Download PHYS 1P21/1P91 Test 3 Solutions 5 June 2014

PHYS 1P21/1P91
Test 3 Solutions
5 June 2014
1. State whether each statement is TRUE or FALSE, and then briefly explain (and if necessary, correct)
each statement. [8 points]
(a) Two objects move vertically upwards at constant speeds, but the speed of Object A is twice as
great as the speed of Object B. This means that the net force on Object A is twice as great as the net
force on Object B.
(b) Car A moves around a circular curve at a constant speed. Car B has the same mass as Car A and
moves at the same constant speed around a circular curve the radius of which is twice as large as for Car
A. This means that the force exerted by the road on Car B is twice as large as the force exerted by the
road on Car A.
(c) If the Moon were moved twice as far from the Earth, then the gravitational force that the Earth
exerts on the Moon would be half as large.
(d) Block A is touching Block B, and both rest on a flat, frictionless surface. The mass of Block
A is greater than the mass of Block B. When Block A is pushed, both blocks accelerate with the same
acceleration. While they accelerate, the force that Block A exerts on Block B is greater than the force
that Block B exerts on Block A.
Solution:
(a) FALSE. Each object moves at a constant speed, and so by Newton’s first law of motion, the net
force on each object is zero.
(b) FALSE. The force exerted by each road on each car is
FA =
mv 2
rA
and
FB =
mv 2
rB
The ratio of the two forces is
FA
=
FB
mv 2
rA
mv 2
rB
2
FA
mv
rB
=
×
FB
rA
mv 2
FA
rB
=
FB
rA
FA
=2
FB
Thus, the force exerted by the road on Car A is twice as large as the force exerted by the road on Car B.
(c) FALSE. The gravitational force between two objects is inversely proportional to the square of
the distance between the objects, so doubling the distance between the objects results in a force that is
one-fourth as large. (You can verify this using a calculation in the same spirit as the one in Part (b).)
(d) FALSE. By Newton’s third law of motion, the force that Block A exerts on Block B has the same
magnitude as the force that Block B exerts on Block A.
2. The two blocks in the figure are at rest on the frictionless surface. Determine the mass of the second
block so that the blocks remain at rest. [4 points]
Solution: Draw a free-body diagram for each block. Then write Newton’s second law of motion for
each block. Because the blocks are in static equilibrium, the net force on each block is zero. This means
that two of the equations that result from your application of Newton’s second law to the blocks are
T − 10g sin 23◦ = 0
and
mg sin 40◦ − T = 0
where T is the tension in the string. Eliminating T from the two previous equations, we obtain
mg sin 40◦ = 10g sin 23◦
m sin 40◦ = 10 sin 23◦
10 sin 23◦
m=
sin 40◦
m = 6.1
Thus, the mass of the block is 6.1 kg.
3. The mass of the Moon is 7.36 × 1022 kg, the mass of the Earth is 5.98 × 1024 kg, and the radius of
the Earth is 6.37 × 106 m. The distance between the centre of the Moon and the centre of the Earth is
3.84 × 108 m.
(a) Determine the gravitational force that the Moon exerts on an object of mass 1.0 kg at the centre
of the Earth. [2 points]
(b) Determine the gravitational force that the Moon exerts on an object of mass 1.0 kg at the point on
the surface of the Earth closest to the Moon. [2 points]
(c) Determine the gravitational force that the Moon exerts on an object of mass 1.0 kg at the point on
the surface of the Earth farthest from the Moon. [2 points]
(d) BONUS (extra credit): Briefly explain the observable consequences of your results in Parts (a), (b),
and (c). [2 bonus points]
Solution: Draw a diagram to help you understand which distances to use in each calculation. Using
Newton’s law of gravity, we obtain
(a)
(6.67 × 10−11 ) (7.36 × 1022 ) (1)
GM m
=
= 3.33 × 10−5 N
Fc =
2
2
8
rc
(3.84 × 10 )
(b)
F1 =
GM m
(6.67 × 10−11 ) (7.36 × 1022 ) (1)
= 3.44 × 10−5 N
=
2
8
6
r12
(3.84 × 10 − 6.37 × 10 )
(c)
F2 =
(6.67 × 10−11 ) (7.36 × 1022 ) (1)
GM m
=
= 3.22 × 10−5 N
2
8
6
r22
(3.84 × 10 + 6.37 × 10 )
(d) The Moon exerts a stronger force on the “near” surface of the Earth than on the “far” surface of
the Earth. One way to interpret this (draw diagrams!) is that the force that the Moon exerts on a mass
at the centre of the Earth can be considered to be the “average” force exerted on masses at all points of
the Earth. Superposed on this average force are additional forces at each point of the Earth; masses at
the “near” surface of the Earth experience an additional force towards the Moon, and masses at the “far”
surface of the Earth experience an additional force away from the Moon. This “stretching force” is the
cause of tides on the Earth.
The Sun also exerts tidal forces on the Earth, so tidal forces on the Earth are superpositions of influences
from the Sun and Moon. The actual height of the tides also depends on local geometry of the coastlines
and bottoms of bodies of water, so the situation is really complicated.
Historically, Newton’s explanation of the tides, which accounted qualitatively for their main features,
was quite a triumph for his theory of gravity, and was influential in its general acceptance.
4. A piece of space junk with a mass of 100 kg moves in a circular orbit at an altitude of 100 km above
the Earth’s surface. (The mass of the Earth is 5.98 × 1024 kg, and the radius of the Earth is 6.37 × 106 m.)
(a) Determine the period of the space junk’s orbit. [1 points]
(b) Determine the speed of the space junk in its orbit. [1 points]
Solution: (a) Using Newton’s second law of motion, we can write
F =
mv 2
r
where m is the mass of the space junk, v is its speed, and r is the radius of its orbit. The force on the
space junk is obtained using Newton’s law of gravity:
F =
GM m
r2
where M is the mass of the Earth. Combining the two previous equations, we obtain
GM m
mv 2
=
r
r2
which simplifies to
v2
GM
= 2
(∗)
r
r
Because the space junk moves in a circular orbit at a constant speed, its speed is related to its period T
by
2πr
v=
T
Inserting this relation into equation (*), we obtain
2
1 2πr
GM
= 2
r
T
r
which simplifies to
2
T =
4π 2
GM
r3
Thus, the period of the space junk’s orbit can be calculated as follows:
3
4π 2
2
T =
6.37 × 106 + 100 × 103
−11
24
(6.67 × 10 ) (5.98 × 10 )
2
T = 2.68 × 107 s2
T = 5178 s
T = 86 min
Thus, the period of the space junk is nearly 1.5 hours.
(b) Using the relation
2πr
v=
T
and the result of Part (a), we can determine the speed of the space junk as follows:
2π × (6.37 × 106 + 100 × 103 )
5178
v = 7852 m/s
v = 7.9 km/s
v = 28, 300 km/h
v=
The space junk is zipping along at quite a speed. The International Space Station orbits at about the same
altitude,1 and therefore at about the same speed, and if there were a head-on collision the results would
be catastrophic. The amount of space junk in orbit around the Earth is large and growing (we humans
are sloppy, aren’t we?), and there have been calls to clean up the mess we created, for good reasons.
It’s also worth noting the high speed of the ISS next time you see a video of an astronaut floating
around in the ISS. It’s striking isn’t it, although we have all experienced similar situations when flying
about in aircraft, driving in cars, or sitting at a desk as the Earth carries us around the galaxy at extremely
high speeds.
1
The ISS altitude is closer to 400 km, so its speed would be slightly less than calculated here; how much less?