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.