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Unformatted text preview: CHAPTER 6
Quick Quizzes
1. (d). We are given no information about the masses of the objects. If the masses are the
same, the speeds must be the same (so that they have equal kinetic energies), and then
p1 = p2. If the masses are not the same, the speeds will be different, as will the momenta,
and either p1 < p2, or p1 > p2, depending on which particle has more mass. Without
information about the masses, we cannot choose among these possibilities. 2. (c). Because the momentum of the system (boy + raft) remains constant with zero
magnitude, the raft moves towards the shore as the boy walks away from the shore. 3. (c) and (e). Because object 1 has larger mass, its acceleration due to the applied force is
smaller, and it takes a longer time interval ∆t to experience the displacement ∆x than does
object 2. Thus, the impulse F∆t on object 1 is larger than that on object 2. Consequently,
object 1 will experience a larger change in momentum than object 2, which tells us that (c)
is true. The same force acts on both objects through the same displacement. Thus, the same
work is done on each object, so that each must experience the same change in kinetic
energy, which tells us that (e) is true. 4. (d). 5. (b). You must conclude that the collision is inelastic because some of the kinetic energy is
carried away by mechanical waves––sound. If the collision were elastic, you would not
hear any clicking sound. 6. (a). 7. (a). Perfectly inelastic—the two “particles”, skater and Frisbee, are combined after the
collision; (b) Inelastic—because the Frisbee bounced back with almost no speed, kinetic
energy has been transformed to other forms; (c) Inelastic—the kinetic energy of the Frisbee
is the same before and after the collision. Because momentum of the skaterFrisbee system
is conserved, however, the skater must be moving after the catch and the throw, so that
the final kinetic energy of the system is larger than the initial kinetic energy. This extra
energy comes from the muscles of the skater. 8. (b). If all of the initial kinetic energy is transformed, then nothing is moving after the
collision. Consequently, the final momentum of the system is necessarily zero. Because
momentum of the system is conserved, the initial momentum of the system must be zero.
Finally, because the objects are identical, they must have been initially moving toward
each other along the same line with the same speed. 173 CHAPTER 6 Problem Solutions 6.1 1
1 ( mv )
1 p2 p2
KE = mv 2 =
=
=
2
2m
2 m 2m 6.2 Assume the initial direction of the ball in the –x direction, away from the net. 2 (a) ( ) Im pu lse = ∆p = m v f − v i = ( 0.0600 k g ) ⎡ 40.0 m s − ( −50.0 m s ) ⎤ giving
⎣
⎦
Impulse = 5.40 kg ⋅ m s = 5.40 N ⋅ s toward the net. ( 1
(b) Work = ∆KE = m v 2 − v i2
f
2 = 6.3 ( 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 k g ⋅ m s
(c) p = ( 75.0 k g ) (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 6.4 (a) Since the ball was thrown straight upward, it is at rest momentarily (v = 0) at its
maximum height. Therefore, p = 0 . 174 CHAPTER 6 2
(b) The maximum height is found from v y = v i2y + 2 ay ( ∆y ) with v y = 0 . 0 = v + 2 ( − g ) ( ∆y ) max . Thus, ( ∆y ) max =
2
iy We need the velocity at ∆y = ( ∆y )max
2 = 2
v iy 2g 2
v iy 4g . 2
, thus v y = v i2y + 2 ay ( ∆y ) gives 2
2
⎛ v iy ⎞ v iy
v iy 15 m s
v = v + 2( − g) ⎜ ⎟ =
, or v y =
.
=
2
2
⎝ 4 g⎠ 2
2
y 2
iy Therefore, p = mv y = 6.5 ( 0.10 kg ) (15 m s ) =
2 1.1 k g ⋅ m s upward. (a) If pball = pbullet ,
then v ball = ( )( ) 3.00 × 10−3 kg 1.50 × 103 m s
mbullet v bullet
=
= 31.0 m s .
0.145 kg
mball (b) The kinetic energy of the bullet is KEbullet ( )( 3.00 × 103 k g 1.50 × 103 m s
1
2
= mbullet v bullet =
2
2 ) 2 = 3.38 × 103 J ( 0.145 kg ) ( 31.0 m s ) = 69.7 J .
1
2
= mball v ball =
2
2
2 while that of the baseball is KEball The bullet has the larger kinetic energy by a factor of 48.4.
6.6 From the impulsemomentum theorem, Thus, F = ( m v f − vi ( ∆t ) ) = ( 55 × 10 3 )( F ( ∆t ) = ∆p = m v f − m v i . ) kg 2 .0 × 102 ft s − 0 ⎛ 1 m s ⎞
⎜ 3.281 ft s ⎟ = 1.7 k N .
0.0020 s − 0
⎝
⎠ 175 CHAPTER 6.7 6 If the diver starts from rest and drops vertically into the water, the velocity just before
impact is found from ( KE + PE ) = ( KE + PE )
g g f i 12
mv impact + 0 = 0 + mgh ⇒ v impact = 2 gh
2
With the diver at rest after an impact time of ∆t, the average force during impact is given
by F= ( m 0 − v impact
∆t ) = m 2 gh
∆t or F = m 2 gh
∆t ( directed upward) . Assuming a mass of 55 kg and an impact time of ~1.0 s, the magnitude of this average
force is F= 6.8 ( 55 kg ) 2 ( 9.80 ) m s 2 (10 m ) 1.0 s ~ 103 N . =770 N , o r ( The speed just before impact is given by KE + PEg ) = ( KE + PE )
g f as i 12
mv impact + 0 = 0 + mgh , or v impact = 2 gh .
2
Taking downward as positive, the impulsemomentum theorem gives the average force
as F= ( ) ( Thus, F = −7.00 × 10 3 N , o r F = 7.00 × 103 N 6.9 ) 2
−m 2 gh − ( 60.0 kg ) 2 9.80 m s (10.0 m )
∆p m 0 − v impact
=
=
=
.
∆t
∆t
∆t
0.120 s ( u p w ard ) . Im pu lse = F ( ∆t ) = ∆p = m ( ∆v ) .
Thus, Im pu lse = m ∆v = ( 70.0 k g ) ( 5.20 m s − 0) = 364 k g ⋅ m s , and F=
or Impulse 364 kg ⋅ m s
=
= 438 kg ⋅ m s 2 ,
∆t
0.832 s F = 438 N d ire ct e d fo rw ard .
176 CHAPTER 6.10 6 From the impulsemomentum theorem, F ( ∆t ) = ∆p = m v f − m v i , the average force
required to hold onto the child is F= ( m v f − vi ( ∆t ) ) = (12 kg ) ( 0 − 60 m i h ) ⎛ 1ms ⎞
3
⎜ 2 .237 m i h ⎟ = − 6.4 × 10 N .
⎝
⎠ 0.050 s − 0 Therefore, the magnitude of the needed retarding force is 6.4 × 103 N , or 1400 lbs. A
person cannot exert a force of this magnitude and a safety device should be used.
6.11 (a) The impulse equals the area under the F versus t graph. This area is the sum of the
area of the rectangle plus the area of the triangle. Thus, Impulse = ( 2 .0 N ) ( 3.0 s ) + ( 1
( 2.0 N ) ( 2.0 s ) = 8.0 N ⋅ s .
2 ) (b) Im pu lse = F ( ∆t ) = ∆p = m v f − v i .
8.0 N ⋅ s = (1.5 k g ) v f − 0, g iv in g v f = 5.3 m s . (c) ( v f = − 2 .0 m s + 6.12 ) Impulse = F ( ∆t ) = ∆p = m v f − v i , so v f = v i + Impulse
.
m 8.0 N ⋅ s
= 3.3 m s .
1.5 kg (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, ⎡ ( 4 .00 N ) ( 2 .00 s ) ⎤
Impulse = 2 ⎢
⎥ + ( 4 .00 N ) (1.00 s ) = 12 .0 N ⋅ s .
2
⎢
⎥
⎣
⎦ ( ) (b) Impulse = F ( ∆t ) = ∆p = m v f − v i , so v f = v i + vf =0+ (c) v f = vi + Impulse
.
m 12.0 N ⋅ s
= 6.00 m s
2.00 k g Impulse
12.0 N ⋅ s
= − 2.00 m s +
= 4.00 m s
m
2.00 kg 177 CHAPTER 6.13 6 (a) The impulse is the area under the curve between 0 and 3.0 s.
Impulse = (4.0 N)(3.0 s) = 12 N ⋅ s . This is: (b) The area under the curve between 0 and 5.0 s is:
Impulse = (4.0 N)(3.0 s) + (2.0 N)(2.0 s) = 8.0 N ⋅ s .
(c) ( ) Impulse = F ( ∆t ) = ∆p = m v f − v i , so v f = v i + Impulse
.
m at 3.0 s: (a) Impulse
12 N ⋅ s
=0+
= 8.0 m s
m
1.50 kg at 5.0 s: 6.14 v f = vi + v f = vi + Impulse
8.0 N ⋅ s
=0+
= 5.3 m s
m
1.50 kg ( ) ∆pwater m v f − v i
(1000 kg ) ( 0 − 20.0 m s ) = − 333 N
=
=
∆t
∆t
60.0 s (b) Fwater = ∆pwater
= −333 N , or
∆t 333 N d ire ct ed o p p o site t o w ater flo w (c) From Newton’s third law,
Fbuilding = − Fwat er = + 333 N , o r 6.15 (a) ∆t = (b) F = (c) 6.16 a= 333 N in d ir e ct io n o f w a t e r flo w 2 (1.20 m )
∆x 2 ( ∆x )
=
=
= 9.60 × 10−2 s
v
v f + v i 0 + 25.0 m s ∆p m ( ∆v ) (1400 kg ) ( 25.0 m s )
=
=
= 3.65 × 105 N
−2
9.60 × 10 s
∆t
∆t
⎛
⎞
25.0 m s
1g
∆v
=
= 260 m s 2 = 260 m s 2 ⎜
= 26.6 g
2⎟
−2
∆t 9.60 × 10 s
⎝ 9.80 m s ⎠ ( ) Choose the positive direction to be from the pitcher toward home plate. 178 CHAPTER (a) ( 6 ) Im pu lse = F ( ∆t ) = ∆p = m v f − v i = ( 0.15 k g ) ⎡( −22 m s ) − ( 20 m s ) ⎤
⎣
⎦ Im pulse = F ( ∆t ) = −6.3 kg ⋅ m s , or
(b) F = 6.3 k g ⋅ m s t o w a r d t h e p it ch e r Impulse −6.3 kg ⋅ m s
=
= −3.2 × 103 N ,
−3
∆t
2 .0 × 10 s or F = 3.2 × 103 N t owa r d t h e p it ch er
6.17 Choose +x in the direction of the initial velocity and +y vertically upward. Consider first
the force components exerted on the water by the roof. ( Fwater ) x =
or ( 20.0 kg ) ⎡( 40.0 m s ) cos 60.0° − 40.0 m s ⎤
∆px m ( ∆v x )
⎣
⎦,
=
=
∆t
∆t
1.00 s ( Fwater ) x = − 400 N ( Fwater ) y = ∆py
∆t = ( ) = ( 20.0 kg ) ⎡( 40.0 m s) sin 60.0° − 0⎤ = 693 N
⎣
⎦ m ∆v y Thus, Fwat er = Fx2 + Fy2 = ∆t 1.00 s ( −400 N ) 2 + ( 693 N ) 2 = 800 N ⎛ 693 N ⎞
⎛ Fy ⎞
and θ = tan −1 ⎜ ⎟ = t an −1 ⎜
⎟ = 120° , so Fwater = 800 N a t 120° .
⎝ Fx ⎠
⎝  400 N ⎠
From Newton’s third law, Froof = − Fwater = 800 N a t 60.0° ,
or Froof = 800 N a t 60.0° b elo w t h e h o r iz o n t a l t o t h e r ig h t . 179 CHAPTER 6.18 6 We shall choose southward as the positive direction.
The mass of the man is m = w
730 N
=
= 74.5 kg . Then, from conservation of
g 9.80 m s 2 momentum, we find ( mman v man + mbook v book ) f = ( mman v man + mbook v book )i (74.5 kg ) v man + (1.2 kg ) ( −5.0 or m s ) = 0 + 0 and v man = 8.1 × 10−2 m s . Therefore, the time required to travel the 5.0 m to shore is t= 6.19 ∆x
5.0 m
=
= 62 s .
v man 8.1 × 10−2 m s Requiring that total momentum be conserved gives ( mclubv club + mball v ball ) f = ( mclubv club + mballv ball )i
or
and 6.20 ( 200 g ) ( 40 m s ) + ( 46 g ) v ball = ( 200 g ) ( 55 m s ) + 0 , v ball = 65 m s . w
30 N
=
= 3.1 kg . We choose the direction of the
g 9.80 m s 2
bullet’s motion to be negative. Then, conservation of momentum gives (a) The mass of the rifle is m = (m v rifle rifle + mbullet v bullet ( ) = (m
f v rifle rifle + mbullet v bullet ) i ) or ( 3.1 kg ) v rifle + 5.0 × 10−3 kg ( −300 m s ) = 0 + 0 and v rifle = 0.49 m s .
(b) The mass of the man plus rifle is m = 730 N
= 74.5 kg . We use the same
9.80 m s 2 ⎛ 5.0 × 10−3 kg ⎞
−2
approach as in (a), to find v = ⎜
⎟ ( 300 m s ) = 2.0 × 10 m s .
⎝ 74.5 kg ⎠ 180 CHAPTER 6.21 6 The velocity of the girl relative to the ice, v gi , is v gi = v gp + v pi where v gp = v elocit y of gir l r ela tiv e t o p la n k , and
v pi = v elocity of p la n k r elat iv e to ice . Since we are given that
v gp = 1.50 m s , this becomes v gi = 1.50 m s + v pi . (1) ⎛ mg ⎞
(a) Conservation of momentum gives mg v gi + mp v pi = 0 , or v pi = − ⎜ ⎟ v gi .
⎝ mp ⎠ (2) ⎛
mg ⎞
Then, Equation (1) becomes ⎜ 1 +
⎟ v gi = 1.50 m s
mp ⎠
⎝
or v gi = 1.50 m s
= 1.15 m s .
⎛ 45.0 k g ⎞
1+ ⎜
⎝ 150 k g ⎟
⎠ (b) Then, using (2) above, ⎛ 45.0 kg ⎞
v pi = − ⎜
(1.15 m s ) = − 0.346 m s
⎝ 150 k g ⎟
⎠ or v pi = 0.346 m s d ir e ct ed o p p o sit e t o t h e g ir l ′s m o tion .
6.22 Consider the thrower first, with velocity after the throw of v t hrower . Applying
conservation of momentum yields ( 65.0 kg ) v thrower + ( 0.0450 kg ) ( 30.0
or m s ) = ( 65.0 k g + 0.0450 k g ) ( 2.50 m s ) , v thrower = 2.48 m s . Now, consider the (catcher + ball), with velocity of v cat cher after the catch. From
momentum conservation, ( 60.0 kg + 0.0450 kg ) v catcher = ( 0.0450 kg ) ( 30.0
or v cat cher = 2.25 × 10−2 m s . 181 m s ) + ( 60.0 k g ) ( 0) , CHAPTER 6.23 6 The ratio of the kinetic energy of the Earth to that of the ball is
2 2
KEE 1 mE v E ⎛ mE ⎞ ⎛ v E ⎞
,
=2
=
2
KEb 1 mbv b ⎜ mb ⎟ ⎜ v b ⎟
⎝ ⎠⎝ ⎠
2 (1) From conservation of momentum, p f = pi = 0 , giving mE v E + mbv b = 0 or m
vE
=− b.
mE
vb
2 m
KEE ⎛ mE ⎞ ⎛ mb ⎞
Equation (1) then becomes
= ⎜ ⎟ ⎜− ⎟ = b .
mE
KEb ⎝ mb ⎠ ⎝ mE ⎠
Using order of magnitude numbers, 6.24 1 kg
KEE mb
~10−25 .
~ 25
=
KEb mE 10 kg (a) Find the velocity of the amoeba after 1.0 s (i.e., after it has ejected 1.0 × 10−13 kg of
water). Using conservation of momentum, (1.0 × 10 −12 ) ( )( ) kg v f + 1.0 × 10−13 kg −1.0 × 10−4 m s = 0 + 0 v f = 1.0 × 10−5 m s . yielding The acceleration has been,
a= v f − vi
∆t = 1.0 × 10−5 m s − 0
= 1.0 × 10−5 m s 2 .
1.0 s (b) The reaction force exerted on the amoeba by the emerging jet is ( )( ) Freaction = ma = 1.0 × 10−12 kg 1.0 × 10−5 m s 2 = 1.0 × 10−17 N .
If the amoeba is to have constant velocity, the net force acting on it must be zero.
Thus, the water must exert a resistance force with magnitude given by
Freact ion − Fresist ance = 0 , or
−17
Fresis ta n ce = Freact ion = 1.0 × 10 N . 182 CHAPTER 6.25 6 From conservation of momentum, () mball ( v ball ) f + m pin v pin
or ( 7.00 kg ) (1.80 f () = mball ( v ball )i + m pin v pin ,
i m s ) + ( 2 .00 k g ) ( 3.00 m s ) = ( 7.00 k g ) ( v ball )i + 0 which gives ( v ball )i = 2.66 m s .
6.26 For each skater, the impulsemomentum theorem gives F= ∆p m ( ∆v ) ( 75.0 kg ) ( 5.00 m s )
=
=
= 3.75 × 103 N .
0.100 s
∆t
∆t Since F < 4500 N , there are no broken bones .
6.27 (a) If M is the mass of a single car, conservation of momentum gives ( 3 M ) v f = M ( 3.00 m s ) + ( 2 M ) (1.20 m s ) , or v f = 1.80 m s
(b) The kinetic energy lost is KElost = KEi − KE f , or KElost = 1
1
1
2
2
2
M ( 3.00 m s ) + ( 2 M ) (1.20 m s ) − ( 3 M ) (1.80 m s )
2
2
2 With M = 2.00 × 104 kg , this yields KElost = 2.16 × 10 4 J . 183 CHAPTER 6.28 6 Let us apply conservation of energy to the block from the time just after the bullet has
passed through until it reaches maximum height in order to find its speed V just after
the collision. 12
1
1
mv i + mgy i = mv 2 + mgy f becomes mV 2 + 0 = 0 + mgy f
f
2
2
2
or V= 2gyf = ( ) 2 9.80 m s 2 ( 0.120 m ) = 1.53 m s Now use conservation of momentum from before until just after the collision in order to
find the initial speed of the bullet, v. (7.0 × 10
from which
6.29 −3 ( ) ) kg v + 0 = (1.5 kg ) (1.53 m s ) + 7.0 × 10−3 kg ( 200 m s ) , v = 5.3 × 102 m s . Let M = mass of ball, m = mass of bullet, v = velocity of bullet, and V = the initial velocity
of the ballbullet combination. Then, using conservation of momentum from just before
to just after collision gives
⎞
v.
( M + m ) V = mv + 0 or V = ⎛
⎜
⎝ M + m⎟
⎠
m 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
= ( M + m ) V 2 + 0 o r hmax =
=
⎜
⎟v.
2
2 g 2 g ⎝ M + m⎠ With the data values provided, this becomes
2 hmax ⎛
⎞
0.030 kg
1
2
=
⎟ ( 200 m s ) = 57 m .
2⎜
2 9.80 m s ⎝ 0.15 kg + 0.030 k g ⎠ ( ) 184 CHAPTER 6.30 6 First, we will find the horizontal speed, v ix , of the block and embedded bullet just after
impact. After this instant, the blockbullet combination is a projectile, and we find the
1
time to reach the floor by use of ∆y = v iy t + ay t 2 , which becomes
2 −1.00 m = 0 + Thus, v ix = ( ) 1
−9.80 m s 2 t 2 , giving
2 t = 0.452 s. ∆x 2.00 m
=
= 4.43 m s .
t
0.452 s Now use conservation of momentum for the collision, with v b = speed of incoming
bullet: ( 8.00 × 10 −3 ) v b = 143 m s .
6.31 ( ) kg v b + 0 = 258 × 103 kg ( 4.43 m s ) , so
(about 320 mph) When Gayle jumps on the sled, conservation of momentum gives ( 50.0 kg + 5.00 kg ) v 2 = ( 50.0 kg ) ( 4.00 m s ) + 0 , or v 2 = 3.64 m s . After Gayle and the sled glide down 5.00 m, conservation of mechanical energy gives ( ) 1
1
2
( 55.0 kg ) v 32 + 0 = 2 ( 55.0 kg ) ( 3.64 m s ) + ( 55.0 kg ) 9.80 m s 2 ( 5.00 m ) ,
2
so v 3 = 10.5 m s . After her Brother jumps on, conservation of momentum yields ( 55.0 kg + 30.0 kg ) v 4 = ( 55.0 kg ) (10.50 m s ) + 0 , and v 4 = 6.82 m s . After all slide an additional 10.0 m down, conservation of mechanical energy gives the
final speed as ( ) 1
1
2
( 85.0 kg ) v 52 + 0 = 2 ( 85.0 kg ) ( 6.82 m s ) + ( 85.0 kg ) 9.80 m s 2 (10.0 m )
2
or v 5 = 15.6 m s . 185 CHAPTER 6.32 6 (a) Conservation of momentum gives mT v fT + mcv fc = mT v iT + mcv ic , or v fT = = ( mT v iT + mc v ic − v fc ) mT ( 9000 kg ) ( 20.0 m s ) + (1200 kg ) ⎡( 25.0 − 18.0)
⎣ m s⎤
⎦ 9000 kg v fT = 20.9 m s Ea st
1
1
⎡1
⎡1
⎤
2⎤
(b) KElost = KEi − KE f = ⎢ mcv i2c + mT v iT ⎥ − ⎢ mcv 2c + mT v 2T ⎥
f
f
2
2
⎣2
⎦ ⎣2
⎦ ( ) ( ) 1
2
= ⎡mc v i2c − v 2c + mT v iT − v 2 ⎤
f
fT ⎦
2⎣ ( ( ) ) 1
= ⎡(1200 kg ) ( 625 − 324) m 2 s 2 + ( 9000 kg ) ( 400 − 438.2) m 2 s 2 ⎤
⎦
2⎣
KElost = 8.68 × 103 J, w h ich becom es in t er n a l en er gy .
Note: If 20.9 m/s were used to determine the energy lost instead of 20.9333, the
answer would be very different. We keep extra digits in all intermediate answers
until the problem is complete.
6.33 First, we use conservation of mechanical energy to find the speed of the block and
embedded bullet just after impact: ( KE + PEs ) f = ( KE + PEs )i
and yields kx 2
V=
=
m+M becom es (150 1
1
( m + M ) V 2 + 0 = 0 + kx 2 ,
2
2 N m ) ( 0.800 m ) ( 0.0120 + 0.100) kg 2 = 29.3 m s Now, employ conservation of momentum to find the speed of the bullet just before
impact: mv + M ( 0) = ( m + M ) V ,
or ⎛ 0.112 kg ⎞
⎛m+ M⎞
v =⎜
( 29.3 m s ) = 273 m s .
⎟V =⎜
⎝m⎠
⎝ 0.0120 k g ⎟
⎠ 186 CHAPTER 6.34 6 (a) Using conservation of momentum, ( Σp ) after = ( Σp )before , gives
⎡( 4.0 + 10 + 3.0) k g ⎤v = ( 4.0 k g ) ( 5.0 m s ) + (10 k g ) ( 3.0 m s ) + ( 3.0 kg ) ( − 4.0 m s )
⎣
⎦
v = + 2 .2 m s , or 2 .2 m s t ow a r d t h e r igh t . Therefore,
(b) No . For example, if the 10kg and 3.0kg 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 k g ) ( − 4.0 m s ) , or v 1 = + 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 m s ) , and v = + 2 .2 m s just as in part (a).
6.35 (a) From conservation of momentum, ( 5.00 g ) v1 f + (10.0 g ) v 2 f = ( 5.00 g ) ( 20.0 cm s ) + 0 (1) Also for an elastic, headon, collision, we have v1i + v1 f = v 2i + v 2 f , which becomes 20.0 cm s + v1 f = v 2 f . (2) Solving (1) and (2) simultaneously yields v1 f = − 6.67 cm s , and v 2 f = 13.3 cm s .
(b) KEi = KE1i + KE2i = ( ) 1
2
5.00 × 10−3 kg ( 0.200 m s ) + 0 = 1.00 × 10−4 J
2 ( )( ) 2
1
1
2
KE2 f = m2 v 2 f = 10.0 × 10−3 kg 13.3 × 102 m s = 8.89 × 10−5 J , so
2
2 KE2 f
KEi = 8.89 × 10−5 J
= 0.889
1.00 × 10−4 J 187 CHAPTER 6.36 6 Using conservation of momentum gives (10.0 g ) v1 f + (15.0 g ) v 2 f = (10.0 g ) ( 20.0 cm s ) + (15.0 g ) ( −30.0 cm s ) (1) For elastic, head on collisions, v1i + v1 f = v 2i + v 2 f which becomes 20.0 cm s + v1 f = − 30.0 cm s +v 2 f . (2) Solving (1) and (2) simultaneously gives v1 f = − 40.0 cm s ,
and v 2 f = 10.0 cm s .
6.37 Conservation of momentum gives ( 25.0 g ) v1 f + (10.0 g ) v 2 f = ( 25.0 g ) ( 20.0 cm s ) + (10.0 g ) (15.0 cm s ) (1) For headon, elastic collisions, we know that v1i + v1 f = v 2i + v 2 f .
Thus, 20.0 cm s + v1 f = 15.0 cm s +v 2 f . (2) Solving (1) and (2) simultaneously yields v1 f = 17.1 cm s , and v 2 f = 22 .1 cm s . 188 CHAPTER 6.38 6 (a) The internal forces exerted by the actor do not change total momentum.
vi
m m m m 4.00 m/s 2.00 m/s
m m m m From conservation of momentum ( 4m ) v i = ( 3m ) ( 2 .00 m s ) +m ( 4.00 m s )
vi = 6 .00 m s + 4.00 m s
= 2.50 m s
4 1
1
2
2
2
(b) Wactor = K f − Ki = ⎡( 3m ) ( 2 .00 m s ) + m ( 4.00 m s ) ⎤ − ( 4m ) ( 2 .50 m s )
⎣
⎦2
2 Wactor ( 2.50 × 10
=
2 4 kg ) ⎡12.0 + 16.0 − 25.0⎤( m s )
⎣
⎦ 189 2 = 3.75 × 10 4 J CHAPTER 6.39 6 We assume equal firing speeds v and equal forces F required for the two bullets to push
wood fibers apart. These forces are directed opposite to the bullets displacements
through the fibers.
When the block is held in the vise, Wnet = KE f − KEi gives ( ) ( ) F 8.00 × 10−2 m cos 180° = 0 − or ( ) 1
7.00 × 10−3 kg v 2 ,
2 ( ) 1
7.00 × 10−3 kg v 2 = 8.00 × 10−2 m F
2 (1) When a second 7.00g bullet is fired into the block, now on a frictionless surface,
conservation of momentum yields
⎛ ×
(1.014 kg ) v f = (7.00 × 10−3 kg ) v + 0 , or v f = ⎜ 7.00 0110
⎝ 1. 4 −3 ⎞
⎟v
⎠ (2) Also, applying the workkinetic energy theorem to the second impact, F d cos 180° = ( ) 1
1
(1.014 kg ) v 2f − 2 7.00 × 10−3 kg v 2
2 (3) Substituting (2) into (3), we obtain
2 −F d = or ⎛ 7.00 × 10−3 ⎞ 2 1
1
2
−3
1.014 kg ) ⎜
(
⎟ v − 2 7.00 × 10 kg v ,
2
⎝ 1.014 ⎠ ( ) 7.00 × 10−3 ⎞
⎡1
⎤⎛
Fd = ⎢ 7.00 × 10−3 k g v 2 ⎥ ⎜ 1 −
1.014 ⎟
⎠
⎣2
⎦⎝ ( ) Finally, substituting (1) into (4) gives
⎛ 7.00 × 10−3 ⎞
, or d = 7.94 × 10−2 m = 7.94 cm
Fd = ⎡ 8.00 × 10−2 m F⎤ ⎜ 1 −
⎣
⎦⎝
1.014 ⎟
⎠ ( ) 190 (4) CHAPTER 6.40 6 First, consider conservation of momentum and write m1v1i + m2v 2i = m1v1 f + m2v 2 f
Since m1 = m2 , this becomes v1i + v 2i = v1 f + v 2 f . (1) ( ) For an elastic headon collision, we also have v 1i − v 2i = − v 1 f − v 2 f , v1i + v 2i = v1 f + v 2 f which may be written as (2) v 2 f = v1i Subtracting Equation (2) from (1) gives (3) v 1 f = v 2i Adding Equations (1) and (2) yields (4) Equations (3) and (4) show us that, under the conditions of equal mass objects striking
one another in a headon elastic collision, the two objects simply exchange velocities.
Thus, we may write the results of the various collisions as
(a) v 1 f = 0 , v 2 f = 1.50 m s (b) v 1 f = − 1.00 m s , v 2 f = 1.50 m s
(c) 6.41 v 1 f = 1.00 m s , v 2 f = 1.50 m s Choose the +xaxis to be eastward and the +yaxis northward.
(a) First, we conserve momentum in the x direction to find (185 kg ) V cos θ = ( 90 kg ) ( 5.0 ⎛ 90 ⎞
m s ) , or V co s θ = ⎜
( 5.0 m s )
⎝ 185 ⎟
⎠ (1) Conservation of momentum in the y direction gives (185 kg ) V sin θ = ( 95 kg ) ( 3.0 ⎛ 95 ⎞
m s ) , or V sin θ = ⎜
( 3.0 m s )
⎝ 185 ⎟
⎠ Divide equation (2) by (1) to obtain tan θ = ( 95) ( 3.0)
, and θ =
( 90) ( 5.0) 32° Then, either (1) or (2) gives V = 2 .88 m s , which rounds to V = 2.9 m s . 191 (2) CHAPTER 6 (b) KElost = KEi − KE f 1
2
2
2
= ⎡( 90 kg ) ( 5.0 m s ) + ( 95 kg ) ( 3.0 m s ) − (185 kg ) ( 2.88 m s ) ⎤
⎣
⎦
2
= 7.9 × 102 J converted into internal energy
6.42 Choose the +xaxis to be eastward and the +yaxis northward.
(a) Conserving momentum in the x direction gives
0 + (10.0 k g ) v 2 x = ( 8.00 k g ) (15.0 m s ) + 0 , or v 2 x = 12 .0 m s . Momentum conservation in the y direction yields ( 8.00 kg ) ( − 4.00 m s ) + (10.0 kg ) v 2y = 0 + 0 , or v 2y = 3. 20 m s. 2
2
After collision, v 2 = v 2 x + v 2 y = 154 m s = 12.4 m s ⎛ 3. 20 ⎞
⎛ v 2y ⎞
θ = tan −1 ⎜ ⎟ = t an −1 ⎜
= 14.9° . Thus, the final velocity of the 10.0kg
⎝ v 2x ⎠
⎝ 12 .0 ⎟
⎠
mass is v 2 = 12.4 m s a t 14.9° N o f E .
and (b) KE f
KElost KEi − KE f
=
= 1−
KEi
KEi
KEi ( ⎡ ( 8.00) ( −4.00) 2 + (10.0) 154
= 1− ⎢
⎢
( 8.00) (15.0) 2 + 0
⎣ ) 2 ⎤
⎥ = 0.0720
⎥
⎦ or 7.20% of the original kinetic energy is lost in the collision.
6.43 Choose the +xaxis to be eastward and the +yaxis northward.
If v i is the initial northward speed of the 3000kg car, conservation of momentum in the
y direction gives 0 + ( 3000 kg ) v i = ( 5000 kg ) ⎡( 5.22 m s ) sin 40.0°⎤ , or v i = 5.59 m s
⎣
⎦
Observe that knowledge of the initial speed of the 2000kg car was unnecessary for this
solution.
192 CHAPTER 6.44 6 We use conservation of momentum for both northward and eastward components.
For the eastward direction: M (13.0 m s ) = 2 M V f co s 55.0°
For the northward direction: M v 2i = 2 M V f sin 55.0°
Divide the northward equation by the eastward equation to find:
v 2i = (13.0 m s ) t a n 55.0° ⎡
⎛ 2.237 m i h ⎞ ⎤
= ⎢(13.0 m s ) ⎜
⎥ t a n 55.0° = 41.5 m i h
⎝ 1 m s ⎟⎥
⎠⎦
⎢
⎣
Thus, the driver of the north bound car was untruthful.
6.45 Choose the xaxis 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 ) co s 30.0° + m v 2 f co s θ , or v 2 f cos θ = 1.25 m s (1) Conservation of momentum in the y direction gives
0 = m ( 4.33 m s ) sin 30.0° + m v 2 f sin θ , or v 2 f sin θ = − 2 .16 m s Dividing (2) by (1) gives ta n θ = (2) − 2.16
= −1.73 and θ = −60.0° .
1.25 Then, either (1) or (2) gives v 2 f = 2.50 m s , so the final velocity of the second ball is v 2 f = 2 .50 m s a t 60.0° . ( 1
1
2
2
(b) KEi = m v 1i + 0 = m ( 5.00 m s ) = m 12 .5 m 2 s 2
2
2 ) 1
1
2
2
KE f = m v 1 f + m v 2 f
2
2
1
1
2
2
= m ( 4.33 m s ) + m ( 2 .50 m s ) = m 12 .5 m 2 s 2
2
2 ( Since KE f = KEi , this is an elastic collision .
193 ) CHAPTER 6.46 6 The recoil speed of the subject plus pallet after a heartbeat is V= ∆x 6.00 × 10−5 m
=
= 3.75 × 10−4 m s .
∆t
0.160 s From conservation of momentum, mv − M V = 0 + 0 , so the mass of blood leaving the
heart is ⎛ 3.75 × 10−4 m s ⎞
⎛V ⎞
m = M ⎜ ⎟ = ( 54.0 kg ) ⎜
= 4.05 × 10−2 kg = 40.5 g .
⎝v⎠
⎝ 0.500 m s ⎟
⎠ 6.47 ( Im pu lse = F ( ∆t ) = ∆p = m v f − v i ) = ( 0.400 kg ) ⎡( − 22 .0 m s ) − 15.0 m s ⎤ = −14.8 kg ⋅ m s
⎣
⎦
Impulse = 14.6 k g ⋅ m s in t h e d ir e ct io n o f t h e fin a l v elocit y o f t h e ba ll
6.48 First, we use conservation of mechanical energy to find the speed of m1 at B just before
1
2
collision. This gives m1 v 1 + 0 = 0 + m1 ghi ,
2
or 2
v1 = 2 g hi = ( ) 2 9.80 m s 2 ( 5.00 m ) = 9.90 m s . Next, we apply conservation of momentum and knowledge of elastic collisions to find
the velocity of m1 at B just after collision.
From conservation of momentum, with the second object initially at rest,
we have m1v1 f + m2v 2 f = m1v1i + 0 , or v 2 f = ( ) m1
v 1i − v 1 f .
m2 (1) For headon elastic collisions, v1 f + v1i = v 2 f + v 2i . Since v 2i = 0 in this case, this becomes v 2 f = v1 f + v1i and combining this with (1) above we obtain
⎛ m − m2 ⎞
⎛ 5.00 − 10.0 ⎞
v1 f = ⎜ 1
⎟ v 1i = ⎜ 5.00 + 10.0 ⎟ ( 9.90 m s ) = − 3.30 m s .
⎝
⎠
⎝ m1 + m2 ⎠ 194 CHAPTER 6 Finally, use conservation of mechanical energy for m1 after the collision to find the
1
2
maximum rebound height. This gives 0 + m1 ghmax = m1v 1 f + 0 ,
2
or 6.49 hmax = 2
v1 f 2g ( − 3.30
= ( m s) 2 9.80 m s 2 2 ) = 0.556 m . Choose the positive direction to be the direction of the truck’s initial velocity.
Apply conservation of momentum to find the velocity of the combined vehicles after
collision: ( 4000 kg + 800 kg ) V = ( 4000 kg ) ( +8.00 m s ) + ( 800 k g ) ( −8.00 m s ) , which yields V = + 5.33 m s . ( ) Use the impulsemomentum theorem, Im pu lse = F ( ∆t ) = ∆p = m v f − v i , to find the
magnitude of the average force exerted on each driver during the collision.
Truck Driver: F= m v f − vi truck ∆t = Car Driver: F= 6.50 m v f − vi
∆t car = ( 80.0 kg ) 5.33 m s − 8.00 m s 0.120 s ( 80.0 kg ) 5.33 m = 1.78 × 103 N s − ( −8.00 m s ) 0.120 s = 8.89 × 103 N If the pendulum bob barely swings through a complete circle, it arrives at the top of the
arc (having risen a vertical distance of 2 l ) with essentially zero velocity.
From conservation of mechanical energy, we find the minimum velocity of the bob at
1
= KE + PEg , or M V 2 = 0 + M g ( 2 l ) . This
the bottom of the arc as KE + PEg
bottom
t op
2
gives V = 2 g l as the needed velocity of the bob just after the collision. ( ) ( ) Conserving momentum through the collision then gives the minimum initial velocity of
the bullet as ( ⎛v⎞
m⎜ ⎟ + M 2 gl
⎝ 2⎠ ) = mv + 0 , or v =
195 4M
m gl . CHAPTER 6.51 6 Note that the initial velocity of the target particle is zero (i.e., v 2i = 0 ).
From conservation of momentum, m1v1 f + m2v 2 f = m1v1i + 0 . (1) For headon elastic collisions, v1 f + v 1i = v 2 f + 0 . (2) Solving (1) and (2) simultaneously yields the final velocities as ⎛ m − m2 ⎞
⎛ 2 m1 ⎞
v1 f = ⎜ 1
⎟ v 1i and v 2 f = ⎜ m + m ⎟ v1i
⎝ m1 + m2 ⎠
⎝1
2⎠
(a) If m1 = 2 .0 g , m 2 = 1.0 g , a n d v 1i = 8.0 m s , then v1 f = 8
32
m s and v 2 f =
ms.
3
3 (b) If m1 = 2 .0 g, m 2 = 10 g , a n d v 1i = 8.0 m s , we find v1 f = − 16
8
m s and v 2 f =
ms.
3
3 (c) The final kinetic energy of the 2.0 g particle in each case is:
2 1
1
⎛8
⎞
2
Case (a): KE1 f = m1v 1 f = 2.0 × 10−3 kg ⎜ m s⎟ = 7.1 × 10−3 J
⎝3
⎠
2
2 ( ) 2 1
1
⎛ 16
⎞
2
m s⎟ = 2.8 × 10−2 J
Case (b): KE1 f = m1v 1 f = 2.0 × 10−3 kg ⎜ −
⎝3
⎠
2
2 ( ) Since the incident kinetic energy is the same in cases (a) and (b), we observe that
the incident particle loses more kinetic energy in case (a) 196 CHAPTER 6.52 ( 6 Use conservation of mechanical energy, KE + PEg ) = ( KE + PE )
g B A the bead at point B just before it collides with the ball. This gives
or v 1i = 2 g yA = ( ) , to find the speed of 1
2
m v 1i + 0 = 0 + m g y A ,
2 2 9.80 m s 2 (1.50 m ) = 5.42 m s . Conservation of momentum during the collision gives ( 0.400 kg ) v1 f + ( 0.600 kg ) v 2 f = ( 0.400 kg ) ( 5.42 m s ) + 0 ,
or v1 f + 1.50 v 2 f = 5.42 m s . (1) For a headon elastic collision, we have v 2 f + v 2i = v1 f + v1i , which gives v 2 f = v 1 f + 5.42 m s . (2) Solving (1) and (2) simultaneously, the velocities just after collision are v1 f = − 1.08 m s and v 2 f = 4.34 m s .
Now, we use conservation of the mechanical energy of the ball after collision to find the
maximum height the ball will reach. This gives
2
v2 f
( 4.34 m s ) = 0.960 m
1
2
M v 2 f + 0 , or y max =
=
2
2 g 2 ( 9.80 m s 2 )
2 0 + M g y max = 197 CHAPTER 6.53 6 We first find the speed of the diver when he reaches the water by using
v 2 = v i2 + 2 ay ( ∆y ) . This becomes ( Fwater
80 kg ) v 2 = 0 + 2 −9.80 m s 2 ( −3.0 m ) , and yields v = − 59 m s w = 784 N The negative sign indicates the downward direction.
Next, we use the impulsemomentum theorem to find the resistive force exerted by the
water as the diver comes to rest. ( ) Im pu lse = Fnet ( ∆t ) = ∆p = m v f − v i , or ( Fwater − 784 N ) ( 2.0 s ) = ( 80 kg ) ⎡0 − ( −
⎣ ) 59 m s ⎤ , which gives
⎦ ⎛ 80 59 ⎞
Fwat er = 784 N + ⎜
N = 1.1 × 103 N
2⎟
⎝
⎠ ( u pw a r d ) . 6.54
7.5 m at rest
12.0 g v 112 100 g Immediately Before Impact g V at rest 112 g At the end Immediately After Impact Using the workkinetic energy theorem from immediately after impact to the end gives: Wnet = Ffriction s cos 180° = KEend − KEafter ,
or, − ⎡µk ( M + m ) g ⎤s = 0 −
⎣
⎦ 1
( M + m ) V 2 and V =
2 2 µk g s . Then, using conservation of momentum from immediately before to immediately after
impact gives mv + 0 = ( M + m ) V , or ⎛ M + m⎞
⎛ M + m⎞
v =⎜
V =⎜
⎝m⎟
⎠
⎝m⎟
⎠ ⎛ 112 g ⎞
2 µk g s = ⎜
⎝ 12 .0 g ⎟
⎠ v = 91 m s 198 ( ) 2 ( 0.650) 9.80 m s 2 ( 7.5 m ) CHAPTER 6.55 6 (a) Using conservation of momentum, ( 60.0 kg + 120 kg ) v f = ( 60.0 kg ) ( 4.00
v f = 1. 33 m s or (b) ( m s) + 0 ,
4.00 m/s
60.00 kg ) ΣFy = n − ( 60.0 kg ) 9.80 m s 2 = 0
120 kg so the normal force is n = 588 N
and fk = µk n = ( 0.400) ( 588 N ) = 235 N
(c) Apply the impulsemomentum theorem to the person: ( Im pu lse = Fnet ( ∆t ) = ∆p = m v f − v i so ∆t = ( m v f − vi
− fk ) ) = ( 60.0 kg ) (1. 33 m s − 4.00 m s) =
−235 N ( 0.681 s ) (d) ∆pperson = m v f − v i = ( 60.0 k g ) (1. 33 m s − 4.00 m s ) = − 160 N ⋅ s ( ) ∆pcart = M v f − 0 = (120 k g ) (1. 33 m s − 0) = + 160 N ⋅ s (e) ⎛ v f + vi ⎞
∆x person = v ( ∆t ) = ⎜
( ∆t )
⎝2⎟
⎠ ⎛ 1. 33 m s + 4.00 m s ⎞
=⎜
⎟ ( 0.681 s ) = 1.82 m
⎝
⎠
2
(f) ⎛ v f + 0⎞
⎛ 1. 33 m s ⎞
∆x cart = v ( ∆t ) = ⎜
⎟ ( 0.681 s ) = 0.454 m
⎟ ( ∆t ) = ⎜
⎝
⎠
2
⎝2⎠ ( 1
(g) ∆KEperson = m v 2 − v i2
f
2
= (h) ∆KEcart = ) ( 60.0 kg ) ⎡ 1. 33 m s 2 − 4.00 m s 2 ⎤ =
)(
)⎦
⎣(
2 − 427 J (120 kg ) ⎡ 1. 33 m s 2 − 0⎤ = 107 J
1
M v2 − 0 =
)⎦
f
⎣(
2
2 ( ) 199 CHAPTER (i) 6.56 6 Equal friction forces act through different distances on person and cart to do
different amounts of work on them. This is a perfectly inelastic collision in which
the total work on both person and cart together is –320 J, which becomes +320 J of
internal energy. (a) Let v 1i and v 2i be the velocities of m1 and m2 just before the collision. Then
conservation of energy gives: v 1i = − v 2i = 2gh = ( ) 2 9.80 m s 2 ( 5.00 m ) = 9.90 m s . (b) From conservation of momentum: ( 2 .00) v 1 f + ( 4.00) v 2 f = ( 2 .00) ( 9.90 m s ) + ( 4.00) ( −9.90 m s ) , or ( 2 .00) v 1 f + ( 4.00) v 2 f = −19.8 m s . (1) For an elastic headon collision, v1 f + v1i = v 2 f + v 2i , giving v1 f + 9.90 m s = v 2 f − 9.90 m s , or v 2 f = v1 f + 19.8 m s
Solving (1) and (2) simultaneously gives v 1 f = − 16.5 m s , and v 2 f = 3.30 m s .
(c) Applying conservation of energy to each block after the collision gives: h1 f = and h2 f = 2
v1 f 2g
2
v2 f 2g ( −16.5 m s )2
= ( 2 9.80 m s 2 = ( 3.30 m s )2 ( 2 9.80 m s 2 ) = 13.9 m ) = 0.556 m 200 (2) CHAPTER 6.57 6 (a) Use conservation of mechanical energy to find the speed of m1 just before collision.
This gives v 1i = 2 g h1 = ( ) 2 9.80 m s 2 ( 2.50 m ) = 7.00 m s . Apply conservation of momentum from just before to just after the collision: ( 0.500 kg ) v1 f + (1.00 kg ) v 2 f = ( 0.500 kg ) (7.00 m s) + 0 , v1 f + 2 v 2 f = 7.00 m s or (1) For a headon elastic collision, v1 f + v1i = v 2 f + v 2i ,
which becomes v1 f − v 2 f = − 7.00 m s . (2) Solving (1) and (2) simultaneously yields v1 f = − 2.33 m s , and v 2 f = 4.67 m s .
(b) Apply conservation of mechanical energy to m1 after the collision to find h1 =
′ 2
v1 f 2g ( − 2.33 m s )2 =
= ( 2 9.80 m s 2 ) 0.277 m . (rebound height) 1
(c) From ∆y = v iy t + ay t 2 , with v iy = 0 , the time for m2 to reach the floor after it flies
2
horizontally off the table is found to be
t= 2 ( ∆y )
ay = 2 ( −2 .00 m )
9.80 m s 2 = 0.639 s . The horizontal distance traveled in this time is
∆x = v ix t = ( 4.67 m s ) ( 0.639 s ) = 2.98 m . (d) After the 0.500 kg mass comes back down the incline, it flies off the table with a
horizontal velocity of 2.33 m/s. The time of the flight to the floor is 0.639 s as found
above and the horizontal distance traveled is
∆x = v ix t = ( 2 .33 m s ) ( 0.639 s ) = 1.49 m . 201 CHAPTER 6.58 6 Use conservation of mechanical energy to find the velocity, v, of Tarzan just as he
reaches Jane. This gives v = 2 g hi = ( ) 2 9.80 m s 2 ( 3.00 m ) = 7.67 m s . Now, use conservation of momentum to find the velocity, V, of Tarzan + Jane just after
the collision. This becomes ( M + m ) V = M v + 0 , or ⎛ 80.0 kg ⎞
⎛M⎞
V =⎜
v =⎜
(7.67 m s ) = 4.38 m s .
⎝ M + m⎟
⎠
⎝ 140 kg ⎟
⎠
Finally, use conservation of mechanical energy from just after he picks her up to the end
of their swing to determine the maximum height, H, reached. This yields ( 4.38 m s ) = 0.980 m
V2
=
2 g 2 ( 9.80 m s 2 )
2 H= 6.59 (a) The momentum of the system is initially zero and remains constant throughout the
motion. Therefore, when m1 leaves the wedge, we must have m2v wedge + m1v block = 0 ,
or ⎛m ⎞
⎛ 0.500 ⎞
v wedge = − ⎜ 1 ⎟ v block = − ⎜
( 4.00 m s ) = − 0.667 m s .
⎝ 3.00 ⎟
⎠
⎝ m2 ⎠
(b) Using conservation of energy as the block slides down the wedge, we have ( KE + PE ) = ( KE + PE )
g i g f or 1
1
2
2
0 + m1 gh = m1v block + m2 v wedge + 0 .
2
2
Thus, h = = ⎛ m2 ⎞ 2 ⎤
1⎡2
⎢v block + ⎜ ⎟ v wedge ⎥
2g ⎣
⎝ m1 ⎠
⎦
1
2
2⎤
⎡
⎛ 3.00 ⎞
4.00 m s ) + ⎜
⎟ ( −0.667 m s ) ⎥ = 0.952 m .
2 ⎢(
⎝ 0.500 ⎠
19.6 m s ⎣
⎦ 202 CHAPTER 6.60 6 (a) Let m be the mass of each cart. Then, if v 0 is the initial velocity of the red cart,
applying conservation of momentum to the collision gives
m v b + m v r = m v 0 + 0 , or v b + v r = v 0 (1) where v b and v r are the velocities of the blue and red carts after collision.
In a headon elastic collision, we have v 2 f + v 2i = v1 f + v1i which reduces to
vb − vr = v0 . (2) Solving (1) and (2) simultaneously gives v r = 0 , and v b = 3.00 m s .
(b) Using conservation of mechanical energy for the blue cartspring system,
( KE + PEs ) f = ( KE + PEs )i becomes 1
1
2
0 + k x 2 = m vb + 0
2
2
or 6.61 x= 0.250 kg
m
vb =
( 3.00 m s ) = 0.212 m .
k
50.0 N m (a) Use conservation of the component of
momentum in the horizontal direction
from just before to just after the cannon
firing. ( Σpx ) f = ( Σpx )i gives m shell ( v shell co s 45.0°) + m cannon v recoil = 0 , or ⎛m
⎞
v recoil = − ⎜ shell ⎟ v shell co s 45.0°
⎝ mcannon ⎠
⎛ 200 kg ⎞
= −⎜
(125 m s ) cos 45.0° = − 3.54 m s
⎝ 5000 kg ⎟
⎠ 203 45.0° CHAPTER 6 (b) Use conservation of mechanical energy for the cannonspring system from right
after the cannon is fired to the instant when the cannon comes to rest. ( KE + PE g + PEs ) = ( KE + PE g f + PEs ) i 12
1
0 + 0 + kx max = mcannon v r2ecoil + 0 + 0
2
2 x max (c) mcannon v r2ecoil
=
=
k ( 5000 kg ) ( 3.54 m s )
2 .00 × 10 4 N m ( 2 = 1.77 m ) Fmax = k x max = 2 .00 × 104 N m (1.77 m ) = 3.54 × 104 N (d) No. The rail exerts a vertical external force (the normal force) on the cannon and
prevents it from recoiling vertically. Momentum is not conserved in the vertical
direction. The spring does not have time to stretch during the cannon firing. Thus,
no external horizontal force is exerted on the system (cannon plus shell) from just
before to just after firing. Momentum is conserved in the horizontal direction
during this interval.
6.62 Conservation of the xcomponent of momentum gives 2
3 ( 3m) v 2 x + 0 = −mv 0 + ( 3m) v 0 , or v 2 x = v 0 . (1) Likewise, conservation of the ycomponent of momentum gives −m v 1y + ( 3m ) v 2 y = 0 , and v1y = 3 v 2 y . (2) Since the collision is elastic, ( KE ) f = ( KE )i , or ( ) 1
1
1
1
2
2
2
2
2
m v 1y + ( 3m ) v 2 x + v 2 y = m v 0 + ( 3m ) v 0
2
2
2
2
Substituting (1) and (2) into (3) yields
2
⎛4 2
2
2⎞
2
.
9 v 2 y + 3 ⎜ v 0 + v 2 y ⎟ = 4 v 0 , or v 2 y = v 0
⎝9
⎠
3 204 (3) CHAPTER 6 (a) The particle of mass m has final speed v1y = 3 v 2 y = v 0 2, and the particle of mass 3m moves at
2
2
v 2 = v 2x + v 2y = 2
42 22
v0 + v0 = v0
.
3
9
9 ⎛ v 2y ⎞
−1 ⎛ 1 ⎞
(b) θ = tan −1 ⎜
⎟ = t an ⎜ 2 ⎟ = 35.3° .
⎝
⎠
⎝ v 2x ⎠
6.63 Let particle 1 be the neutron and particle 2 be the carbon nucleus. Then, we are given
that m 2 = 12 m1 .
(a) From conservation of momentum m2v 2 f + m1v1 f = m1v1i + 0 . Since m 2 = 12 m1 , this
reduces to 12v 2 f + v1 f = v1i . (1) For a headon elastic collision, v 2 f + v 2i = v1 f + v1i .
Since v 2i = 0 , this becomes v 2 f − v1 f = v1i . (2) Solve (1) and (2) simultaneously to find v1 f = − 11
2
v 1i , and v 2 f = v 1i .
13
13 1
2
The initial kinetic energy of the neutron is KE1i = m1v 1i , and the final kinetic energy
2
of the carbon nucleus is
1
1
48
⎛ 4 2 ⎞ 48 ⎛ 1
2
2⎞
KE2 f = m2 v 2 f = (12 m1 ) ⎜
v 1i ⎟ =
KE .
⎜ m1v 1i ⎟ =
⎝ 169 ⎠ 169 ⎝ 2
⎠ 169 1i
2
2 The fraction of kinetic energy transferred is KE2 f
KE1i = 48
= 0.28 .
169 (b) If KE1i = 1.6 × 10−13 J , then KE2 f = ( ) 48
48
KE1i =
1.6 × 10−13 J = 4.5 × 10−14 J .
169
169 The remaining energy 1.6 × 10−13 J − 4.5 × 10−14 J = 1.1 × 10−13 J stays with the
neutron.
205 CHAPTER 6.64 6 Choose the positive xaxis in the direction of the initial velocity of the cue ball. Let v i be
the initial speed of the cue ball, v c be the final speed of the cue ball, v T be the final
speed of the target, and θ be the angle the target’s final velocity makes with the xaxis.
Conservation of momentum in the xdirection gives
m v T co s θ + m v c co s 30.0° = 0 + m v i , or v T cos θ = v i − v c co s 30.0° (1) From conservation of momentum in the ydirection,
m v T sin θ − m v c sin 30.0° = 0 + 0 , or v T sin θ = v c sin 30.0° (2) Since this is an elastic collision, kinetic energy is conserved, giving 1
1
1
2
2
2
2
m v T + m v c = m v i2 , or v T = v i2 − v c
2
2
2
(b) To solve, square equations (1) and (2). Then add the results to obtain
2
2
v T = v i2 − 2 v i v c cos 30.0° + v c . Substitute this into equation (3) and simplify to find
v c = v i co s 30.0° = ( 4.00 m s ) co s 30.0° = 3.46 m s .
2
Then, equation (3) yields v T = v i2 − v c , or vT = ( 4.00 m s ) − ( 3.46 m s ) = 2.00 m s .
2 2 (a) With the results found above, equation (2) gives ⎛v ⎞
⎛ 3.46 ⎞
sin θ = ⎜ c ⎟ sin 30.0° = ⎜
sin 30.0° = 0.866 , or θ = 60.0° .
⎝ 2.00 ⎟
⎠
⎝ vT ⎠
Thus, the angle between the velocity vectors after collision is φ = 60.0° + 30.0° = 90.0° . 206 (3) CHAPTER 6.65 The deceleration of the incident block is a = − 6 µk ( mg )
fk
=−
= − µk g .
m
m Therefore, v 2 = v i2 + 2 a( ∆x ) gives the speed of the incident block just before collision as
2
v = v 0 − 2 µk g d . Conservation of momentum from just before to just after collision gives m v 1 + ( 2m ) v 2 = mv , or 2 v 2 + v 1 = v . (1) where v1 and v2 are the speeds of the two blocks just after collision.
Since this is a headon elastic collision, v 2 f + v 2i = v1 f + v1i ,
v 2 − v1 = v . which becomes (2) Adding equations (1) and (2) yields v 2 = 2
22
v=
v 0 − 2 µk g d .
3
3 Note that the mass canceled in the calculation of the deceleration above. Thus, the
second block will have the same deceleration after collision as the incident block had
before. Then, v 2 = v i2 + 2 a( ∆x ) with v f = 0 gives the stopping distance for the second
f 2
block as 0 = v 2 + 2 ( − µk g ) D , or D= 2
v2 2 µk g = (v
9µ g
2 k 2
0 ) − 2 µk g d = 2
2 v0
4d
−
.
9µk g 9 207 CHAPTER 6 Answers to Even Numbered Conceptual Questions
2. Only if the collision is perfectly headon. If the two objects collide even slightly off center,
a glancing collision will occur and the final velocities will be along lines other than that of
the initial motion. 4. No. Only in a precise headon 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. 6. The skater gains the most momentum by catching and then throwing the frisbee. 8. Kinetic energy can be written as p2
. Thus, even through the particles have the same
2m
kinetic energies their momenta may be different due to a difference in mass. 10. The resulting collision is intermediate between an elastic and a completely inelastic
collision. Some energy of motion is transformed as the pieces buckle, crumple, and heat
up during the collision. Also, a small amount is lost as sound. The most kinetic energy is
lost in a headon collision, so the expectation of damage to the passengers is greatest. 12. The less massive object loses the most kinetic energy in the collision. 14. The superhero is at rest before the toss and the net momentum of the system is zero. When
he tosses the piano, say toward the right, something must get an equal amount of
momentum to the left to keep the momentum at zero. This something recoiling to the left
must be Superman. He cannot stay at rest. 16. The passenger must undergo a certain momentum change in the collision. This means that
a certain impulse must be exerted on the passenger by the steering wheel, the window, an
air bag, or something. By increasing the time during which this momentum change occurs,
the resulting force on the passenger can be decreased. 18. A certain impulse is required to stop the egg. But, if the time during which the momentum
change of the egg occurs is increased, the resulting force on the egg is reduced. The time is
increased as the sheet billows out as the egg is brought to a stop. The force is reduced low
enough so that the egg will not break. 208 CHAPTER 6 Answers to Even Numbered Problems
2. (a) 5.40 N·s (b) –27.0 J 4. (a) 0 (b) 1.1 k g ⋅ m s 6. 1.7 k N 8. 7.00 × 103 N u p w a r d 10. An average force of 6.4 × 103 N ( ≈ 1400 lbs) would be required to hold the child. 12. (a) 12.0 N·s (b) 6.00 m s 14. (a) − 333 N (b) 333 N directed opposite to water flow (c) 333 N in direction of water flow (a) 6.3 k g ⋅ m s toward the pitcher (b) 3.2 × 103 N toward the pitcher (b) 2.0 × 102 m s 16. 18. 62 s 20. (a) 22. v thrower = 2 .48 m s , v cat cher = 2.25 × 10 −2 m s 24. (a) 1.0 × 105 m s 2 (b) 1.0 × 1017 N 0.49 m s 26. F = 3.75 × 103 N , no broken bones 28. 5.3 × 102 m s 30. 143 m s 32. (a) 20.9 m s Ea st (b) 34. (a) 2 .2 m s toward the right (b) No 36. − 40.0 cm s (10.0g object), + 10.0 cm s (15.0g object) 38. (a) 2 .50 m s (b) 209 8.68 × 103 J into internal energy 3.75 × 104 J (c) 4.00 m s CHAPTER 40. (a) 0, 1.50 m s (b) (c)
(a) 12 .4 m s at 14.9° N of E 44. No, his speed was 41.5 m i h . 46. 40.5 g 48. 0.556 m 50. ⎛4M ⎞
v min = ⎜
⎟
⎝m⎠ 52. 0.960 m above the level of point B 54. 91 m s 56. (a) 9.90 m s , − 9.90 m s (c) − 1.00 m s , 1.50 m s 1.00 m s , 1.50 m s 42. 6 13.9 m, 0.556 m (b) 7.20 % gl (b) 58. (a) v red = 0, v blue = 3.00 m s 62. (a) vm = v0 64. (a) 90.0° (b) 0.212 m (b) 35.3° (b) 3.46 m s ( cu e ba ll ) , 2.00 m s ( t a r g e t ) 0.980 m 60. − 16.5 m s , 3.30 m s 2 , v 3m = v 0 2
3 210 211 ...
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 Spring '05
 Dr.Ha
 Physics, Mass

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