217 Momentum and Collisions SOLUTIONS TO PROBLEMS Section 8.1 Linear Momentum and Its Conservation P8.1 m = 3.00 kg , r v = 3.00 ˆ i ! 4.00 ˆ j ( ) ms (a) r p = m r v = 9.00 ˆ i ! 12.0 ˆ j ( ) kg " ms Thus, p x = 9.00 kg ! ms and p y = !12.0 kg " ms . (b) p = p x 2 + p y 2 = 9.00 ( ) 2 + 12.0 ( ) 2 = 15.0 kg ! ms ! = tan "1 p y p x # $ % & ' ( = tan "1 "1.33 ( ) = 307° P8.4 (a) The momentum is p = mv , so v = p m and the kinetic energy is K = 1 2 mv 2 = 1 2 m p m ! " # $ % & 2 = p 2 2m . (b) K = 1 2 mv 2 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 " mv i , the average force required to hold onto the child is F = mv f ! v i ( ) "t ( ) = 12 kg ( ) 0 ! 60 mi h ( ) 0.050 s ! 0 1 m s 2.237 mi h # $ % & ' ( = !6.44 ) 10 3 N . Therefore, the magnitude of the needed retarding force is 6.44 ! 10 3 N , or 1 400 lb. A person cannot exert a force of this magnitude and a safety device should be used.
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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
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.
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: