this wave?
v = גf
ג = v/f
= (3.3 × 10-1 m/s) (5.0 × 10-1 Hz)
= 0.66 m
200
A man runs an average velocity of 1.30 m/s south for 98.0 s, and then walks at an average velocity of 0.45 m/s in the same direction for 90.0 s. what is the average velocity of the man during his total time over travel?
0.893 m/s south
200
A 925kg car accelerates uniformly from rest to a velocity of 25.0m/s south in 10.0s. What is the net force acting on the car during this acceleration?
a=(vf-vo) /t
=(25m/s - 0)/10s
=2.50m/s2
200
A 2.20 N object is held 2.20 m above the floor for 10.0 s. How much work is done on the onject?
W = Fd = (2.20 N)(0) = 0
200
Length contraction: A space vehicle that is 35.0 m long is traveling at a constant speed of 2.55 × 108 m/s past an observer standing on earth. How long does the space vehicle appear to the observer?
L = Lo × √[1-(v2 - c2)]
= (35.0 m) × √[1-(2.55 × 108 m/s)2 /(3.00 × 108 m/s)2] = 18.4 m
200
If a ray of light makes an angle of 58° with a mirror, what is the angle between the incident ray and the reflected ray?
∠i = 90° - 58° = 32°
∠r = ∠i
∴ angle = 32° + 32° = 64°
300
An object accelerates uniformly from rest to a velocity of 12.0 m/s west in 3.40 s. Calculate the acceleration of the object?
3.53m/s west
300
A net force of 6.6N east acts on a 9.0kg object. If this object accelerates uniformly from rest to a velocity of 3.0m/s east,
a) How far did the object travel while accelerating?
b) What is the time of acceleration?
(a) F=ma
a =F/m
=6.6N/9.0kg
=0.73m/s2 east
vf2=vo2+2ad
(3.0m/s)2=2(0.73m/s2)d
d=6.1m east

(b) a=vf-vo/t
0.73m/s2=3.0m/s-0/t
t=4.1s
300
A 2.00 × 102 kg object is pushed to the top of an incline. If the force applied along the incline is 6.00 × 102 N, what is the potential energy of the object when it is at the top of the incline with respect to the bottom of the incline?
Ep = mgh = (2.00 × 102 )(9.80 m/s2)(6.0 m) = 1.2 × 104 J
300
Mass: Calculate the rest mass of an object that has a relativistic mass of 5.50 × 10-10 kg at 0.800c.
m = m0 / √[1-(v2/c2)]
m0 = m × √[1-(v2/c2)]
= (5.50 × 10-10 kg) (√[1-0.8002])
= 3.30 × 10-10 kg
300
An object 3.0 cm tall is placed 6.0 cm in front of a mirror. If a virtual image is produced is 1.0 cm tall, what is the focal length of the mirror? What kind of mirror is used?
hi / ho = di / do

(1.0 cm / 3.0 cm) = - (di / 6.0)
di = - 2.0 cm

(1/f= 1/do + 1/di) = (1/6.0cm)+1/ (-2.0cm)
f= - 3.0 cm (convex mirror)
400
An object accelerates uniformly from rest at a rate of 1.9 m/s2 right for 5.0s, Find
a) the displacement
b) the final velocity
c) the distance travelled
d) the final speed
a) 24 m right
b) final velocity: 9.5 m/s right
c) magnitude of displacement: 24 m
d) magnitude of velocity: 9.5 m/s
400
If the gravitation force between two object of equal mass is 2.30×10-8N when the objects are 10.0m apart, what is the mass of each object?
Fg = Gm1m2/r2
m2= Fgr2/G
= 3.45×104 kg2
m = 1.86×102 kg
400
A 3.0 kg object is travelling at a constant speed of 7.5 m/s. What is the kinetic energy of this object?
Ek = ½mv2 = ½(3.0 kg)(7.5 m/s)2 = 84 J
400
Velocity: A space vehicle toward the earth at a velocity of 0.80c emits a light flash. What is the velocity of this light flash as seen by an observer on earth?
u = v + u’ / 1+ (vu’ / c2)
= 0.80c + c / 1+ [(0.80c)(c) / (c2)]
= c
400
Monochromatic light has a wavelength of 6.0 × 10-7 m in air and 5.0 × 10-7 m in a clear liquid. What is the index of refraction of the clear liquid?
גa/ג1= n1 / na

(6.0 × 10-7 m) / (5.0 × 10-7 m)= n1 / 1.00
n1 = 1.2
500
A runner hopes to complete the 10,000-m run in less than 30 min. After exactly 27.0 min, there are still 1100 m to go. The runner must then accelerate at 0.20m/s2 for how many seconds in order achieve the desired time?
3.1 s
500
A mass, M, is released from rest on an incline that makes a 42° angle with the horizontal. In 3 s, the mass is observed to have gone a distance of 3 m. What is the coefficient of kinetic friction between the mass and the surface of the incline?
We know that the mass accelerates from rest uniformly in 3 s and goes a distance of 3 m. Thus we can say that
d = ½ at2
Substituting the known number gives us an acceleration of 0.67 m/s2. On an incline, the normal force is given by mg cosθ, and so the friction can be expressed as f = μmg cosθ. Once again, this force opposes the downward force of gravity parallel to the incline, given by -mg sinθ. In the downward direction, these two forces are added together and set equal to -ma. Thus we write for this case:
μ(M)(9.8) cos 42 - (M)(9.8)sin 42 = -(M)(0.67)
The masses all cancel out, and solving for μ gives 0.8 as the coefficient of kinetic friction.
500
A 0.75-kg sphere is dropped through a tall column of liquid. When the sphere has fallen a distance of 2.0 m, is is observed to have a velocity of 2.5 m/s.
How much work was done by the frictional "viscosity" of the liquid?
The change in potential energy is a measure of the initial energy and equals (0.75)(9.8)(2) = 14.7 J. After the sphere has fallen 2 m, its velocity is 5 m/s, so the kinetic energy is given by
Ek = ½(0.75)(5)2 = 9.375 J.
The work done by friction is due to a nonconservative force that is equal to the difference between the final and initial energies:
Ef - Ei = 9.365 J - 14.7 J = -5.325 J
500
What is the speed of an electron whose mass is 10 000 times its rest mass?
1.5 m/s less than c.
500
Light is incident on a piece of flint glass (n = 1.66) from the air in such a way that the angle fo refraction is exactly half the angle of incidence. What are the values of the angles of incidence and the angle of refraction?
We want the angle of refraction to be equal to half the angle of incidence. This means that i = 2r. Now, since the light ray is initially in air (n = 1.00), Snell's law can be written:
sin i / sin r = n (glass)
Substituting our requirement that i = 2r gives
sin 2r / sin r = 1.66
Now recall the trigonometric identity: sin 2θ = 2 sin θ cos θ
Thus (2 sin r cos r)/(sin r) = 1.66
and 2 cos r = 1.66.
Therefore r = 34° and i = 68°.