Physics 110 Homework Solutions Week #3 - Tuesday
Friday, April 9, 2010
Chapter 5
Questions
- None
Multiple-Choice
- None
Problems
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, April 12, 2010
Chapters 2
Questions
2.8
The VW bug and the semi truck experience the same magnitude of the force by
Newton’s 3rd law. Since the VW bug has a smaller mass it experiences the greater
acceleration.
Multiple-Choice
2.2
B
2.3
B
2.5
D
2.27 D
2.28 A
2.29 C
Problems
2.6 From Newton’s second law the net force is given as the product of the mass and the
acceleration. We need to calculate the acceleration of the car traveling from rest to 60
mph in 6s. We first convert 60 mph in to meters per second and use an equation of
motion to determine the acceleration. Thus we have
and
. Therefore the force is F = ma =
1400kg * 4.48 m/s2 = 6270 N.
2.18 From Newton’s 2nd law we have:
2.19 The force per unit area is the pressure.
2.24 A water strider
a. From the data given we have that there are 6 feet and the insect has a mass of
0.01g. Thus a single leg needs to support a mass of 1.67x10-3 g. Thus the vertical
force is due to surface tension caused by the weight of the leg and we have
.
b. Since the insect travels at constant velocity then the total resistive force is divided
among each of the legs, so that the resistive force on a leg is 1x10-6 N / 6 legs =
1.67x10-7 N.