Download University of KwaZulu-Natal Pietermaritzburg

University of KwaZulu-Natal
Pietermaritzburg Campus
School of Physics
Physics 131: tutorial week 5 Additional mechanics
Gravitation and projectile motion
Surname....................................................
Initials..................
Prac day & group......................................
Unless otherwise stated in the question, take the acceleration due to gravity, g = 9, 8 ms−2
and the universal gravitational constant G = 6, 67 × 10−11 Nm2 kg−2 .
1. The earth’s mass is approximately 80 times as large as the moon’s mass and the
moon’s orbit radius is approximately 3, 8 × 105 km. Calculate the gravitational force
of attraction between the two. (ME = 6, 0 × 1024 kg)
(2, 1 × 1020 N)
Use the universal law of gravitation with m1 = ME , m2 = ME /80 and r = 3, 8 × 108 m. Then
F
=
6, 0 × 1024
2
× 6.67 × 10−11
2
80 × (3, 8 × 108 )
=
2, 1 × 1020 N.
2. Calculate the radius of the earth, given that its mass is 6, 0 × 1024 kg.
For an object at the earth’s surface,
GmME
Weight = mg =
.
r2
Hence
1
1
GME 2
6, 67 × 10−11 × 6, 0 × 1024 2
r =
=
g
9, 8
=
6, 4 × 106 m.
3. Calculate the acceleration due to gravity on the surface of the moon, given that
the mass and radius of the moon are approximately 7, 4 × 1022 kg and 1, 7 × 103 km
respectively.
(1,7 ms−2 )
For an object at the moon’s surface (mass of the moon = MM ),
GmMM
weight = mg =
.
r2
Hence
GMM
6, 67 × 10−11 × 7, 4 × 1022
= 1, 7 ms−2 .
g =
=
2
2
r
(1, 7 × 106 )
4. Determine the acceleration due to gravity for a satellite orbit at a height of 200 km
above the earth’s surface. Take the mass of the earth ME = 6, 0 × 1024 kg and the
radius of the earth as r = 6, 38 × 106 m.
(9,2 ms−2 )
For an object at a height h above the earth’s surface,
GmME
weight = mg =
,
r2
where r = RE + h. Hence
GME
6, 67 × 10−11 × 6, 0 × 1024
g =
=
=
2
(RE + h)
(6, 38 × 106 + 200 × 103 )2
9, 2 ms−2 .
5. An aircraft flying horizontally with a speed of 200 ms−1 drops an object from a height
of 1960 m above level ground. Calculate (a) the position of the object 3,0 seconds
after release, (b) the complete time of flight of the object and (c) the velocity of the
object (in magnitude and direction) when it hits the ground.
(a) 600 m ahead of and 44,1 m below point of release.
(b) 20 seconds.
(c) 280 ms−1 at 44◦ 25’ below the horizontal.
- 200 ms−1
................................................
.......
...................
...............
............
...........
.........
.........
......
......
......
......
......
.....
.....
.....
.....
....
.....
....
....
....
....
....
....
....
....
....
.
h = 1960 m
θ
R
?
w
(a) Take the direction downwards as negative, then after 3 seconds
sx = vt = 200 × 3 = 600 m.
sy = ut + 21 at2 = 0 + 21 (−9, 8) × 32 = −44, 1 m.
(b) Consider the vertical motion and use s = ut + 21 at2 with u = 0 ms−1 , then
s = 21 at2
1
21
2s
2 × (−1960) 2
and t =
=
= 20 s.
a
−9, 8
(c) The loss of potential energy equals the change in kinetic energy:
mgh = 21 m vf2 − vi2 ,
Hence
1
1
vf = 2gh + vi2 2 = 2 × 9, 8 × 1960 + 2002 2 = 280 ms−1 .
The angle with which the object hits the ground is found from
200
cos θ =
280
which gives
θ = 44, 4◦ = 44◦ 25′
6. A body is projected with a speed of 19,6 ms−1 at an angle of 30◦ above the horizontal
from the top of a tower 39,2 m high. Calculate (a) the maximum height the body
reaches above the ground, (b) the time taken for the body to reach the ground
and (iii) the angle subtended by the tower at the point of impact with the ground.
(assume the ground to be horizontal and neglect air resistance.)
(44,1 m; 4 s; 30◦ )
*
........................................................
..................
................................
..................
..............
...........
..............
...........
..........
.
.
.
.
.
.
.
.
.
.......
.....
.
◦
.
.
.......
.
.
.
.
30
......
.......
......
......
.....
.....
.....
.....
.....
....
....
....
....
....
....
....
....
....
....
...
...
...
...
...
...
...
...
...
..
h
39,2 m
θ
R
-
(a) The height h indicated in the diagram can be determined from v 2 = u2 + 2as with uy =
u sin 30◦ and vy = 0:
−u2y
−9, 82
=
= 4, 9 m.
h =
2a
2 × (−9, 8)
The maximum height above the ground is therefore 4, 9 + 39, 2 = 44, 1 m.
(b) We need to determine the time taken from the top of the tower to the maximum height
and the time from the maximum height to the ground. From the top of the tower we use
v = u + at with v = 0 ms−1 and uy as above.
v−u
−19, 6 × sin 30◦
t =
=
= 1 s.
a
−9, 8
For the downward motion of the body, we use s = ut + 21 at2 with u = 0 ms−1 .
12
1
2 × 44, 1 2
2s
=
= 3 s.
t =
a
9, 8
Therefore the total time is 4 s.
(c) The angle required is θ in the diagram. The displacement along the horizontal is
s = vt = 19, 6 × cos 30◦ × 4 = 67, 9 m.
Hence
39, 2
,
tan θ =
67, 9
which gives θ = 30◦ .
7. (a) Liquid squirted from a pipe can travel a maximum distance of 500 m over level
ground. Ignoring any air drag, calculate the velocity u with which the liquid
stream is projected.
(70 ms−1 )
(b) If the liquid stream is projected horizontally with the same speed from a point
490 m above the ground, calculate the new range (as measured from a point on
the ground directly below the point of projection).
(700 m)
(a) A maximum distance implies the liquid is squirted at 45◦ to the horizontal. Hence
ux = uy = v cos 45◦ .
For motion in the horizontal direction, using s = vt we find,
500
.
t =
u cos 45◦
For the motion in the vertical direction we use s = ut + 21 at2 with s = 0 (since the liquid
is projected over level ground). Hence
0 = u cos 45◦ + 21 (−9, 8) t.
Solving these equations, we obtain u = 70 ms−1 .
(b) Considering the motion in a vertical direction, we can calculate the time from s = ut +
1 2
−1
.
2 at with uy = 0 ms
1
12
2 × (−490) 2
2s
=
= 10 s.
t =
g
−9, 8
The range may now be determined from s = vt using v = 70 ms−1 (calculated above) and
t = 10 s. Hence s = 700 m.
8. A body is projected horizontally from a height of 2,5 m. After covering a horizontal
distance of 12 m, it is at a height of 1,0 m. (a) What was its initial velocity? Calculate
also (b) the magnitude and direction of its velocity just before it hits the ground.
(21,8 ms−1 ; 22,9 ms−1 18◦ below the horizontal)
-
1,5 m
................................................
.....................
................
.............
............
.........
.........
.......
......
.......
......
......
......
......
.....
.....
.....
....
....
....
....
....
....
....
....
....
...
12 m
1,0 m
θ
w
For this problem it is convenient to treat motion downwards as positive.
(a) We first determine the time it takes for the body to fall a distance of 1,5 m. For the vertical
component of the motion, uy = 0 ms−1 , hence using s = ut + 12 at2 ,
1
12
2 × 1, 5 2
2s
=
= 0, 553 s
t =
g
9, 8
The time t above is the time taken for the body to travel 12 m in the horizontal direction.
Using s = ux t, we find
12
s
=
= 21, 7 ms−1 .
ux =
t
0, 553
(b) The magnitude of the velocity with which the body hits the ground may be calculated by
first finding the total time, the vertical component of the velocity and hence the resultant of
the vertical and horizontal components. Alternatively, we can use the work–energy theorem,
with the work done by the gravitational force equal to the change in kinetic energy. i.e.
mgh = 12 m vf2 − vi2 .
This gives
1
1
vf = 2gh + vi2 2 = 2 × 9, 8 × 2, 5 + 21, 72 2 = 22, 8 ms−1 .
The angle θ with which the body hits the ground can be obtained from
21, 7
,
cos θ =
22, 8
which gives θ = 18◦ below the horizontal.
9. A body is projected downward at an angle of 30◦ to the horizontal, with an initial
speed of 40 ms−1 , from the top of a tower 150 m high. What will be the vertical
component of its velocity when it strikes the ground? In what time will it strike the
ground? How far from the tower will it strike the ground? At what angle with the
horizontal will it strike?
(58 ms−1 ; 3,9 s; 135 m; 59◦ )
Consider motion downwards as positive. Then uy = 40 sin 30◦ ms−1 , ay = g = 9, 8 ms−2 and
sy = 150 m. Using v 2 = u2 + 2as for the vertical motion,
1
vy = 202 + 2 × 9, 8 × 150 2 = 58 ms−1 .
Use v = u + at to find the time:
59 − 20
v−u
=
= 3, 9 s.
t =
a
9, 8
√
−1
◦
The horizontal
√ component of the velocity ux = 40 cos 30 = 20 3 ms . Therefore
s = vt = 20 3 × 3, 9 = 135 m.
The angle below the horizontal is obtained from
59
√ ,
tan θ =
20 3
which yields θ = 59◦ .
True or false questions
TRUE
1. The earth pulls on the moon with a bigger force than the moon
pulls the earth.
2. At a height above the earth’s surface equal to the earth’s radius,
the acceleration due to gravity is only one–quarter of its value
on the earth’s surface.
3. A ball rolled rapidly over the edge of a table will take longer to
hit the floor than one simply dropped from the same height.
4. In the ‘monkey and the bushman’ demonstration, the speed
of the dart is immaterial, within broad limits. (Consider the
simple case where the blowpipe, when pointing at the monkey,
is horizontal.)
FALSE