Physics 110 Homework Solutions Week #2
Wednesday, September 16, 2009
Chapter 2
Questions
2.13 Since the force of gravity varies inversely as the square of the distance between the
objects, we have that if the distance between the objects triples the force decreases
by a factor of 9. Thus the force is 2pN.
2.14 If the mass is doubled the gravitational field g doubles. However the distance is
also doubled so the force must decrease by a factor of 4. Combining these two
effects we have that the gravitational field must decrease by a factor of 2.
2.15 From the law of gravity the acceleration of a planet about the sun can be used to
determine the mass of the sun. The acceleration is given by
where the
separation is assumed to be known. If we can calculate the acceleration (which we
will when we study circular motion) we can determine the mass of the sun.
Multiple-Choice
- None
Problems
2.15 The orbital speed is given by
2.16 The gravitational field on the moon is given as
2.17 The force is determined from the law of gravity:
towards the Sun.
Chapter 3
Questions
- None
Multiple-Choice
3.5
C
.
Problems
3.11 Throwing a ball vertically down
a. Taking the vertically upward as the positive y-direction we
have
vertically
down.
b. The time of flight of the ball is given
by
.
c. At the maximum height the vertical velocity is zero. Thus the maximum
height is found
from
.
3.13 Dropping a rock
a. Assuming that the positive y-direction is vertically up, the time needed is
obtained from the trajectory
equation.
b. The acceleration is due to gravity and is 9.8 m/s2 vertically down. The
velocity just before impact is given
by
, or 34.3 m/s vertically down.
3.14 Dropping a rock
a. Assuming that the positive y-direction is upwards, when the rock is dropped
from rest and hits the water 4s later, we find the cliff’s height from the
trajectory equation.
b. Since the time is longer, the ball must have been launched vertically upwards
and the initial speed is given from the trajectory equation using the time to hit,
5s.
. The
sign of the velocity tells us that the choice of initial velocity direction was
correct.
Thursday, September 17, 2009
Questions
- None
Multiple-Choice
- None
Problems
- None
Friday, September 18, 2009
Chapter 5
Questions
- None
Multiple-Choice
5.1
C
5.2
D
5.3
A
5.4
C
5.5
A
5.6
B
Problems
5.3 Map of New York State
a. For BGM to WTR we have 135 N vs Alb to PLT 5 N; for KGN to SYR we
have125 NW vs NYC to BGM 125 NW; and WTR to ROC 85 S vs WRT to PLT
101 NE.
b. We have for the different cities BGM to JMS to BUF to ROC to WTR to SYR
distances of (120 + 75 + 60 +
+ 60) = 400 and 75 N along with ALB to
SYR to BGM to KGN to BGM distances of (100 + 75 + 100 + 100 ) = 375 and
125 SW
c. NYC  BGM and KGN  SYR, both are 125 NW.
5.13 Throwing a ball
a) vx = vo = constant ; vy2 = voy2 + 2gy = 2gy so vy = sqrt[2(9.8)(25)] = 22.1 m/s
down; we have that v = (20, -22.1) or the magnitude of v is sqrt[(20)2 + (22.1)2]
= 29.8 m/s and the direction of the velocity is given by tan θ = 22.1/20, or θ =
47.9o below the horizontal
b) Using y = yo + voyt + ½ ayt2, we have 0 = 25 + 0 –1/2 (9.8) t2 and solving for t, we
have t = 2.26 s
c) x = voxt = (20)(2.26) = 45.2 m
d) 9.8 m/s2 downward
5.15 B-2 Bomber
1
a) Δy = – gt2; Δx = vit. Combine the equations eliminating t:
2
.
b) The plane has the same velocity as the bomb in the x direction. Therefore, the
plane will be 10600m directly above the bomb when it hits the ground.
5.21 A lacrosse goalie
a. The time of flight is given as
.
b. The horizontal displacement is given
by
c. The speed is given as
.
. The minimum is when the ball is at its highest
point so that viy is zero and the velocity is
therefore
.
d. The magnitude of the impact velocity is given as
,
where vfx = 8.2 m/s and
. Thus
the velocity is
e. The velocity is given as
. Again we have vfx = 8.2
m/s and the magnitude of the final y-velocity
is
. We are told in the
problem that the ball is on its way back down when caught. Thus the final yvelocity is -3.6m/s. Thus the velocity
is
.
5.30 Lava Bombs
a. In order to determine the speed of the lava bomb we use our horizontal and
vertical trajectory equations and eliminate time between them. This gives the
vertical position of the lava bomb in terms of its horizontal position. Thus we will
have only one equation and one unknown.
b. To determine the time of flight of the lava bomb we use our horizontal trajectory
equation since we know the horizontal displacement and now the initial speed.
Thus we have
c. We calculate the final velocity of the lava bomb from our horizontal and vertical
velocity equations of motion. Thus we have
d. The acceleration is due to gravity and is 9.8m/s2 vertically down.
Monday, September 21, 2009
Questions
- None
Multiple-Choice
- None
Problems
- None