Physics 192 Solutions to Mastering Physics Week 2

Physics 192 Solutions to Mastering Physics Week 2 P9.15. Prepare: This is a problem with no external forces so we can use the law of conservation of momentum.
Solve: The total momentum before the bullet hits the block equals the total momentum after the bullet passes through the
block so we can write
mb (vb )i  mbl (vbl )i  mb (vb )f  mbl (vbl )f 
(3.0  103 kg)(500 m/s)  (2.7 kg)(0 m/s)  (3.0  103 kg)(220 m/s)  (2.7 kg)(v bl )f .
We can solve for the final velocity of the block: (vbl )f  0.31 m/s .
Assess: This is reasonable since the block is about one thousand times more massive than the bullet and its change in
speed is about one thousand times less.
P9.19. Prepare: We will define our system to be archer arrow. The force of the archer (A) on the arrow (a) is equal
to the force of the arrow on the archer. These are internal forces within the system. The archer is standing on frictionless


ice, and the normal force by ice on the system balances the weight force. Thus Fext  0 on the system, and momentum is
conserved. The initial momentum pix of the system is zero, because the archer and the arrow are at rest. The final moment
pfx must also be zero.
Solve:
We have MAvA  mava  0 kg m/s. Therefore,
vA 
ma va (0.04 kg)(60 m/s)

  0.48 m/s
mA
50 kg
The archer’s recoil speed is 0.48 m/s.
Assess: It is the total final momentum that is zero, although the individual momenta are nonzero. Since the arrow has
forward momentum, the archer will have backward momentum.
P9.22. Prepare: There is no external force on the player in the horizontal direction. Momentum is conserved in the
horizontal direction. At the highest point in his leap the player has no velocity. Consider the player and the ball as the
system.
Solve: The momentum of the system before the catch is entirely due to the motion of the ball. Momentum is conserved
in the system. See the following diagram.
After the player catches the ball, they move together with a common final velocity. Writing the mass of the player as mP ,
and the momentum of the ball as mB (vBx ), the momentum conservation equation is
(mP  mB )(vx )f  mB (vBx )i
Solving for (vx )f ,
(v x ) f 
mB (vBx )i (0.140 kg)(28 m/s)

 5.5 cm/s
mP  mB
71 kg  0.140 kg
The player moves with a speed of 5.5 cm/s in the same direction the ball was originally moving.
Assess: This result seems reasonable, since the mass of the ball is so small relative to the mass of the player.
P9.26. Prepare: This problem deals with the conservation of momentum in two dimensions in an inelastic collision. We
will thus use Equation 9.14.
Solve:


The conservation of momentum equation pbefore  pafter is
m1 (v1x )i  m2 (v2x )i  (m1  m2 )(vx )f
Substituting in the given values,
m1 (v1 y )i  m2 (v2 y )i  (m1  m2 )(v y )f
(.02 kg)(3.0 m/s)  0 kg  m/s  (.02 kg  .03 kg)vf cos 
0 kg m/s  (.03 kg)(2.0 m/s)  (.02 kg  .03 kg)vf sin   vf cos   1.2 m/s
vf sin   1.2 m/s  vf  (1.2 m/s) 2  (1.2 m/s) 2  1.7 m/s
  tan 1
vy
vx
 tan 1 (1)  45 
The ball of clay moves 45 north of east at 1.7 m/s.
Assess: Noting that the magnitude of momenta of both particles have the same value 0.6 kg  m/s. This information, along
with the fact that they are coming at each other at 90 degrees, may be used to reason out that the outgoing path will be
right in the middle, that is, at 45 degrees as shown above.
P9.33. Prepare: The disk is a rotating rigid body. Please refer to Figure P9.33. The angular velocity  is 600 rpm 
600  2/60 rad/s  20 rad/s. From Table 7.4, the moment of inertial of the disk about its center is (1/2) MR2, which
can be used with L = I to find the angular momentum.
Solve:
I
1
1
MR 2  (2.0 kg)(0.020 m)2  4.0  104 kg  m 2
2
2
Thus, L  I  (4.0  104 kg m2)(20 rad/s)  0.025 kg m2/s. If we wrap our right fingers in the direction of the disk’s
rotation, our thumb will point in the –x direction. Consequently,

L  (0.025 kg  m 2/s, into page)
Assess:!Don’t forget the direction of the angular momentum because it is a vector quantity.
P9.34. Prepare: Once the diver has left the diving board her angular momentum is conserved. Her weight acts at her
center of gravity, so there is no net external torque about her center of gravity. Equation 9.23 applies.
Solve:
Using Equation 9.23,
I ii  I f f
Solving for f / i ,
f I i 14 kg  m 2
 
 3.5
i I f 4.0 kg  m 2
Her final angular velocity is 3.5 times her initial angular velocity.
Assess: This result makes sense. Her final angular velocity is greater than her initial angular velocity.
P9.73.!Prepare: Since there are no external torques, we can use the conservation of angular momentum to solve this
1
problem. We need to use the moment of inertia of a rotating disk I disk  MR 2 in order to write the final angular
2
momentum of the merry-go-round: ( Lm )f  I diskf . The final angular momentum of Joey is given by ( LJ )f  m(vJ )f R.
Solve: Before Joey begins running, both he and the merry-go-round are at rest, so the total angular momentum is 0. Let us
say that he runs counterclockwise so that his final angular momentum is positive. Then the merry-go- round must rotate
clockwise so that its final angular momentum is negative. In this way the angular momenta of Joey and the merry-goround will still add to 0. The equation for conservation of angular momentum is
1

0 kg  m 2 / s  m(vJ )f R   MR 2  f 
2

1
0 kg  m 2 / s  (36 kg)(5.0 m/s)(2.0 m)    (200 kg)(2.0 m)2f
2
This equation can be solved for the angular velocity: f  0.90 rad/s.
Assess: The negative sign is as expected considering that the total angular momentum is zero. For this to be true, the two
bodies, the merry-go-round and Joey, must move in opposite directions around the axis.
P9.75. Prepare: Define the system to be Disk A plus Disk B so that during the time of the collision there are no net
torques on this isolated system. This allows us to use the law of conservation of angular momentum.
After the collision the combined object will have a moment of inertia equal to the sum of the moments of inertia of Disk A
and Disk B and it will have one common angular speed f (which we seek).
We must remember to call counterclockwise angular speeds positive and clockwise angular speeds negative.
Known
MA  2.0 kg
RA  0.40 m
(A )i  30 rev/s
M B  2.0 kg
RB  0.20 m
(B)i  30 rev/s
Find
f
Preliminarily compute the moments of inertia of the disks:
1
1
I A  M A RA2  (2.0 kg)(0.40 m) 2  0.16 kg  m 2
2
2
1
1
2
I B  MB RB  (2.0 kg)(0.20 m) 2  0.040 kg  m 2
2
2
Solve:
 Li   Lf
( LA)i  (LB)i  ( LA  B)f
IA (A)i  IB (B)i  ( IA  I B )f
Solving for f gives
f 
IA (A)i  IB (B)i
(0.16 kg  m 2 )( 30 rev/s)  (0.040 kg  m 2 )(30 rev/s)

  18 rev/s
IA  IB
0.16 kg  m 2  0.040 kg  m 2
That is, the angular speed is 18 rev/s and the direction is clockwise.
Assess: Since the two disks were rotating in opposite directions at the same speed we expect the angular speed
afterwards to be less than the original speeds, and indeed it is.

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