Download PHYS 1P21/1P91 Test 3 Solutions 30 May 2013

PHYS 1P21/1P91
Test 3 Solutions
30 May 2013
1. State whether each statement is TRUE or FALSE, and then briefly explain (and if necessary, correct)
each statement. [8 points]
(a) For an object moving in a circle at a constant speed, no net force acts on the object.
(b) If car B travels around a curve at twice the speed as car A, then the force exerted by the road on
car B through the curve 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) Basil has a powerful magnet that he places in front of his car, hanging from a stick that he holds
from inside the car. He says the magnet will attract the car, pulling it forward.
Solution:
(a) FALSE. Even though the speed is constant, the velocity changes because the direction of the motion
changes. Thus, there must be acceleration. The acceleration points towards the centre of the circle, and
so the net force on the object also points towards the centre of the circle.
(b) FALSE. The centripetal acceleration is proportional to the square of the speed, and so the force
causing the acceleration is also proportional to the square of the speed. Thus, the force on car B is four
times as large as the force on car A.
(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 decrease in the
force by a factor of four.
(d) FALSE. By Newton’s third law, the force that the magnet exerts on the car is equal to and opposite
the force that the car exerts on the magnet. Because the car and magnet are connected, the net force on
the system is zero. (Another way to say this is that the forces acting between the car and the magnet are
internal to the car + magnet system.)
2. Determine the acceleration of the blocks if there is no friction between the upper block and the table
top. The pulley is massless, the string is massless and does not stretch, and there is no friction between
the string and the pulley. [6 points]
Solution: Draw free body diagrams for each block. Choose positive directions to be up and to the
right for the 2-kg block, and down for the 1-kg block; in this way, the accelerations of the blocks are the
same, and so the same symbol a can be used for each acceleration.
The equations that result from applying Newton’s second law to each free body diagram are:
T = 2a
n − 2g = 0
g − T = (1)a
The second equation is correct but not very relevant; substituting the expression for T from the first
equation into the third equation and solving for a, we obtain
g − (2a) = a
g = 3a
1
a= g
3
a = 3.3 m/s2
3. The moon has a mass of 7.36 × 1022 kg and a radius of 1.74 × 106 m.
(a) Determine the gravitational force that the moon exerts on an object of mass 5.3 kg at its surface.
[3 points]
(b) Determine the acceleration due to gravity at the moon’s surface. [3 points]
Solution:
(a)
Gm1 m2
r2
(6.67 × 10−11 ) (7.36 × 1022 ) (5.3)
F =
(1.74 × 106 )2
F = 8.6 N
F =
(b)
Gm1 m2
r2
(6.67 × 10−11 ) (7.36 × 1022 )
g=
(1.74 × 106 )2
g = 1.62 m/s2
m2 g =
4. Phobos, a moon of Mars, orbits at a distance of 9400 km from the centre of Mars, and has a period
of 7 hours and 40 minutes. Determine the mass of Mars. [5 points]
Solution: The gravitational force between Mars and Phobos is
F =
GM m
r2
where M is the mass of Mars and m is the mass of Phobos. Because the centripetal acceleration of Phobos
is
v2
r
we can use Newton’s second law to write
GM m
v2
=
m
·
r2
r
Solving for M , we obtain
v2r
G
The orbital speed of Phobos can be obtained from the data given, as
M=
v=
2πr
T
where T is the period of the orbit. Thus,
v2r
G
(2πr)2 r
M=
GT 2
4π 2 r3
M=
GT 2
M=
Using the given data, we obtain
4π 2 (9.4 × 106 )
G (460 × 60)2
M = 6.5 × 1023 kg
M=
3