momentum and collisions -...
TRANSCRIPT
217
Momentum and Collisions
SOLUTIONS TO PROBLEMS Section 8.1 Linear Momentum and Its Conservation P8.1 m = 3.00 kg ,
rv = 3.00i ! 4.00 j( ) m s
(a)
rp = mrv = 9.00i ! 12.0 j( ) kg "m s
Thus, px = 9.00 kg !m s
and
py = !12.0 kg "m s .
(b) p = px2 + py
2 = 9.00( )2 + 12.0( )2 = 15.0 kg !m s
! = tan"1 py
px
#
$%&
'(= tan"1 "1.33( ) = 307°
P8.4 (a) The momentum is p = mv , so v = p
m and the kinetic energy is
K =
12
mv2 =12
m pm
!"#
$%&
2=
p2
2m.
(b) K =
12
mv2 implies v = 2K
m, so
p = mv = m 2K
m= 2mK .
Section 8.2 Impulse and Momentum
*P8.6 From the impulse-momentum theorem, F !t( ) = !p = mv f " mvi , the average force required to hold onto the child is
F =
m v f ! vi( )"t( )
=12 kg( ) 0 ! 60 mi h( )
0.050 s ! 01 m s
2.237 mi h#$%
&'(= !6.44 ) 103 N .
Therefore, the magnitude of the needed retarding force is 6.44 ! 103 N , or 1 400 lb. A person cannot exert a force of this magnitude and a safety device should be used.
218 Momentum and Collisions
P8.7 (a) I = Fdt! = area under curve
I = 1
21.50 ! 10"3 s( ) 18 000 N( ) = 13.5 N #s
(b) F =
13.5 N !s1.50 " 10#3 s
= 9.00 kN
(c) From the graph, we see that
Fmax = 18.0 kN
FIG. P8.7
P8.9
!rp =
rF!t
!py = m v fy " viy( ) = m v cos 60.0°( ) " mv cos 60.0° = 0
!px = m "v sin 60.0° " v sin 60.0°( ) = "2mv sin 60.0°= "2 3.00 kg( ) 10.0 m s( ) 0.866( )= "52.0 kg #m s
Fave =!px
!t="52.0 kg #m s
0.200 s= "260 N
FIG. P8.9
Section 8.3 Collisions P8.13 (a) mv1i + 3mv2i = 4mv f where m = 2.50 ! 104 kg
v f =
4.00 + 3 2.00( )4
= 2.50 m s
(b) K f ! Ki =
12
4m( )v f2 !
12
mv1i2 +
12
3m( )v2i2"
#$%&'= 2.50 ( 104( ) 12.5 ! 8.00 ! 6.00( ) = !3.75 ( 104 J
P8.14 (a) The internal forces exerted by the actor do
not change the total momentum of the system of the four cars and the movie actor
4m( )vi = 3m( ) 2.00 m s( ) + m 4.00 m s( )
vi =6.00 m s + 4.00 m s
4= 2.50 m s
FIG. P8.14
(b) Wactor = K f ! Ki =
12
3m( ) 2.00 m s( )2 + m 4.00 m s( )2"#
$% !
12
4 m( ) 2.50 m s( )2
Wactor =
2.50 ! 104 kg( )2
12.0 + 16.0 " 25.0( ) m s( )2 = 37.5 kJ
(c)
The event considered here is the time reversal of the perfectly inelastic collision in theprevious problem. The same momentum conservation equation describes both processes.
Chapter 8 219
P8.16 v1 , speed of m1 at B before collision.
12
m1v12 = m1gh
v1 = 2 9.80( ) 5.00( ) = 9.90 m s
v1 f , speed of m1 at B just after collision.
v1 f =
m1 ! m2m1 + m2
v1 = !13
9.90( ) m s = !3.30 m s
At the highest point (after collision)
FIG. P8.16
m1ghmax =
12
m1 !3.30( )2 hmax =
!3.30 m s( )2
2 9.80 m s2( )= 0.556 m
P8.19 (a) According to the Example in the chapter text, the fraction of total kinetic energy transferred
to the moderator is
f2 =
4m1m2
m1 + m2( )2
where m2 is the moderator nucleus and in this case, m2 = 12m1
f2 =
4m1 12m1( )13m1( )2
=48
169= 0.284 or 28.4%
of the neutron energy is transferred to the carbon nucleus.
(b) KC = 0.284( ) 1.6 ! 10"13 J( ) = 4.54 ! 10"14 J
Kn = 0.716( ) 1.6 ! 10"13 J( ) = 1.15 ! 10"13 J
220 Momentum and Collisions
P8.20 We assume equal firing speeds v and equal forces F required for the two bullets to push wood fibers apart. These equal forces act backward on the two bullets.
For the first, Ki + !Emech = K f 12
7.00 ! 10"3 kg( )v2 " F 8.00 ! 10"2 m( ) = 0
For the second, pi = pf 7.00 ! 10"3 kg( )v = 1.014 kg( )v f
v f =
7.00 ! 10"3( )v1.014
Again, Ki + !Emech = K f : 12
7.00 ! 10"3 kg( )v2 " Fd = 12
1.014 kg( )v f2
Substituting for v f ,
12
7.00 ! 10"3 kg( )v2 " Fd = 12
1.014 kg( ) 7.00 ! 10"3 v1.014
#
$%&
'(
2
Fd = 1
27.00 ! 10"3( )v2 "
12
7.00 ! 10"3( )21.014
v2
Substituting for v, Fd = F 8.00 ! 10"2 m( ) 1 " 7.00 ! 10"3
1.014#
$%&
'( d = 7.94 cm
P8.21 At impact, momentum of the clay-block system is conserved, so:
mv1 = m1 + m2( )v2 After impact, the change in kinetic energy of the clay-block-surface
system is equal to the increase in internal energy:
12
m1 + m2( )v22 = f f d = µ m1 + m2( ) gd
12
0.112 kg( )v22 = 0.650 0.112 kg( ) 9.80 m s2( ) 7.50 m( )
v22 = 95.6 m2 s2 v2 = 9.77 m s
12.0 ! 10"3 kg( )v1 = 0.112 kg( ) 9.77 m s( ) v1 = 91.2 m s
FIG. P8.21
Section 8.4 Two-Dimensional Collisions P8.27 By conservation of momentum for the system of the two billiard
balls (with all masses equal),
5.00 m s + 0 = 4.33 m s( )cos 30.0° + v2 fx
v2 fx = 1.25 m s
0 = 4.33 m s( )sin 30.0° + v2 fy
v2 fy = !2.16 m srv2 f = 2.50 m s at ! 60.0°
FIG. P8.27 Note that we did not need to use the fact that the collision is perfectly elastic.
Chapter 8 221 P8.29
m1rv1i + m2
rv2i = m1 + m2( ) rv f : 3.00 5.00( ) i ! 6.00 j = 5.00rv
rv = 3.00i ! 1.20 j( ) m s
(b) E =
12
m1v12 +
12
m2v22 +
12
m3v32
E =12
5.00 ! 10"27( ) 6.00 ! 106( )2 + 8.40 ! 10"27( ) 4.00 ! 106( )2 + 3.60 ! 10"27( ) 12.5 ! 106( )2#$%
&'(
E = 4.39 ! 10"13 J
Section 8.5 The Center of Mass P8.33 Let A1 represent the area of the bottom row of squares,
A2 the middle square, and A3 the top pair.
A = A1 + A2 + A3M = M1 + M2 + M3M1A1
=MA
A1 = 300 cm2 , A2 = 100 cm2 , A3 = 200 cm2 , A = 600 cm2
M1 = M A1A
!"#
$%&=
300 cm2
600 cm2 M =M2
M2 = M A2A
!"#
$%&=
100 cm2
600 cm2 M =M6
M3 = M A3A
!"#
$%&=
200 cm2
600 cm2 M =M3
FIG. P8.33
xCM =x1M1 + x2M2 + x3M3
M=
15.0 cm 12 M( ) + 5.00 cm 1
6 M( ) + 10.0 cm 13 M( )
MxCM = 11.7 cm
yCM =12 M 5.00 cm( ) + 1
6 M 15.0 cm( ) + 13 M( ) 25.0 cm( )
M= 13.3 cm
yCM = 13.3 cm
222 Momentum and Collisions
Section 8.6 Motion of a System of Particles P8.37 (a)
rvCM =mi
rv i!M
=m1
rv1 + m2rv2
M
=2.00 kg( ) 2.00i m s " 3.00 j m s( ) + 3.00 kg( ) 1.00i m s + 6.00 j m s( )
5.00 kg
rvCM = 1.40i + 2.40 j( ) m s
(b) rp = MrvCM = 5.00 kg( ) 1.40i + 2.40 j( ) m s = 7.00i + 12.0 j( ) kg !m s
Additional Problems
* P8.48 Using conservation of momentum from just before to just after the impact of the bullet with the block:
mvi = M + m( )v f
or vi =
M + mm
!"#
$%&
v f . (1) The speed of the block and embedded bullet just after
impact may be found using kinematic equations:
d = v f t and h = 1
2gt2 .
Thus, t = 2h
g and
v f =
dt= d g
2h=
gd2
2h.
FIG. P8.48
Substituting into (1) from above gives vi =
M + mm
!"#
$%&
gd2
2h.
Chapter 8 223 P8.51 (a) The initial momentum of the system is zero, which
remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have
m2vwedge + m1vblock = 0
or 3.00 kg( )vwedge + 0.500 kg( ) +4.00 m s( ) = 0
so vwedge = !0.667 m s (b) Using conservation of energy for the block-wedge-
Earth system as the block slides down the smooth (frictionless) wedge, we have
FIG. P8.51
Kblock +Usystem!" #$i+ Kwedge!" #$i
= Kblock +Usystem!" #$ f+ Kwedge!" #$ f
or 0 + m1gh[ ] + 0 = 1
2m1 4.00( )2 + 0!
"#$%&+
12
m2 '0.667( )2 which gives h = 0.952 m .
P8.59 A picture one second later differs by showing five extra kilograms of sand moving on the belt.
(a) !px
!t=
5.00 kg( ) 0.750 m s( )1.00 s
= 3.75 N
(b) The only horizontal force on the sand is belt friction,
so from pxi + f !t = pxf this is
f =
!px
!t= 3.75 N
(c) The belt is in equilibrium:
Fx! = max : +Fext ! f = 0 and Fext = 3.75 N
(d)
W = F!r cos" = 3.75 N 0.750 m( )cos0° = 2.81 J
(e) 12!m( )v2 =
12
5.00 kg 0.750 m s( )2 = 1.41 J
(f)
Friction between sand and belt converts half of the input work into extra internal energy.