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PHYS-207 Honors section LL, HW7 Solutions
9.61. Let m1 be the mass of the cart that is originally moving, v1i be its velocity before the collision, and v1f
be its velocity after the collision. Let m2 be the mass of the cart that is originally at rest and v2f be its
velocity after the collision. Conservation of linear momentum gives m1v1i  m1v1 f  m2 v2 f . Similarly, the
total kinetic energy is conserved and we have
1
1
1
m1v12i  m1v12f  m2 v22 f .
2
2
2
Solving for v1 f and v2 f , we obtain:
m1  m2
2m1
v1i , v2 f 
v1i
m1  m2
m1  m2
m v  m2v2i
The speed of the center of mass is vcom  1 1i
.
m1  m2
(a) With m1 = 0.34 kg, v1i  1.2 m/s and v1 f  0.66 m/s , we obtain
v1 f 
m2 
v1i  v1 f
 1.2 m/s  0.66 m/s 
m1  
 (0.34 kg)  0.0987 kg  0.099 kg.
v1i  v1 f
 1.2 m/s  0.66 m/s 
(b) The velocity of the second cart is:
v2 f 


2m1
2(0.34 kg)
v1i  
 (1.2 m/s)  1.9 m/s.
m1  m2
 0.34 kg  0.099 kg 
(c) From the above, we find the speed of the center of mass to be
vcom 
Note: In solving for
m1v1i  m2 v2i (0.34 kg) (1.2 m/s)  0

 0.93 m/s.
m1  m2
0.34 kg  0.099 kg
vcom , values for the initial velocities were used. Since the system is isolated with no
external force acting on it,
vcom remains the same after the collision, so the same result is obtained if
values for the final velocities are used. That is,
vcom 
m1v1 f  m2v2 f
m1  m2

(0.34 kg) (0.66 m/s)  (0.099 kg)(1.9 m/s)
 0.93 m/s.
0.34 kg  0.099 kg
9.75. We orient our +x axis along the initial direction of motion, and specify angles in the “standard” way
— so  = +60° for the proton (1), which is assumed to scatter into the first quadrant and  = –30° for the
target proton (2), which scatters into the fourth quadrant (recall that the problem has told us that this is
perpendicular to ). We apply the conservation of linear momentum to the x and y axes, respectively.
m1v1  m1v1 cos   m2v2 cos 
0  m1v1 sin   m2v2 sin .
We are given v1 = 500 m/s, which provides us with two unknowns and two equations, which is sufficient
for solving. Since m1 = m2 we can cancel the mass out of the equations entirely.
(a) Combining the above equations and solving for v2 we obtain
v2 
v1 sin 
(500 m/s) sin(60)

 433 m/s.
sin (   )
sin (90)
We used the identity sin cos – cos sin = sin (– ) in simplifying our final expression.
(b) In a similar manner, we find
v1 sin 
(500 m/s) sin(30)

 250 m/s .
sin (   )
sin (90)
v1 
10.2. (a) The second hand of the smoothly running watch turns through 2  radians during

2
 0.105 rad/s.
60
(b) The minute hand of the smoothly running watch turns through 2 radians during

2
 175
.  103 rad / s.
3600
60 s . Thus,
3600 s . Thus,
(c) The hour hand of the smoothly running 12-hour watch turns through 2 radians during 43200 s. Thus,

2
 145
.  104 rad / s.
43200
10.13. The wheel has angular velocity 0 = +1.5 rad/s = +0.239 rev/s at t = 0, and has constant value of
angular acceleration  < 0, which indicates our choice for positive sense of rotation. At t1 its angular
displacement (relative to its orientation at t = 0) is 1 = +20 rev, and at t2 its angular displacement is 2 =
+40 rev and its angular velocity is 2  0 .
(a) We obtain t2 using Eq. 10-15:
1
2(40 rev)
 335 s
0  2  t2  t2 
2
0.239 rev/s
2
which we round off to t2  3.4 10 s .
2 
(b) Any equation involving  can be used to find the angular acceleration; we select Eq. 10-16.
1
2
 2  2t2   t22 
 
2(40 rev)
 7.12  104 rev/s 2
2
(335 s)
which we convert to  = – 4.5  10–3 rad/s2.
(c) Using
t1 
 1   0t1  21 t12 (Eq. 10-13) and the quadratic formula, we have
0  02  21


(0.239 rev/s)  (0.239 rev/s) 2  2(20 rev)( 7.12 104 rev/s 2 )
7.12 104 rev/s 2
which yields two positive roots: 98 s and 572 s. Since the question makes sense only if t1 < t2 we conclude
the correct result is t1 = 98 s.
10.30. (a) The tangential acceleration, using Eq. 10-22, is
c
h
at  r  14.2 rad / s2 (2.83 cm)  40.2 cm / s2 .
(b) In rad/s, the angular velocity is  = (2760)(2/60) = 289 rad/s, so
ar   2 r  (289 rad / s) 2 (0.0283 m)  2.36  103 m / s2 .
(c) The angular displacement is, using Eq. 10-14,

 2 (289 rad/s)2

 2.94 103 rad.
2 2(14.2 rad/s 2 )
Then, using Eq. 10-1, the distance traveled is
s  r  (0.0283 m) (2.94  103 rad)  83.2 m.
10.47. Two forces act on the ball, the force of the rod and the force of gravity. No torque about the pivot
point is associated with the force of the rod since that force go through the pivot point.
As can be seen from the diagram, the component of the force of gravity that is perpendicular to the rod is
mg sin . If  is the length of the rod, then the torque associated with this force has magnitude
  mg sin   (0.75)(9.8)(1.25) sin 30  4.6 N  m .
For the position shown, the torque is counterclockwise.
10.51. (a) We use constant acceleration kinematics. If down is taken to be positive and a is the acceleration
of the heavier block m2, then its coordinate is given by y 
a
1
2
at 2 , so
2 y 2(0.750 m)

 6.00  102 m / s2 .
2
2
t
(5.00 s)
Block 1 has an acceleration of 6.00  10–2 m/s2 upward.
(b) Newton’s second law for block 2 is m2 g  T2  m2 a , where m2 is its mass and T2 is the tension. Thus,
T2  m2 ( g  a)  (0.500 kg)  9.8 m/s 2  6.00 10 2 m/s 2   4.87 N.
(c) Newton’s second law for block 1 is m1 g  T1  m1a, where T1 is the tension force on the block. Thus,
T1  m1 ( g  a)  (0.460 kg)  9.8 m/s 2  6.00 10 2 m/s 2   4.54 N.
(d) Since the cord does not slip on the pulley, the tangential acceleration of a point on the rim of the
pulley must be the same as the acceleration of the blocks, so

a 6.00  102 m / s2

 120
. rad / s2 .
R
5.00  102 m
(e) The net torque acting on the pulley is
  (T2  T1 ) R . Equating this to I we solve for I:
2
T2  T1  R  4.87 N  4.54 N   5.00 10 m 

I

 1.38 102 kg  m2 .

10.63. We use

1.20 rad/s 2
to denote the length of the stick. Since its center of mass is
initial potential energy is
1
2
/2
from either end, its
mg, where m is its mass. Its initial kinetic energy is zero. Its final potential
energy is zero, and its final kinetic energy is
1
2
I 2 , and conservation of energy yields
1
1
mg
mg  I 2   
.
2
2
I
The free end of the stick is a distance  from the rotation axis, so its speed as it hits the floor is
mg3
v   
.
I
2
Using Table 10-2 and the parallel-axis theorem, the rotational inertial is I  13 m , so
c
hb g
v  3g  3 9.8 m / s2 1.00 m  5.42 m / s.