# Chap Problem Assignment 17 Chap 6 CQ 6 4

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```					Assignment 17: Chap 6: CQ 6.4              and   Prob 6.30, 40, 48

Conceptual Question
CQ 6.4
If two objects collide and one is initially at rest, is it possible for both objects to be at rest
after the collision? Explain.

No. Only in a precise head-on collision with equal and opposite momentum can both
balls wind up at rest. Yes. In the second case, assuming equal masses for each ball,
if Ball 2, originally at rest, is struck squarely by Ball 1, then Ball 2 takes off with the
velocity of Ball 1. Then Ball 1 is at rest.

Problems:
Problem 6.30
An 8 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of
height 1.0 m The bullet remains lodged in the block after impact and the block lands 2.0
m from the bottom of the table. Determine the initial speed of the bullet.

Solution:
First, we will find the horizontal speed, v0x , of the block and embedded bullet just after
impact. After this instant, the block-bullet combination is a projectile, and we find the
1
time to reach the floor by use of y  v0yt ay t , which becomes
2

2

1. m  0 
00
1
2
         
9. m s2 t , giving
80      2
t = 0.452 s

x 2. m
00
Thus, v0x            4. m s
43
452
t 0. s

Now use conservation of momentum for the collision, with vb = speed of incoming
bullet:

8.  10
00      3
                      
kg vb  0  258  10-3 kg  4. m s , so
43

vb  143 m s                       (about 320 mph)
Problem 6.40
A billiard ball rolling across the table at 1.5 m/s makes a head on collision with an
identical ball. Find the speed of each ball after the collision a) when the second ball is
initially at rest, b) when the second ball is moving toward the first ball at a speed of 1.0
m/s and c) when the second ball is moving away from the first ball at a speed of 1.0 m/s.

Solution:

First, consider conservation of momentum and write

m 1v1i  m 2v2i  m 1v1 f  m 2v2 f

Since m 1  m 2 , this becomes v1i  v2i  v1 f  v2 f .                                (1)


For an elastic head-on collision, we also have v1i  v2i   v1 f  v2 f , 
which may be written as                          v1i  v2i  v1 f  v2 f                (2)

Subtracting Equation (2) from (1) gives                      v2 f  v1i                 (3)

Adding Equations (1) and (2) yields                          v1 f  v2i                 (4)

Equations (3) and (4) show us that, under the conditions of equal mass objects
striking one another in a head-on elastic collision, the two objects simply
exchange velocities. Thus, we may write the results of the various collisions as

(a)   v1f  0 , v2 f  1. m s
50

(b) v1 f   1. m s , v2 f  1. m s
00              50

(c)   v1 f  1. m s , v2 f  1. m s
00              50
Problem 6.48
Consider the frictionless track shown in figure p6.48. A block of mass m1 = 5 kg is
released from A. It makes a head on collision at B with a block of mass m2 = 10 kg that is
initially at rest. Calculate the maximum height to which m1 rises after the collision.

Solution:
First, we use conservation of mechanical energy to find the speed of m 1 at B just before
1
collision. This gives m 1 v1  0  0  m 1ghi,
2

2

or       v1 
2
2 ghi                     
2 9. m s2  5. m   9. m s
80        00       90

Next, we apply conservation of momentum and knowledge of elastic collisions
to find the velocity of m 1 at B just after collision.

From conservation of momentum, with the second object initially at rest,

we have m 1v1 f  m 2v2 f  m 1v1i  0 , or v2 f 
m1
m2

v1i  v1 f                          (1)

For head-on elastic collisions, v1 f  v1i  v2 f  v2i . Since v2i  0 in this case, this
becomes v2 f  v1 f  v1i and combining this with (1) above we obtain

 m  m2         5.  10. 
00    0
v1 f   1       v1i   5.  10.   9. m s   3. m s
 m1  m2        00     0
90          30

Finally, use conservation of mechanical energy for m 1 after the collision to find
1
the maximum rebound height. This gives 0  m 1ghm ax  m 1v1 f  0
2

2

or       hm ax 
2
v1 f

  3. m s
30
2

 0. m
556
2g        
2 9. m s2
80          
Assignment 18: Chap 7: Prob 2, 4, 6, 10

Problems:
Problem 7.2:
A wheel has a radius of 4.1 m. How fart does a point on the circumference travel if
the wheel is rotated through angles of 30 degree, 30 rad, and 30 rev respectively?

Solution:
The distance traveled is s r , where  is in radians.

      r 
For 30°,          s r   4. m   30 
     1                       2. m
1
      180  

For30 radians,           s r   4. m  30 r   1.  102 m
         2 r  
For30 revolutions,       s r   4. m   30 r 
     1           ev              7.  102 m
7
         1r 
ev  

Problem 7.4:
A potter's wheel moves from rest to an angular speed of 0.2 rev/s in 30 s. Find its
angular acceleration in radians per second per section.

Solution:
 f  i
We use                 and find
t
0. r s 0  2 r 
                     4.  10-2 r s2
30 s    1r 
ev 

Problem 7.6
A centrifuge in a medical laboratory rotates at an angular speed of 3600 rev/min.
When switched off, it rotates through 50 revs before coming to rest. Find the
constant angular acceleration of the centrifuge.

Solution:
r  2 r   1 m i 
 i  3600                      377 r s
m i  1 r   60. s
n      ev     0

 2 r 
  50. r 
 1r 
ev 

w   i 0   377 r s
2
2
Thus,                            226 r s2
2       2 314 r 
Problem 7.10
The tub of a washer goes into it spin dry cycle starting from rest and reaching an
angular speed of 5 rev/s in 8 s. At this point, a person doing the laundry opens the
lid, and the washer turns off. The tub slows to rest in 12 s. Through how many
revolutions does the tub turn during the entire 20 s interval? Assume constant
acceleration.

Solution:
We will break the motion into two stages: (1) an acceleration period and (2) a
deceleration period.

The angular displacement during the acceleration period is

  f   i      5. r s 2 r 1 r   0
1   avt           t                             8. s  126 r
     2        
              2           


and while decelerating,

  f   i     0   5. r s 2 r 1 r  
2             t                             12 s  188 rad
     2        
              2            


 1r 
ev
The total displacement is      =1  2  126  188 r  
 2 r 
Assignment 19: Chap 7: Conceptual Question 13 Problems 19, 23, 43

CQ 7.13
A pail of water can be whirled in a vertical circular path such that no water is
spilled. Why does the water remain in the pail even when the pail is upside down

The tendency of water is to move in a straight line path tangent to the circular path
followed by the container. As a result, at the top of the circular path, the water
presses against the bottom of the pail, and the normal force exerted by the pail on
the water provide the radial force required to keep the water moving in its circular
path.

Problem 7:19
A 55 kg skater is moving at 4 m/s when she grads the loose end of a rope, the
opposite end of which is tied to a pole. She then moves in a circle of radius 0.8 m
around the pole. a) determine the force exerted by the horizontal rope on her arms.
b) Compare this force with her weight.

Solution:
(a) From Fr  m ac , we have

 v2   55. kg 4. m s
2
0       00
T  m  t                      1.  103 N  1. kN
10           10
 r           800
0. m

(b) The tension is l ger than her weight by a factor of
ar

T       1.  103 N
10
                   2. tm es
04 i
0      
m g  55. kg 9. m s2
80        
Problem 7:23
A 50 kg child stands at the rim of a merry go round of radius 2 m rotating with an
angular speed of 3 rad/s a) What is the child's centripetal acceleration? B) What is
the minimum force between her feet and the floor of the carousel that is required to
keep her in the circular path? c) What minimum coefficient of static friction is
required? Is the answer you found reasonable? In other words, is she likely to stay
on the merry go round?

Solution:
ac  r 2   2. m   3. r s  18. m s2
2

c                           
(b) F  m ac   50. kg 18. m s2  900 N
0       0                
(c) We know the centripetal acceleration is produced by the force of friction.
Therefore, the needed static friction force is f  900 N . Also, the normal
s

force is n  m g  490 N . Thus, the minimum coefficient of friction required
 f
s m ax       900 N
is s                           84
= 1.
n            490 N

So large a coefficient of friction is unreasonable, and she will not be able to
stay on the merry-go-round.
Problem 7:43
An athlete swings a 5 kg ball horizontally on the end of a rope. The ball moves in a
circle of radius 0.8 m at an angular speed of 0.5 rev/s. What are the a) tangential
speed of the ball and b) its centripetal acceleration? c) If the maximum tension the
rope can withstand before breaking is 100 N, what is the maximum tangential speed
the ball can have?

Solution:
500 r s  r s
The angular velocity of the ball is   0.    ev     ad

(a)   vt  r   0. m   r s  2. m s

2
vt
 r 2   0. m        r s  7. m s2
2
(b) ac                   800            ad     90
r

(c) During parts a) and b) the ball hangs down at some angle as it swings
around.
T

L

θ
m
0.8 m                           ac
W

From summing the forces horizontally                         and vertically
 F  mac                                               F  m(0)
T cos   mac                                             T sin  W  0
T cos   mac                                        T sin   mg
T cos   5kg *7.9m / s 2  39.5 N                  T sin   mg  5kg *9.81m / s 2  49 N

Solve for the angle:
T sin    49 N
tan                  1.24          arc tan(1.24)  51.1o
T cos  39.5 N
From this the length of the rope can be found.

0.8m                                   0.8m
 cos               L                   1.27m
L                                  cos 51.1o
When swinging the rope faster, this angle will change and the radius at which the
ball swings will be farther out…Assume the length of the rope remains the same.

Then,
100N

1.27m

θ=?
m
r                               ac
W

From summing the forces vertically                    and horizontally
 F  m(0)                                       F  mac
T sin  W  0                                  T cos   mac
mg 49                                                v2
sin           0.49                           T cos   m
T   100                                               r
T cos 
  arcsin(0.49)  29.3o                                 v2          r
m

so
r
 cos 29.3o
1.27
(100 N ) cos(29.3o ) 
r  1.11m                    v2                           (1.11m)  12m 2 / s 2
5kg

v  12m2 / s2  3.47m / s
Assignment 20: Chap 7   Conceptual Questions 4 and 6   Problems 29, 36

CQ 7:4
Explain why the

CQ 7:6

Problem 7:29

Solution:

Problem 7:36

Solution:

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