Physics 103
8.1.
IDENTIFY and SET UP:
Assignment 8
p  mv. K  12 mv 2 .
EXECUTE: (a) p  (10,000 kg)(120 m/s)  120 105 kg  m/s
(b) (i) v 
p 120  105 kg  m/s

 600 m/s. (ii)
m
2000 kg
vSUV 
8.2.
1 m v2
2 T T
2
 12 mSUVvSUV
, so
mT
10,000 kg
vT 
(120 m/s)  268 m/s
mSUV
2000 kg
EVALUATE: The SUV must have less speed to have the same kinetic energy as the truck than to have the same
momentum as the truck.
IDENTIFY: Each momentum component is the mass times the corresponding velocity component.
SET UP: Let  x be along the horizontal motion of the shotput. Let  y be vertically upward. vx  v cos ,
v y  v sin  .
EXECUTE: The horizontal component of the initial momentum is
px  mvx  mv cos  (730 kg)(150 m/s)cos400  839 kg  m/s.
The vertical component of the initial momentum is
p y  mv y  mv sin   (730 kg)(150 m/s)sin400  704 kg  m/s.
EVALUATE: The initial momentum is directed at 400 above the horizontal.
8.3.
IDENTIFY and SET UP:
p  mv. K  12 mv 2 .
2
EXECUTE: (a) v 
p2
p
 p
.
and K  12 m   
2m
m
m
(b) Kc  K b and the result from part (a) gives
pc2
p2
mb
0145 kg
 b . pb 
pc 
pc  190 pc . The
2mc 2mb
mc
0040 kg
baseball has the greater magnitude of momentum. pc /pb  0526.
(c) p2  2mK so pm  pw gives 2mm Km  2mw K w . w  mg , so wm Km  ww K w .
8.4.
w 
 700 N 
Kw   m  Km  
 K m  156 K m .
w
 450 N 
 w
The woman has greater kinetic energy. Km /K w  0641.
EVALUATE: For equal kinetic energy, the more massive object has the greater momentum. For equal momenta,
the less massive object has the greater kinetic energy.
IDENTIFY: For each object p  mv and the net momentum of the system is P  p A  pB . The momentum
vectors are added by adding components. The magnitude and direction of the net momentum is calculated from
its x and y components.
SET UP: Let object A be the pickup and object B be the sedan. vAx  140 m/s, v Ay  0. vBx  0,
vBy  230 m/s.
EXECUTE: (a) Px  pAx  pBx  mAvAx  mBvBx  (2500 kg)( 140 m/s)  0  350 104 kg  m/s
Py  p Ay  pBy  mAvAy  mBvBy  (1500 kg)(  230 m/s)  345 104 kg  m/s
(b) P  Px2  Py2  491104 kg  m/s. From Figure 8.4, tan  
Px 350  104 kg  m/s

and   454. The
Py 345  104 kg  m/s
net momentum has magnitude 491104 kg  m/s and is directed at 454 west of north.
1
EVALUATE: The momenta of the two objects must be added as vectors. The momentum of one object is west
and the other is north. The momenta of the two objects are nearly equal in magnitude, so the net momentum is
directed approximately midway between west and north.
8.7.
Figure 8.4
IDENTIFY: The average force on an object and the object’s change in momentum are related by Eq. 8.9. The
weight of the ball is w  mg.
SET UP: Let  x be in the direction of the final velocity of the ball, so v1x  0 and v2 x  250 m/s.
EXECUTE: ( Fav ) x (t2  t1)  mv2 x  mv1x gives ( Fav ) x 
8.13.
mv2 x  mv1x (00450 kg)(250 m/s)

 562 N.
t2  t1
200  10 –3 s
w  (00450 kg)(980 m/s2 )  0441 N. The force exerted by the club is much greater than the weight of the
ball, so the effect of the weight of the ball during the time of contact is not significant.
EVALUATE: Forces exerted during collisions typically are very large but act for a short time.
IDENTIFY: The force is constant during the 1.0 ms interval that it acts, so J  F t. J  p2  p1m(v2 v1 ).
SET UP: Let  x be to the right, so v1x  500 m/s. Only the x component of J is nonzero, and
J x  m(v2 x  v1x ).
EXECUTE: (a) The magnitude of the impulse is J  F t  (250 103 N)(100 10–3 s)  250 N  s. The
direction of the impulse is the direction of the force.
250 N  s
J
 500 m/s  625 m/s. The stone’s velocity has
(b) (i) v2 x  x  v1x . J x  250 N  s. v2 x 
200 kg
m
magnitude 6.25 m/s and is directed to the right. (ii) Now J x  250 N  s and
250 N  s
 500 m/s  375 m/s. The stone’s velocity has magnitude 3.75 m/s and is directed to the right.
200 kg
EVALUATE: When the force and initial velocity are in the same direction the speed increases and when they are
in opposite directions the speed decreases.
IDENTIFY: Since drag effects are neglected there is no net external force on the system of squid plus expelled
water and the total momentum of the system is conserved. Since the squid is initially at rest, with the water in its
v2 x 
8.19.
cavity, the initial momentum of the system is zero. For each object, K  12 mv 2 .
SET UP: Let A be the squid and B be the water it expels, so mA  650 kg  175 kg  475 kg. Let  x be the
direction in which the water is expelled. v A2 x  250 m/s. Solve for vB 2 x .
EXECUTE: (a) P1x  0. P2 x  P1x , so 0  mAv A2 x  mBvB 2 x .
vB 2 x  
m Av A2 x
(475 kg)(250 m/s)

 679 m/s.
mB
175 kg
(b) K 2  K A2  K B 2  12 mAv A2 2  12 mBvB2 2  12 (475 kg)(250 m/s) 2  12 (175 kg)(679 m/s) 2  552 J. The initial
8.21.
kinetic energy is zero, so the kinetic energy produced is K2  552 J.
EVALUATE: The two objects end up with momenta that are equal in magnitude and opposite in direction, so the
total momentum of the system remains zero. The kinetic energy is created by the work done by the squid as it
expels the water.
IDENTIFY: Apply conservation of momentum to the system of the two pucks.
SET UP: Let  x be to the right.
2
EXECUTE: (a) P1x  P2 x says (0250 kg)vA1  (0250 kg)(0120 m/s)  (0350 kg)(0650 m/s) and
v A1  0790 m/s.
(b) K1  12 (0250 kg)(0790 m/s) 2  00780 J.
K 2  12 (0250 kg)(0120 m/s) 2  12 (0350 kg)(0650 m/s) 2  00757 J and K  K2  K1  00023 J.
8.22.
EVALUATE: The total momentum of the system is conserved but the total kinetic energy decreases.
IDENTIFY: Since road friction is neglected, there is no net external force on the system of the two cars and the
total momentum of the system is conserved. For each object, K  12 mv 2 .
SET UP: Let A be the 1750 kg car and B be the 1450 kg car. Let  x be to the right, so vA1x  150 m/s,
vB1x  110 m/s, and vA2 x  0250 m/s. Solve for vB 2 x .
EXECUTE: (a) P1x  P2 x . mAvA1x  mBvB1x  mAvA2 x  mBvB 2 x . vB 2 x 
m Av A1x  mB vB1x  m Av A2 x
.
mB
(1750 kg)(150 m/s)  (1450 kg)( 110 m/s)  (1750 kg)(0250 m/s)
 0409 m/s.
1450 kg
After the collision the lighter car is moving to the right with a speed of 0.409 m/s.
vB 2 x 
(b) K1  12 mAv A21  12 mB vB21  12 (1750 kg)(150 m/s)2  12 (1450 kg)(110 m/s)2  2846 J.
K 2  12 m Av A2 2  12 mBvB2 2  12 (1750 kg)(0250 m/s)2  12 (1450 kg)(0409 m/s) 2  176 J.
8.33.
The change in kinetic energy is K  K2  K1  176 J  2846 J  2670 J.
EVALUATE: The total momentum of the system is constant because there is no net external force during the
collision. The kinetic energy of the system decreases because of negative work done by the forces the cars exert
on each other during the collision.
IDENTIFY: Since drag effects are neglected there is no net external force on the system of two fish and the
momentum of the system is conserved. The mechanical energy equals the kinetic energy, which is K  12 mv 2
for each object.
SET UP: Let object A be the 15.0 kg fish and B be the 4.50 kg fish. Let  x be the direction the large fish is
moving initially, so vA1x  110 m/s and vB1x  0. After the collision the two objects are combined and move
with velocity v 2 . Solve for v2 x .
EXECUTE: (a) P1x  P2 x . mAvA1x  mBvB1x  (mA  mB )v2 x .
v2 x 
mAv A1x  mB vB1x (150 kg)(110 m/s)  0

 0846 m/s.
m A  mB
150 kg  450 kg
(b) K1  12 mAv A21  12 mBvB21  12 (150 kg)(110 m/s)2  908 J.
K 2  12 (mA  mB )v22  12 (195 kg)(0846 m/s)2  698 J.
8.46.
K  K2  K1  2 210 J . 2.10 J of mechanical energy is dissipated.
EVALUATE: The total kinetic energy always decreases in a collision where the two objects become combined.
IDENTIFY: No net external horizontal force so Px is conserved. Elastic collision so K1  K2 and can use Eq.
8.27.
SET UP:
Figure 8.46
EXECUTE: From conservation of x-component of momentum:
3
mAvA1x  mBvB1x  mAvA2 x  mBvB 2 x
mAvA1  mBvB1  mAvA2 x  mBvB 2 x
(0150 kg)(080 m/s)  (0300 kg)(220 m/s)  (0150 kg)vA2 x  (0300 kg)vB 2 x
360 m/s  vA2 x  2vB 2 x
From the relative velocity equation for an elastic collision Eq. 8.27:
vB 2 x  vA2 x  (vB1x  vA1x )  (220 m/s  080 m/s)  300 m/s
300 m/s  vA2 x  vB 2 x
Adding the two equations gives 060 m/s  3vB 2 x and vB 2 x  020 m/s. Then
8.47.
vA2 x  vB 2 x  300 m/s  2 320 m/s
The 0.150 kg glider (A) is moving to the left at 3.20 m/s and the 0.300 kg glider (B) is moving to the left at
0.20 m/s.
EVALUATE: We can use our v A2 x and vB 2 x to show that Px is constant and K1  K2
IDENTIFY: When the spring is compressed the maximum amount the two blocks aren’t moving relative to each
other and have the same velocity V relative to the surface. Apply conservation of momentum to find V and
conservation of energy to find the energy stored in the spring. Since the collision is elastic, Eqs. 8.24 and 8.25
give the final velocity of each block after the collision.
SET UP: Let  x be the direction of the initial motion of A.
EXECUTE: (a) Momentum conservation gives (200 kg)(200 m/s)  (120 kg)V and V  0333 m/s. Both
blocks are moving at 0.333 m/s, in the direction of the initial motion of block A. Conservation of energy says the
initial kinetic energy of A equals the total kinetic energy at maximum compression plus the potential energy
U b stored in the bumpers: 12 (200 kg)(200 m/s) 2  U b  12 (120 kg)(0333 m/s) 2 and U b  3.33 J.
 m  mB 
 200 kg  100 kg 
(b) v A2 x   A
 v A1x  
 (200 m/s)  133 m/s. Block A is moving in the
m

m
120 kg


B
 A
 x direction at 1.33 m/s.
8.51.
 2m A 
2(200 kg)
vB 2 x  
(200 m/s)  + 0667 m/s. Block B is moving in the  x direction at 0.667
 v A1x 
120 kg
 mA  mB 
m/s.
EVALUATE: When the spring is compressed the maximum amount the system must still be moving in order to
conserve momentum.
IDENTIFY: Apply Eq. 8.28.
SET UP: mA  0300 kg, mB  0400 kg, mC  0200 kg.
EXECUTE: xcm 
xcm 
mA xA  mB xB  mC xC
.
mA  mB  mC
(0300 kg)(0200 m)  (0400 kg)(0100 m)  (0200 kg)(0300 m)
 00444 m.
0300 kg  0400 kg  0200 kg
ycm 
mA y A  mB yB  mC yC
.
mA  mB  mC
(0300 kg)(0300 m)  (0400 kg)( 0400 m)  (0200 kg)(0600 m)
 00556 m.
0300 kg  0400 kg  0200 kg
EVALUATE: There is mass at both positive and negative x and at positive and negative y and therefore the
center of mass is close to the origin.
IDENTIFY: Apply Eqs. 8.28, 8.30 and 8.32. There is only one component of position and velocity.
SET UP: mA  1200 kg, mB  1800 kg. M  mA  mB  3000 kg. Let  x be to the right and let the origin be
ycm 
8.54.
at the center of mass of the station wagon.
m x  mB xB 0  (1800 kg)(400 m)

 240 m.
EXECUTE: (a) xcm  A A
mA  mB
1200 kg  1800 kg
4
The center of mass is between the two cars, 24.0 m to the right of the station wagon and 16.0 m behind the lead
car.
(b) Px  mAv A, x  mB vB, x  (1200 kg)(120 m/s)  (1800 kg)(200 m/s)  504  104 kg  m/s.
(c) vcm, x 
8.55.
mAvA, x  mBvB, x
mA  mB

(1200 kg)(12.0 m/s)  (1800 kg)(20.0 m/s)
 16.8 m/s.
1200 kg  1800 kg
(d) Px  Mvcm x  (3000 kg)(168 m/s)  504 104 kg  m/s, the same as in part (b).
EVALUATE: The total momentum can be calculated either as the vector sum of the momenta of the individual
objects in the system, or as the total mass of the system times the velocity of the center of mass.
IDENTIFY: Use Eq. 8.28 to find the x and y coordinates of the center of mass of the machine part for each
configuration of the part. In calculating the center of mass of the machine part, each uniform bar can be
represented by a point mass at its geometrical center.
SET UP: Use coordinates with the axis at the hinge and the  x and  y axes along the horizontal and vertical
bars in the figure in the problem. Let ( xi , yi ) and ( xf , yf ) be the coordinates of the bar before and after the
vertical bar is pivoted. Let object 1 be the horizontal bar, object 2 be the vertical bar and 3 be the ball.
m x  m2 x2  m3 x3 (400 kg)(0750 m)  0  0
EXECUTE: xi  1 1

 0333 m.
m1  m2  m3
400 kg  300 kg  200 kg
8.58.
yi 
m1 y1  m2 y2  m3 y3 0  (300 kg)(0900 m)  (200 kg)(180 m)

 0700 m.
m1  m2  m3
900 kg
xf 
(400 kg)(0750 m)  (300 kg)(0900 m)  (200 kg)(180 m)
 0366 m.
900 kg
yf  0. xf  xi  0700 m and yf  yi  0700 m. The center of mass moves 0.700 m to the right and 0.700
m upward.
EVALUATE: The vertical bar moves upward and to the right so it is sensible for the center of mass of the
machine part to move in these directions.
(a) IDENTIFY and SET UP: Apply Eq. 8.28 and solve for m1 and m2 .
EXECUTE:
m1  m2 
ycm 
m1 y1  m2 y2
m1  m2
m1 y1  m2 y2 m1(0)  (050 kg)(60 m)

 125 kg and m1  075 kg.
ycm
24 m
EVALUATE: ycm is closer to m1 since m1  m2 .
(b) IDENTIFY and SET UP: Apply a=dv/dt for the cm motion.
dv
EXECUTE: acm = cm =(15 m/s3 )tiˆ.
dt
(c) IDENTIFY and SET UP: Apply Eq. 8.34.
EXECUTE:  F  Ma (125 kg)(15 m/s3 )tiˆ
ext
cm
At t  30 s,  Fext  (125 kg)(15 m/s3 )(30 s)iˆ  (56 N)iˆ
EVALUATE: vcm-x is positive and increasing so acm-x is positive and Fext is in the  x - direction There is no
motion and no force component in the y - direction
8.59.
IDENTIFY: Apply  F 
dP
to the airplane.
dt
d n
(t )  nt n 1. 1 N  1 kg  m/s2
dt
dP
=[(150 kg  m/s3 )t ]i  (025 kg  m/s 2 ) j . Fx  (150 N/s)t , Fy  025 N, Fz  0.
EXECUTE:
dt
EVALUATE: There is no momentum or change in momentum in the z direction and there is no force component
in this direction.
SET UP:
5