Download 3 - Homework Ans

AP Physics – Gravity – 3 ans
1. A girl on a skateboard throws a bag of sand away from herself, giving it a speed of 12.0 m/s. The girl/skateboard's
mass is 37.2 kg and the mass of the sand is 22.9 kg. What is the final velocity gained by the lass?
0  m1v1 f  m2 v2 f
m1v1 f  m2 v2 f
v1 f 
m2 v2 f
m1
m

  22.9 kg  12.0 
s


37.2 kg

 7.39
m
s
2. A 5.25 kg ball travelling at 4.25m/s collides head on with a 2.00 kg ball traveling in the opposite direction at 3.55
m/s. If the final velocity of the second ball is 5.15 m/s in the opposite direction, (a) what is the final velocity of
the first ball? (B) What is the change in kinetic energy?
(a)
v '1 
m1v1  m2v2  m1v '1  m2 v '2
v '1 
m1v1  m2 v2  m2 v '2
m1
 5.25 kg   4.25 ms    2.0 kg   3.55 ms    2.0 kg   5.15 ms 


(b) K  K o  K

5.25 kg

1
1
K o  m1v12  m2v2 2
2
2



0.936
2
1
m 1
m


  5.25 kg   4.25    2 kg   3.55 
2
s 2
s


2
K o  60.02 J
1
m 1
m


K o   5.25 kg   0.936    2 kg   5.15 
s 2
s
2


K  60.02 J  28.82 J

m
s
2
2
 28.82 J
31.2 J
3. A 65.0 kg astronaut at rest in space throws a 23.0 kg oxygen tank which travels at a speed of 6.25 m/s. With what
velocity does the astronaut start to move through space?
0  m1v1 f  m2 v2 f
m1v1 f  m2 v2 f
v1 f 
m2 v2 f
m1
m

  23.0 kg   6.25 
s


65.0 kg

 2.21
m
s
4. You playfully toss a 1725 g tomato at your sibling. The tomato, traveling at 25.0 m/s, smashes into the young
juvenile. (a) If the time of the interaction is 0.350 sec, what is the average force on the child? (b) What is the
impulse? (c) If the tomato bounced off with no loss in kinetic energy (and the sibling remained motionless) what
would be the force on the child?
m

1.725 kg  25.0 
mv
s

(a) F t  mv F 


123 N
t
 0.350 s 
m
kg  m

 1.725 kg  25.0  
43.1
s
s

m
m

1.725 kg  25.0    25.0 
mv
s
s

(c) F 


246 N
t
 0.350 s 
5. Jupiter has a mass about 300 times bigger than the earth’s mass. So you should weigh 300 times more on Jupiter,
right? Wrong. You would actually only weigh about three times as much on Jupiter as you do on earth. How
come?
6. The following are true/false questions.
a. There is no gravity in outer space - like on board the space shuttle. (F)
b. Newton discovered gravity. (F) (he defined it with an equation)
c. The earth’s gravity does not extend out to the outer planets, and certainly not to next nearest star. (F)
d. The force of the earth/s gravity on objects in orbit is near zero. (F)
e. The crescent shape of the moon is caused by the earth’s shadow. (F)
f. The pull of the moon’s gravity causes the tides. The pull of the sun does not cause any significant tides.
Therefore the moon exerts a larger force on the earth than does the sun. (calculate the difference and find out)
(b) F t  mv
7. What is the force pulling you towards the center of the earth if you are on its surface and weigh 500 N?
500N
8. What happens to the force of gravity between two masses if you halve the distance between them?
It quadruples (x4)
9. An asteroid revolves around the sun at a distance of 5.35 x 1011 m. The Sun's mass is 1.99 x 1030 kg (a) What is its
orbital velocity? (b) What is the period of its orbit (in years)?
m1v 2 Gm1m

r
r2
(a) FC  FG
v
Gm
r
v
(b)

6.67 x 1011
2.481 x 108
x
v
t
x
t
v
m2
s2
Gm
r
v
kg  m  m 2
1.99 x 1030 kg
s 2  kg 2


2 r

Gm
r
t
2
2  r 2


t  2
Gm
r
 5.35 x 1011 m 
1
 5.35 x 1011 m 
m
s
1.58 x 104
2

r3
Gm
3
T  2
2

11 kg  m m 
1.99 x 1030 kg
 6.67 x 10
2
2

s
kg


 2
11.54 x 1014 s 2

 2 3.40 x 107 s


 1 day 
1h
 2.47 x 103 day
T  2 3.40 x 107 s 
3 
1 
 3.6 x 10 s  2.4 x 10 h 


1 yr
1
T  2.47 x 103 day 
6.77 yr
  0.677 x 10 yr 
2
x
day
3.65
10


10. The mass of Pluto was totally unknown until the fairly recent discovery of a moon orbiting the planet. How come
it took so long to figure out Pluto’s mass? Because Pluto is a satellite in orbit around the sun. Knowing the orbital
radius of Pluto or its period does not let you know its mass because an object at that radius and that period would
follow a certain path no matter its mass (that’s why you don’t need the mass of the satellite to know its orbital
information, only the mass of the thing it is circling). So once you find something orbiting Pluto, you take that data
and find out the object in the middle (PLUTO)
11. A satellite is in orbit - like 595 km above the earth's surface. (a) What is its orbital velocity? Earth's radius is 6.37
x 106 m, earth's mass is 5.98 x 1024 kg. (b) What is the centripetal acceleration acting on the satellite?

(a)

r  6.37 x 106 m  0.595 x 106 m  6.965 x 106 m
2

11 kg  m m 
5.98 x 1024 kg
 6.67 x 10
2
2

s
kg

v 
6
6.965 x 10 m
v
Gm
r
v
m2
5.727 x 107 2
s

m2
 57.27 x 106 2
s


7.57 x 103
m
s
3
v 2  7.57x10 m/s 

 8.23 m/s 2
(b) ac 
r
6.965x106 m
2
12. A 2.35 kg ball is traveling at 5.30 m/s to the north. It glances off of a 2.75 kg ball that is at rest. The first ball ends
up traveling to the west at 3.16 m/s. What is the final speed of the 2.75 kg ball?
x : 0  m2 v2 fx  m1v1 fx
v2 fx 
 m1v1 fx
m2

m 
1
m

v2 fx  2.35 kg  3.16  
  2.70


2.75
s
kg
s



y : m1v1iy  m2 v2 f y
2
v2  v2 f x  v2 f y
m1v1iy
v2 f y 

m2
 2.35 kg   5.30 ms 

2.75 kg
2
m 
m

v2   2.70    4.529 
s 
s

2

 4.529
2

5.27
m
s
m
s
13. A 1.50 kg block is pushed into a spring (k = 345 N/m) a distance of 25.0 cm. When the block is released it slides
along a smooth surface and then up a ramp (elevation angle is
m
32.0). (a) What is the velocity of the block when it is released from
the spring? If the length of the ramp is 1.20 m, (b) does the block
q
slide off the end of the ramp? If it does not, how far up the ramp
does it go?
(a)
1 2 1 2
kx  mv
2
2
vx
k
m
  0.25 m 
345
kg  m
s2  m
1.50 kg

3.79
m
s
2
(b)
1
mgh  mv 2
2
h  d sin 
h
m

 3.79 s 



m
2  9.8 2 
s 

v2
2g

 1.20 m sin 32o

 0.733 m
 0.636 m
Flies off!
14. A 2.50 kg stone ball hangs from two pieces of line as shown. The long line has a length of 76.0 cm. Find: (a) the
tension in each of the cords. The short string breaks and the rock swings to the right. (b) find the speed of the ball
at the lowest point in its travel.
78.0
40.0
(a) T1 sin   T2 sin   mg  0
T1 cos 1  T2 cos  2  0 T1  T2
cos  2
cos 1
 T2
cos 40o
cos 78o
cm
76.0
 3.684 T2
T1 sin 1  T2 sin  2  mg
 3.684 T2  sin 1  T2 sin  2  mg
m

2.50 kg  9.8 2 
s 

T2 
o
3.684 sin 78  sin 40o
T1  3.684  5.77 N 
T1  0.2479 T2

T2 
mg
3.684 sin 1  sin  2
t1
78.0

t2
q1
q2 40.0
5.77 N
21.3 N
 0.2479  445 N  
110. N
mg