Download File - Meissnerscience.com

SPH 4U Exam Review Questions
1. An adult exerts a force of 26 N to pull a child across the snow on a sled. The pull-rope attached to the sled is at an angle of 21
above the horizontal. Ignoring friction, how much work is done in pulling the sled 250 m?
2. Which of the pairs of masses below will have the greatest force of attraction between them?
Mass 1
Mass 2
Distance between the masses
A
32 kg
25 kg
20 cm
B
80 kg
20 kg
40 cm
C
90 kg
10 kg
15 cm
3. What is the spring constant of an archery bow that is pulled back 0.710 m with a force of 133 N?
4. A physics student, with a mass of 75 kg, is standing on a scale inside an elevator. The scale is calibrated in newtons. What will
the scale read if the elevator accelerates upward at 1.7 m/s2?
5. A batter hits the baseball so that the velocity of the ball is 48.1 m/s at an angle of 17° to the horizontal.
a) A player in the field is standing 76.0 away from the batter. Where will the ball land relative to the fielder?
b) The fielder is standing still and then starts to run as soon as the ball is hit. What acceleration will the fielder need to make
the catch?
6. A potato was launched from a potato cannon with a velocity of 34.4 m/s at an angle of 41.0° above the horizontal. The potato
landed on the roof of a house that was 112 m away. Determine the height of the house.
7. The Moon has a mass of 7.34 x 1022 kg and a radius of 1.74 x 106 m. Calculate the acceleration due to gravity on the surface of
the Moon.
8. The force of gravitational attraction between two physics textbooks is 1.5 x 10–8 N. If each of the textbooks has a mass of 1.50
kg, determine how far apart the textbooks would be.
9. A 60 g tennis ball moving at 25 m/s[E] is hit by a racquet and returned at 35 m/s[W]. If the interaction force between the ball
and the racquet is 400 N, find
a) the change of momentum of the ball
b) the time of contact between the ball and the racket
10. How much force is needed to hold all the electrons in 1 g of hydrogen at a distance of 1.00 m from its remaining protons? (1 g
of hydrogen contains 6.02 x 1023 atoms.)
11. Each side of a vertical right-angled isosceles triangle measures 1.87 cm. Three charged objects are placed at the vertices
(corners) of the triangle. Each of the two objects at the vertices of the base has a charge of +5.00 mC. The object at the top
vertex (right angle) of the triangle has a charge of +4.74 mC. What is the net force on the object at the top vertex?
12. A helium atom without its electrons is called an alpha particle. What angle does an alpha particle make with a 0.030 T magnetic
field, if the alpha particle has a velocity of 3.00 x 10 4 m/s and a force of 1.12 x 10-16 N on it?
13. The separation between two slits is 1.75 x 10-5 and a screen is 0.56 m from the slits. Monochromatic violet light with a
wavelength of 4.30 x 10-7 passes through the slits.
a) How far from the central band will the first bright band appear on the screen?
b) What is the separation between successive bright bands?
c) What is the separation between successive dark bands?
d) How far from the central band will the tenth bright band appear on the screen?
e) What is the separation between the first and tenth bright bands?
14. Before the introduction of cable and satellite TV broadcasting, a roof-mounted TV antenna consisted of many horizontal metal
bars with lengths equal to one quarter of a wavelength of a TV signal. If the frequency of a Canadian TV station is 200 MHz,
find the appropriate length of the metal bars on the receiving antenna.
15. Imagine that Earth has been squeezed to the size of a marble with a radius of 0.89 cm.
a) What is the escape speed for an object on the surface of the marble-sized Earth?
b) Can light escape from the surface of the marble-sized Earth?
c) Give a common name for the marble-sized Earth.
G = 6.673 x 10–11 N.m2/kg2, MEarth = 5.978 x 1024 kg, rEarth = 6.378 x 106 m
16. You speak on the telephone from Toronto to Vancouver, a distance of 4000 km.
a) What kind of electromagnetic waves are used to transmit the communication?
b) What is the speed of these waves?
c) How long does it take for the waves to travel from Toronto to Vancouver?
17. Each side of an equilateral triangle measures 3.61 cm. A charge of -0.34 mC is placed at each vertex. What is the potential
difference at the midpoint of the base of the triangle?
18. The lifetime of a kaon particle, measured at rest in a laboratory, is 1.2 x 10 -8 s. At what speed must the kaon particle travel to
have its lifetime measured as 2.0 x 10-8 s?
19. An electron is moving at 0.995c, parallel to a metre stick. How long is the metre stick in the electron’s frame of reference?
20. The density of a body, as measured by an observer relative to which the body moves, increases with its speed. Give two reasons
why.
21. Suppose that the Michelson-Morley experiment could be adapted to test the relative speed of source and medium, using sound
or water instead of light as the wave energy. Would there be the same null result? Explain.
22. An observer on Earth measures the elapsed time between two events on board a spaceship, moving at a relativistic speed, to be
1.5 s. A crew member on board measures the elapsed time to be 1.0 s. The length of the spaceship is measured to be 150 m by
the crew member. What is the length of the spaceship, as measured by the Earth observer?
1] 6.1
x 103 J 2] FA = 1.3 x 10–6 N 3] k = 187 N/m 4] 8.6 x 102 N 5] 56 m
6] 6.29 m 7] 1.62 N/kg
8] 0.1 m 9b] 0.009 s 10] 8.35 x 1019 11] 8.50 x 102 N[Up] 12] 22.9 13a] 1.4 cm 13b] 13 cm 14] 0.4 m
15a] “c”
16] 0.013 ms 17] -7.8 x 108 V 18] 0.8c 19] 10 cm 22] 100 m
Exam Review Answers
1. ANS:
The only force that is acting on a satellite is the force of gravity attracting the satellite to Earth. This force
of attraction, along with the perpendicular velocity of the satellite, maintains the satellite’s position above
Earth’s surface. The satellite is falling toward Earth, but the speed of the satellite causes it to be in
constant free fall around Earth’s surface.
2. ANS:
Work = (force)(cosine of the angle)(distance)
= (26 N)(cos 21°)(250 m)
= 6068
= 6.1 x 103 J
3. ANS:
Cosmic rays do not fit into the electromagnetic spectrum, because they are not electromagnetic waves.
Cosmic rays are actually high-energy particles flying through outer space.
4. ANS:
No. The work function of a metal applies only to surface electrons. Electrons at deeper levels require
greater amounts of energy to be liberated from the metal.
5. ANS:
FA = 1.3 x 10–6 N
6. ANS:
F = kx
k = 187 N/m
7. ANS:
a) Kinetic energy =
mv2
=
(m)(at)2
=
(5)(2t)2
= 10t2
b)
c) Work done = 90 J
8. ANS:
a) Answers might vary, since the orientation of the wire and the direction of the current were not
specified. One possibility is shown in the diagram.
b) The pattern of magnetic field lines remains the same, so the strength of the magnetic field does not
change, but the direction of the magnetic field at each point is reversed.
9. ANS:
Determine the net force acting on the student.
Fnet = ma
= (75 kg)(1.7 m/s2)
= 128 N
To determine the reading on the scale, calculate the normal force.
Fnet = FN - Fg
FN = Fnet + Fg
= 128 N + (75 kg)(9.81 N/kg)
= 863 N
= 8.6 x 102 N
10. ANS:
a) Determine the range of the ball.
x = 82.7 m – 76.0 m = 6.7 m
b) Find the time during which the baseball is in the air.
v1 = 0, d = 6.7 m, a = ?
d = v(t) +
a(t2)
11. ANS:
To determine the height (dv) of the house, first consider the horizontal components of the potato (use up
as positive).
vh = + (34.4 m/s)(cos 41°)
vh = + 26.0 m/s, dh = 112 m, t = ?
dh = vh(t)
Consider the vertical components of the potato to determine the height.
vv = + (34.4 m/s)(sin 41°)
vv = + 22.6 m/s, t = 4.31 s, ag = –9.81 m/s2, dv = ?
dv = vvi(t) +
a(t2)
= (22.6 m/s)(4.31 s) +1/2(–9.81 m/s2)(4.31 s)2
= 6.29 m
12. ANS:
To determine the acceleration due to gravity on the surface of the Moon, use the force of gravity formula
(Fg = mg), and the force of universal gravitational attraction formula (Fg = Fg)
13. ANS:
Use Kepler’s constant for the Sun (k = 3.31 x 1018 m3/s2), or if this is not known, the asteroid’s orbital
period can be calculated by using Earth’s data.
14. ANS:
15. ANS:
a) The change of momentum = (0.060)(–35 – 25) = 3.6 kg.m/s[W].
b) The time of contact =
16. ANS:
17. ANS:
= 0.009 s.
Vector Sum =
= 8.50 x 102 N[Up]
18. ANS:
19. ANS:
a)
b) From part (a), bright bands are separated by 1.4 cm.
c) Dark bands are separated by the same distance as bright bands, which is 1.4 cm.
d) The tenth bright band is a distance of 10 x 1.4 cm = 14 cm from the central band.
e) The distance between the first and tenth bright bands is 9 x 1.4 cm = 13 cm.
20. ANS:
The length of the metal bars is:
21. ANS:
a)
b) No.
c) Black hole
22. ANS:
a) microwaves
b) 3 x 108 m/s
c) Time required
23. ANS:
V = VAV + VBV + VCV
= -3.39 x 108 V + (-3.39 x 108 V) + (-9.79 x 107 V)
= -7.8 x 108 V
24. ANS:
There is no effect. Since the frequency of the incident light is less than the cutoff frequency, no electrons
are ejected, regardless of the intensity of the light. Thus, the current is zero. Because there are no ejected
electrons, their kinetic energy does not exist either.
25. ANS:
26. ANS:
27. ANS:
28. ANS:
1) Mass increases with speed.
2) Length decreases with speed. Since the “sideways” dimensions do not change, volume must decrease
also.
29. ANS:
No, there would not be the same null result. For sound or water waves, the vector sum of the speeds of the
wave source and the medium would apply. Therefore, the relative speed of wave source and medium
would be found.
30. ANS:
The length dimension is shortened for the Earth observer by the same factor that the time dimension is
lengthened. Since this factor is 1.5, the Earth observer measures the length of the spaceship to be