** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #2b
(20 Points Total)
CHAPTER 2b
GROUP B SOLUTIONS
16. A rock thrown straight up climbs for 2.50 s, then falls to the ground. Neglecting air
resistance, with what velocity did the rock strike the ground?
This problem requires the relation a 
Given: a  g  9.8
v
m
to solve it, where a  g  9.8 2
t
s
m
and t  2.5s
s2
m
m
 2.5s  24.5
2
s
s
This velocity represents the speed the rock would have if it were to free fall from rest for 2.5 s.
In freefall the rock will take the same time to go up as it does to come down. The rock will strike
the ground at the same speed as if it were dropped from rest and allowed 2.5 s to fall.
v  g  t  9.8
The rock strikes the ground at a velocity of 24.5 m/s.
17. An object is observed to fall from a bridge, striking the water below 2.5 s later.
a. With what velocity did it strike the water?
This problem requires the relation a 
Given: a  g  9.8
v
m
to solve it, where a  g  9.8 2
t
s
m
and t  2.5s
s2
v  g  t  9.8
m
m
 2.5s  24.5
2
s
s
The rock strikes the water at a velocity of 24.5 m/s.
b. What was its average velocity during the fall?
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #2b
(20 Points Total)
This problem requires that you know the definition of average velocity. Since the initial
velocity is 0 m/s and the final velocity is 24.5 m/s, the average is just half of 24.5 m/s.
The average velocity during the fall is 12.25 m/s.
c. How high is the bridge?
This problem is most easily solved using the relation x 
1
m
a  t 2 , where a  g  9.8 2 .
2
s
m
and t  2.5s
s2
1
1
1
m
2
x  a  t 2  g  t 2   9.8 2  2.5s   30.6m
2
2
2
s
Given a  g  9.8
The height of the bridge is 30.6 m.
18. A ball dropped from a window strikes the ground 2.00 s later. How high is the window above
ground?
1
m
This problem is most easily solved using the relation x  a  t 2 , where a  g  9.8 2 .
2
s
m
Given a  g  9.8 2 and t  2.0s
s
1
1
1
m
2
x  a  t 2  g  t 2   9.8 2  2.0s   19.6m
2
2
2
s
The window is 19.6 m high.
19. Find the resulting acceleration from a 300 N force that acts on an object with a mass of
3,000 kg.
a. This is a Newton’s 2ND Law problem: F  ma . Solve for a.
F  ma
a
F
300 N
m

 0.1 2
m 3,000 kg
s
b. The object will accelerate at 0.1 m/s2.
20. What is the momentum of a 30.0 kg shell fired from a cannon with a velocity of 500 m/s?
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #2b
(20 Points Total)
a. Use the definition of momentum to solve this problem: p  mv
p  mv  30 kg  500
m
kg  m
 15,000
s
s
b. The momentum of the shell is 15,000 kgm/s.
21. What is the momentum of a 39.2 N bowling ball with a velocity of 7.00 m/s?
a. Use Newton’s 2nd Law and the definition of momentum to solve this problem:
F  ma and p  mv
F  ma
F F 39.2 N
m   
 4.00 kg
a g 9.8 m
s2
m
kg  m
p  mv  4 kg  7.00  28.0
s
s
b. The momentum of the shell is 28.0 kgm/s.
22. A 30.0 kg shell fired from a 2,000 kg cannon with a velocity of 500 m/s. What is the
resulting velocity of the cannon?
a. This is a momentum conservation problem. We need to use the relation m1v1  m2 v2 ,
where m1 = 2,000.0 kg, m2 = 30.0 kg, v2 = 500 m/s and solve for v1.
m1v1  m2 v2
v1 
m2 v 2

m1
m
s  7.5 m
2,000 kg
s
30.0 kg  500
b. The cannon will recoil at 7.5 m/s in the opposite direction of the shell.
23. An 80.0 kg man is standing on a frictionless ice surface when he throws a 2.00 kg book at
10 m/s. With what velocity does the man move across the ice?
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #2b
(20 Points Total)
a. This is a momentum conservation problem. We need to use the relation m1v1  m2 v2 ,
where m1 = 80.0 kg, m2 = 2.00 kg, v2 = 10 m/s and solve for v1.
m1v1  m2 v2
v1 
m2 v 2

m1
m
s  0.25 m
80 kg
s
2.00 kg 10
b. The man will move across the ice at 0.25 m/s in the opposite direction of the book.
24. What is the weight of a 5.00 kg backpack? What is the acceleration of the backpack if a
net force of 10.0 N is applied?
a. The first part is a Newton’s 2ND Law problem: F  ma , where a=g.
F  ma  mg  5.00 kg  9.8
m
 49 N
s2
b. The backpack weights 49 N.
c. The second part of the question is also a Newton’s 2ND Law problem: F  ma . Solve
for a.
F  ma
F
10 N
m
a 
 2.00 2
m 5.00 kg
s
d. The backpack will accelerate at 2.00 m/s2.
25. What is the net force required to accelerate a 20.0 kg object to 10.0 m/s2?
a. This is a Newton’s 2ND Law problem: F  ma . Solve for F.
F  ma  20.0 kg 10
b. The net force required is 200 N.
m
 200 N
s2
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #2b
(20 Points Total)
26. What forward force must the ground apply to the foot of a 60.0 kg person to result in the
acceleration of 1 m/s2?
a. This is a Newton’s 2ND Law problem: F  ma . Solve for F.
F  ma  60.0 kg 1
m
 60.0 N
s2
b. The net force required is 60 N.
27. A 1,000 kg car accelerates uniformly to double its speed from 36.0 km/hr in 5.00 s. What
net force acted on this car?
a. This is a Newton’s 2ND Law problem: F  ma . Solve for F. First we must determine
v
the acceleration using a 
. A unit conversion is needed to convert 36 km/hour into
t
m/s.
b.
36
km
km 1,000 m   1 hr 
m
 36


 10


hr
hr  1 km   3600 s 
s
m
m
 10
v
s
s 2m
a


t
t
5s
s2
m
F  ma  1,000 kg  2 2  2,000 N
s
v f  vi
20
c. The net force required is 2,000 N.
28. A net force of 3,000 N accelerates a car from rest to 36.0 km/hr in 5.00 s. What is the
mass of the car? What is the weight of the car?
a. This is a Newton’s 2ND Law problem: F  ma . Solve for m, but first, find a
v
using a 
.
t
36
km
km 1,000 m   1 hr 
m
 36


 10


hr
hr  1 km   3600 s 
s
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #2b
(20 Points Total)
m
v
s 2m
a

t
5s
s2
a. F  ma
10
m
F 3,000 N

 1,500 kg
m
a
2 2
s
b. The mass of the car is 1,500 kg.
c. The weight of the car is given by F  mg  1,500 kg  9.8
m
 14,700 N . Thus the car
s2
weights 14,700 N.
29. How much does a 60.0 kg person weight?
a. The weight of the person is given by F  mg  60.0 kg  9.8
m
 588 N . Thus the
s2
person weights 588 N.
30. What tension must a 50.0 cm length of a string support in order to whirl an attached
1,000.0 g stone in a circular path at 5.00 m/s?
a. This is a circular motion problem so that you must use the equation for centripetal force
mv 2
F 
. The tension asked for is just the force, and we need only convert to
r
MKS units.
2
m

1.0 kg   5.00 
2
mv
s


 50 N
b. F 
r
0.5 m
c. The tension in the string is 50 N.
31. A 200.0 kg astronaut and equipment move with a velocity of 2.00 m/s toward an orbiting
spacecraft. How long will the astronaut need to fire a 100.0 N rocket backpack to stop the
motion relative to the spacecraft?
a. This is an impulse problem where we have to drain away the momentum of the
astronaut by applying an impulse. The change in momentum we need is p  m  v ,
and the working relation is F  t  p . Solve for t .
** SOLUTIONS Jaquin’s Sections **
SCI111 Homework Assignment #2b
(20 Points Total)
F  t  p
32. F  t  m  v
t 
m  v

F
200 kg  2.00
100 N
m
s  4s
a. The astronaut must fire the backpack rockets for 4 seconds to bring themselves to a
stop.