Physics 20 Homework Set 6
Chapter 6 Problems 2,3,12,13,16,21,24,27,32,34,43
6.2
Assume the initial direction of the ball in the –x direction, away from the net.
(a)
I  p  m  v f  vi    0.0600 kg  40.0 m s   50.0 m s  giving
I  5.40 kg  m s  5.40 N  s toward the net.
(b) Work  KE 

6.3
1
m  v 2f  vi2 
2
 0.0600 kg   40.0
2
2
m s    50.0 m s  
   27.0 J
2
Use p  mv :
(a)
p  1.67 1027 kg 5.00 106 m s   8.35 1021 kg  m s
(b)
p  1.50  102 kg 3.00  102 m s  4.50 kg  m s
(c)
p   75.0 kg 10.0 m s   750 kg  m s






(d) p  5.98 1024 kg 2.98  104 m s  1.78  1029 kg  m s
6.12
(a) Impulse = area under curve = (two triangular areas of altitude 4.00 N and
base 2.00 s) + (one rectangular area of width 1.00 s and height of 4.00 N.)
Thus,
(b)
  4.00 N  2.00 s  
I  2
   4.00 N 1.00 s   12.0 N  s
2


I  Fav  t   p  m  v f  vi  , so v f  vi 
vf  0 
(c)
v f  vi 
I
m
12.0 N  s
 6.00 m s
2.00 kg
I
12.0 N  s
 2.00 m s 
 4.00 m s
m
2.00 kg
6.13
(a) The impulse is the area under the curve between 0 and 3.0 s.
This is:
I  (4.0 N)(3.0 s)  12 N  s
(b) The area under the curve between 0 and 5.0 s is:
I  (4.0 N)(3.0 s)  (2.0 N)(2.0 s)  8.0 N  s
(c)
6.16
I  Fav  t   p  m  v f  vi  , so v f  vi 
I
m
at 3.0 s:
v f  vi 
I
12 N  s
 0
 8.0 m s
m
1.50 kg
at 5.0 s:
v f  vi 
I
8.0 N  s
 0
 5.3 m s
m
1.50 kg
Choose the positive direction to be from the pitcher toward home plate.
(a)
I  Fav  t   p  m  v f  vi    0.15 kg   22 m s    20 m s 
I  Fav  t   6.3 kg  m s or
(b) Fav 
6.3 kg  m s toward the pitcher
I
6.3 kg  m s

 3.2  103 N
t
2.0  103 s
or Fav  3.2  103 N toward the pitcher
6.21
The velocity of the girl relative to the ice, vGI , is vGI  vGP  vPI where
vGP  velocity of girl relative to plank, and
vPI  velocity of plank relative to ice. Since we are given that
vGP  1.50 m s , this becomes
vGI  1.50 m s  vPI
(1)
m 
(a) Conservation of momentum gives mG vGI  mP vPI  0 , or vPI    G  vGI
 mP 
 m 
Then, Equation (1) becomes 1  G  vGI  1.50 m s
 mP 
or vGI 
1.50 m s
 1.15 m s
 45.0 kg 
1 

 150 kg 
(b) Then, using (2) above,
 45.0 kg 
vPI   
 1.15 m s    0.346 m s
 150 kg 
or vPI  0.346 m s directed opposite to the girls motion
6.24
For each skater, the impulse-momentum theorem gives
Fav 
p m  v   75.0 kg  5.00 m s 


 3.75  103 N
t
t
0.100 s
Since Fav  4500 N , there are no broken bones
(2)
6.27
Let M = mass of ball, m = mass of bullet, v = velocity of bullet, and V = the initial
velocity of the ball-bullet combination. Then, using conservation of momentum
from just before to just after collision gives
 m 
or V  
v
 M m
 M  m V  mv  0
Now, we use conservation of mechanical energy from just after the collision until
the ball reaches maximum height to find
0   M  m  g hmax 
2
1
V2
1  m  2
2
M

m
V

0
or
h




max

 v
2
2g 2g  M m
With the data values provided, this becomes
2
hmax
6.32


1
0.030 kg
2

  200 m s   57 m
2 
2 9.80 m s  0.15 kg  0.030 kg 


(a) Using conservation of momentum,  p after   p before , gives
 4.0  10  3.0 kg v   4.0 kg 5.0 m s   10 kg 3.0 m s   3.0 kg   4.0 m s 
Therefore,
(b)
v   2.2 m s , or 2.2 m s toward the right
No . For example, if the 10-kg and 3.0-kg mass were to stick together first,
they would move with a speed given by solving
13 kg  v1  10 kg  3.0
m s    3.0 kg   4.0 m s  , or v1   1.38 m s
Then when this 13 kg combined mass collides with the 4.0 kg mass, we have
17 kg  v  13 kg 1.38 m s    4.0 kg 5.0
just as in part (a).
m s  , and v   2.2 m s
6.34
Using conservation of momentum gives
10.0 g  v1 f  15.0 g  v2 f  10.0 g  20.0
cm s   15.0 g  30.0 cm s 
(1)
For elastic, head on collisions, v1i  v1 f  v2i  v2 f which becomes
20.0 cm s  v1 f   30.0 cm s+v2 f
(2)
Solving (1) and (2) simultaneously gives v1 f   40.0 cm s ,
and v2 f  10.0 cm s
6.43
Choose the x-axis to be along the original line of motion.
(a) From conservation of momentum in the x direction,
m  5.00 m s   0  m  4.33 m s  cos30.0  m v2 f cos
or v2 f cos  1.25 m s
(1)
Conservation of momentum in the y direction gives
0  m  4.33 m s  sin 30.0  m v2 f sin  , or v2 f sin    2.16 m s
Dividing (2) by (1) gives tan  
 2.16
 1.73 and   60.0
1.25
Then, either (1) or (2) gives v2 f  2.50 m s , so the final velocity of the second
ball is v2 f  2.50 m s at -60.0

1
1
2
(b) KEi  mv12i  0  m 5.00 m s   m 12.5 m2 s2
2
2
KE f 


1
1
m v12f  m v22 f
2
2

1
1
2
2
m  4.33 m s   m  2.50 m s   m 12.5 m2 s2
2
2
Since KE f  KEi , this is an elastic collision

(2)