Download 11. Two blocks of masses m and 3m are placed on a frictionless

11. Two blocks of masses m and 3m are placed on a frictionless, horizontal surface. A
light spring is attached to the more massive block, and the blocks are pushed
together with the spring between them (Fig. P9.11). A cord initially holding the
blocks together is burned; after that happens, the block of mass 3m moves to the
right with a speed of 2.00 m/s. (a) What is the velocity of the block of mass m? (b)
Find the system’s original elastic potential energy, taking m = 0.350 kg. (c) Is the
original energy in the spring or in the cord? (d) Explain your answer to part (c). (e)
Is the momentum of the system conserved in the bursting-apart process? Explain
how that is possible considering (f) there are large forces acting and (g) there is no
motion beforehand and plenty of motion afterward?
Figure P9.11
SOL. (a)
For the system of two blocks Dp = 0, or pi = p f . Therefore,
0  mυm   3m  2.00 m/s 
(b)
1 2 1 2 1
kx  mυM   3m  υ32M
2
2
2
1
3
2
2
 0.350 kg  6.00 m/s    0.350 kg  2.00 m/s 
2
2
 8.40 J

(c)
The original energy is in the spring.
(d)
A force had to be exerted over a displacement to compress the spring,
transferring energy into it by work. The cord exerts force, but over no
displacement.
(e)
System momentum is conserved with the value zero.
(f)
The forces on the two blocks are internal forces, which cannot change the
momentum of the system— the system is isolated.
(g)
Even though there is motion afterward, the final momenta are of equal
magnitude in opposite directions so the final momentum of the system is
still zero.
29. A tennis ball of mass 57.0 g is held just above a basketball of mass 590 g. With
their centers vertically aligned, both balls are released from rest at the same time, to
fall through a distance of 1.20 m, as shown in Figure P9.29. (a) Find the magnitude
of the downward velocity with which the basketball reaches the ground. (b) Assume
that an elastic collision with the ground instantaneously reverses the velocity of the
basketball while the tennis ball is still moving down. Next, the two balls meet in an
elastic collision. To what height does the tennis ball rebound?
Figure P9.29
SOL.
(a)
The speed υ of both balls just before the basketball reaches the ground may be
found from υ2yf = υ2yi + 2ay Dy as
υ=
=
(b)
υ2yi + 2a y Dy =
0 + 2 ( - g ) (- h ) =
2 gh
2 ( 9.80 m/s 2 ) (1.20 m ) = 4.85 m/s
Immediately after the basketball rebounds from the floor, it and the tennis
ball meet in an elastic collision. The velocities of the two balls just before
collision are
for the tennis ball (subscript t): υti  υ
and for the basketball (subscript b): υbi = + υ
We determine the velocity of the tennis ball immediately after this elastic
collision as follows:
Momentum conservation gives mt υtf + mb υbf = mt υti + mb υbi
or mt υtf  mb υbf   mb  mt  υ
[1]
From the criteria for a perfectly elastic collision: υti  υbi    υtf  υbf
or υbf  υtf  υtti  υbi  υtf  2υ

[2]
Substituting equation [2] into [1], mt υtf  mb  υtf  2υ    mb  mt  υ
or the upward speed of the tennis ball immediately after the collision is
 3m  mt 
 3mb  mt 
υtf   b
υ  
 2 gh
 mt  mb 
 mt  mb 
The vertical displacement of the tennis ball during its rebound following
the collision is given by υ2yf = υ2yi + 2ay Dy as
y 
υ 2yf  υ 2yi

2a y
 1

2g

0  υ 2yf
2g 
  3mb  mt 

 m  m 
t
b 

2
 2 gh 
2
 3mb  mt 

 h
 mt  mb 
 3  590 g    57.0g  
Substituting, y  
 1.20m   8.41m
 57.0g  590g 
2
30. As shown in Figure P9.30, a bullet of mass m and speed υ passes completely
through a pendulum bob of mass M. The bullet emerges with a speed of υ/2. The
pendulum bob is suspended by a stiff rod (not a string) of length ℓ, and negligible
mass. What is the minimum value of υ such that the pendulum bob will barely
swing through a complete vertical circle?
Figure P9.30
SOL. Energy is conserved for the bob-Earth system between bottom and top of the swing. At
the top the stiff rod is in compression and the bob nearly at rest.
Ki 
i
υb2  4 g
1
Mυb2  0  0  Mg 2
2
so υb  2 g
 Kf 
f
:
ANS. FIG. P9.30
Momentum of the bob-bullet system is conserved in the collision:

υ
mυ  m  M 2 g
2
  υ= 4mM
g
44. The mass of the blue puck in Figure P9.44 is 20.0% greater than the mass of the
green puck. Before colliding, the pucks approach each other with omenta of equal
magnitudes and opposite directions, and the green puck has an initial speed of 10.0
m/s. Find the speeds the pucks have after the collision if half the kinetic energy of
the system becomes internal energy during the collision.
Figure P9.44
SOL. The initial momentum of the system is 0. Thus,
(1.20m)υ Bi  m(10.0 m/s) and υBi = 8.33 m/s
From conservation of energy,
1
1
1
m(10.0 m/s) 2  (1.20m)(8.33m/s) 2  m(183m 2 /s 2 )
2
2
2
1
1
11

K f  m(υG ) 2  (1.20m)(υ B ) 2   m(183m 2 /s 2 ) 
2
2
22

Ki 
or
G2  1.20υ2B  91.7 m2 /s2
From conservation of momentum,
[1]
mυG  (1.20m)υ B
[2]
Solving [1] and [2] simultaneously, we find
(1.20υ B ) 2  1.20υ 2B  91.7 m 2 /s 2
υ B  (91.7 m 2 /s 2 / 2.64)1/ 2
which gives
υB = 5.89 m/s (speed of blue puck after collision)
and
υG = 7.07 m/s (speed of green puck after collision)
50. A water molecule consists of an oxygen atom with two hydrogen atoms bound to
(Fig. P9.50). The angle between the two bonds is 106°. If the bonds are 0.100 nm
long, where is the center of mass of the molecule?
Figure P9.50
SOL. We use a coordinate system centered in the oxygen (O) atom, with the x axis to the right
and the y axis upward. Then, from symmetry,
xCM  0
and
yCM 
mi yi
mi


1


 15.999 u +1.008 u +1.008 u 
 [0  (1.008 u)(0.100 nm) cos53.0
 (1.008 u)(0.100 nm) cos53.0]
The center of mass of the molecule lies on the dotted line shown in ANS. FIG.
P9.50, 0.006 73 nm below the center of the O atom.
ANS. FIG. P9.50
71. A 1.25-kg wooden block rests on a table over a large hole as in Figure P9.71. A
5.00-g bullet with an initial velocity υi is fired upward into the bottom of the block
and remains in the block after the collision. The block and bullet rise to a maximum
height of 22.0 cm. (a) Describe how you would find the initial velocity of the bullet
using ideas you have learned in this chapter. (b) Calculate the initial velocity of the
bullet from the information provided.
Figure P9.71
SOL. (a) Momentum of the bullet-block system is conserved in the collision, so you can relate
the speed of the block and bullet right after the collision to the initial speed of
the bullet. Then, you can use conservation of mechanical energy for the
bullet-block-Earth system to relate the speed after the collision to the
maximum height.
(b)
Momentum is conserved by the collision. Find the relation between the speed
of the bullet υi just before impact and the speed of the bullet + block υ just
after impact:
p1i  p 2i  p1 f  p 2 f  m1υ1i  m2 υ2i  m1υ1 f  m2 υ2i
mυi  M (0)  mυ  Mυ  (m  M )

υi 
(m  M )
υ
m
For the bullet-block-Earth system, total energy is conserved. Find the relation
between the speed of the bullet-block υ and the height h the block climbs to:
Ki  U i  K f  U f
1
(m  M )υ2  0  (m  M ) gh  υ  2 gh
2
Combining our results, we find
υi 
mM
m
 1.255kg 
2
2 gh  
 2  9.80 m/s   0.220 m 
 0.00500 kg 
75. Two gliders are set in motion on a horizontal air track. A spring of force constant k
is attached to the back end of the second glider. As shown in Figure P9.75, the first
glider, of mass m1, moves to the right with speed υ1, and the second glider, of mass
m2, moves more slowly to the right with speed υ2. When m1 collides with the spring
attached to m2, the spring compresses by a distance xmax, and the gliders then move
apart again. In terms of υ1, υ2, m1, m2, and k, find (a) the speed v at maximum
compression, (b) the maximum compression xmax, and (c) the velocity of each glider
after m1 has lost contact with the spring.
Figure P9.75
SOL. (a)
When the spring is fully compressed, each cart moves with same velocity υ.
Apply conservation of momentum for the system of two gliders
pi  p f :
(b)
m11  m2 υ2  (m1  m2 )υ  υ 
m1υ1  m2 υ2
m1  m2
Only conservative forces act; therefore, ∆E = 0.
1
1
1
1
m1υ12  m1υ22   m1  m2  υ2  kxm2
2
2
2
2
Substitute for υ from (a) and solve for xm.


1
xm2  
[ m  m2  m1υ12   m1  m2  m2 υ22
 k  m  m    1
1
2 

  m1υ1    m2 υ2   2m1m2 υ1υ2 ]
2
xm 
(c)

m1m2 υ12  υ22  2υ1υ2
k  m1  m2 
2
   υ  υ


1

m1m2

k  m1  m2  
2
m1υ1 + m2v2 = m1υ1f + m2υ2f
Conservation of momentum: m1 (υ1 − υ1f) = m2 (υ2f − υ2)
1
1
1
1
m1υ12  m2 υ22  m1υ12 f  m2 υ22 f
2
2
2
2
Conservation of energy:



which simplifies to: m1 υ12  υ12f  m2 υ22 f  22

[1]
Factoring gives m1 υ1  υ1 f
 υ  υ   m  υ
1
1f
2
2f

 υ2    υ 2 f  υ 2 
and with the use of the momentum equation (equation [1]),
this reduces to
υ1 + υ1f = υ2f + υ2
or
υ1f = υ2f + υ2 − υ1
[2]
Substituting equation [2] into equation [1] and simplifying yields
υ2 f 
2m1υ1   m2  m1  υ2
m1  m2
Upon substitution of this expression for into equation [2], one finds
υ1 f 
 m1  m2  υ1  2m2 υ2
m1  m2
Observe that these results are the same as two equations given in the chapter
text for the situation of a perfectly elastic collision in one dimension.
Whatever the details of how the spring behaves, this collision ends up being
just such a perfectly elastic collision in one dimension.
79. A 0.400-kg blue bead slides on a frictionless, curved wire, starting from rest at
point
A
in Figure P9.79, where h = 1.50 m. At point
B
, the blue bead collides
elastically with a 0.600-kg green bead at rest. Find the maximum height the green
bead rises as it moves up the wire.
Figure P9.79
SOL. We will use the subscript 1 for the blue bead and the subscript 2 for the green bead.
Conservation of mechanical energy for the blue bead-Earth system,
Ki + Ui = Kf + Uf, can be written as
1 2
mυ1  0  0 m g h
2
where υ1 is the speed of the blue bead at point B just before it collides with the
green bead. Solving for υ1 gives
υ1  2 gh  2  9.80 m/s  (1.50 m)  5.42 m/s
2
Now recall Equations 9.21 and 9.22 for an elastic collision:
 m  m2 
 2m2 
υ1 f   1
 υ1i  
 υ 2i
 m1  m2 
 m1  m2 
 2m2 
 m1  m2 
υ2 f  
 υ1i  
 υ2i
m

m
m

m
2 
2 
 1
 1
For this collision, the green bead is at rest, so υ2i = 0, and Equation 9.22
simplifies to
 2m2 
 m1  m2 
 2m2
υ2 f  
 υ1i  
 υ2i  
 m1  m2 
 m1  m2 
 m1  m2

 υ1i

Plugging in gives


2(0.400 kg)
υ2 f  
 (5.42 m/s)  4.34 m/s
0.400
kg

0.600
kg


Now, we use conservation of the mechanical energy of the green bead after
collision to find the maximum height the ball will reach. This gives
0  m2 g ymax 
Solving for ymax gives ymax 
22 f
2g
1
m222 f  0
2

 4.34 m/s 

2
2 9.80 m/s 2

 0.960 m
89. A 5.00-g bullet moving with an initial speed of υi = 400 m/s is fired into and passes
through a 1.00-kg block as shown in Figure P9.89. The block, initially at rest on a
frictionless, horizontal surface, is connected to a spring with force constant 900
N/m. The block moves d = 5.00 cm to the right after impact before being brought to
rest by the spring. Find (a) the speed at which the bullet emerges from the block and
(b) the amount of initial kinetic energy of the bullet that is converted into internal
energy in the bullet-block system during the collision.
Figure P9.89
SOL. (a)
We find the speed when the bullet emerges from the block by using momentum
conservation:
mυi  MVi  mυ
The block moves a distance of 5.00 cm. Assume for an approximation that
the block quickly reaches its maximum velocity, Vi, and the bullet kept going
with a constant velocity, υ. The block then compresses the spring and stops.
After the collision, the mechanical energy is conserved in the block-spring
system:
1
1
MVi 2  kx 2
2
2
Vi 
υ=

 900 N/m   5.00 102 m 
2
1.00 kg
mi  MV
m
 5.00 10
3
 1.50 m/s

kg (400 m/s)  (1.00 kg)(1.50 m/s)
5.00 103 kg
υ  100 m/s
(b) Identifying the system as the block and the bullet and the time interval from
just before the collision to just after the collision,
K  Eint  0 gives
1
1
1

Eint  K    mυ2  MVi 2  mυi2 
2
2
2

Then
1
Ei n t   ( 0 . 0 0 5 0 0 k g ) ( 1 20 0 m / s )
2
1

 ( 1 . 0 0 k g ) ( 1 . 520 m / s )
2

1
2
 (0.005 kg)(400 m/s)

2
 374 J
ANS. FIG. P9.89