Physics I
Homework X
Chapter 10: 10, 15, 31, 36, 50, 56
CJ
10.10. Model: Model the ball as a particle undergoing rolling motion with zero
rolling friction. The sum of the ball’s kinetic and gravitational potential energy, therefore,
does not change during the rolling motion.
Visualize:
Solve: Since the quantity K
Ug does not change during rolling motion, the energy
conservation equations apply. For the linear segment the energy conservation equation
K 0  Ug0  K1  Ug1 is
1
1
1
1 2
1
mv1  mgy1  mv02  mgy0  m v12  mg(0 m)  m(0 m/s)2  mgy0  mv12  mgy0
2
2
2
2
2
For the parabolic part of the track,
K1  Ug1  K2  Ug2
is
1 2
1
1
1
1
mv1  mgy1  mv22  mgy2  mv12  mg(0 m)  m(0 m/s)2  mgy2  mv12  mgy2
2
2
2
2
2
Since from the linear segment we have 21 m v12  mgy0 , we get mgy0  mgy2 or y2  y0  1.0 m. Thus,
the
ball
rolls
up to exactly the same height as it started from.
Assess: Note that this result is independent of the shape of the path followed by the
ball, provided there is no rolling friction. This result is an important consequence of
energy conservation.
10.15. Model: Assume that the spring is ideal and obeys Hooke’s law. We also model
the 5.0 kg mass as a particle.
Visualize: We will use the subscript s for the scale and sp for the spring.
Solve: (a) The scale reads the upward force
second law gives
 (F
Fs on m
that it applies to the mass. Newton’s
)  Fs on m  w  0  Fs on m  w  mg  (5.0 kg)(9.8 m/s2 )  49 N
on m y
(b) In this case, the force is
 (F
)  Fs on m  Fsp  w  0  20 N  ky  mg  0
on m y
 k  (mg  20 N)/ y  (49 N  20 N)/0.02 m  1450 N/m
(c) In this case, the force is
10.31. Model: For an energy diagram, the sum of the kinetic and potential energy is a
constant.
Visualize:
Since the particle oscillates between x
x
particle is zero at these points. That is, for these values of x, E
U
5.0 J, which
defines the total energy (TE) line. The distance from the potential energy (PE) curve to
the TE line is the particle’s kinetic energy. These values are transformed as the position
changes, but the sum K U does not change.
Solve: The equation for total energy E
maximum when U is minimum. We have
U
K means K
E
U, so that K is
10.36. Model: For the marble spinning around the inside of a smooth pipe, the sum of
the kinetic and gravitational potential energy does not change.
Visualize:
We use a coordinate system with the origin at the bottom of the pipe, that is, y1
radius
(R)
of
the
pipe
is
10 cm, and therefore ytop y2
R
, measured from the
bottom
of
the
circle,
y
R R cos.
Solve: (a) The energy conservation equation

K2  Ug2  K1  Ug1
is
1 2
1
mv2  mgy2  mv12  mgy1
2
2
 v2  v12  2g( y1  y2 )  (3.0 m/s)2  2(9.8 m/s2 )(0 m  0.20 m)  2.25 m/s
(b) Expressing the energy conservation equation as a function of
K ( )  Ug ( )  K1  Ug1 
:
1 2
1
mv ( )  mgy( )  mv12  0 J
2
2
 v    v12  2gy( )  v12  2gR(1  cos )
Using v1
g
2
, and R
v( )  9  1.96(1  cos ) (m/s)
(c) The accompanying figure shows a graph of v for a complete revolution
(0    360).
Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble’s potential energy
increases and kinetic energy decreases as it gets toward the top of the circle. At the top,
its speed is 2.25 m/s. This is reasonable since some of the kinetic energy has been
transformed into the marble’s potential energy.
10.50. Model: Since there is no friction, the sum of the kinetic and gravitational
potential energy does not change. Model Julie as a particle.
Visualize:
We place the coordinate system at the bottom of the ramp directly below Julie’s starting
position. From geometry, Julie launches off the end of the ramp at a 30º angle.
1
1
K1  Ug1  K 0  Ug0  mv12  mgy1  mv02  mgy0
2
2
v0  0 m/s, y0  25 m, and y1  3 m, the above equation simplifies to
Solve: Energy conservation:
Using
1 2
mv1  mgy1  mgy0  v1  2g( y0  y1 )  2(9.8 m/s2 )(25 m  3 m)  20.77 m/s
2
We can now use kinematic equations to find the touchdown point from the base of the
ramp. First we’ll consider the vertical motion:
1
1
y2  y1  v1y (t2  t1 )  ay (t2  t1 )2 0 m  3 m  ( v1 sin30)(t2  t1 )  ( 9.8 m / s2 )(t2  t1 ) 2
2
2
(20.77 m/s)sin30
(3 m)
(t2  t1 )2 
(t2  t1 ) 
0
2
(4.9 m/s )
(4.9 m/s2 )
(t2  t1 )2  (2.119 s)(t2  t1 )  (0.6122 s2 )  0  ( t2  t1 )  2.377 s
For the horizontal motion:
1
x2  x1  v1x (t2  t1 )  ax (t2  t1 )2
2
x2  x1  (v1 cos30)(t2  t1 )  0 m  (20.77 m/s)(cos30)(2.377 s)  42.7 m
Assess: Note that we did not have to make use of the information about the circular arc
at the bottom that carries Julie through a 90 turn.
10.56. Model: Model Lisa (L) and the bobsled (B) as particles. We will assume the
spring) is conserved. Furthermore, during the collision, as Lisa leaps onto the bobsled,
m is conserved. We will also assume the
spring to be an ideal one that obeys Hooke’s law.
Visualize:
We place the origin of our coordinate system directly below the bobsled’s initial position.
Solve: (a) Momentum conservation in Lisa’s collision with bobsled states
p1  p0 ,
or
(mL  mB )v1  mL (v0 )L  mB (v0 )B  (mL  mB )v1  mL (v0 )L  0
 mL 


40 kg
 v1  
 (v0 )L  
 (12 m/s)  8.0 m/s
 40 kg  20 kg 
 mL  mB 
The energy conservation equation:
K2  Us2  Ug2  K1  Us1  Ug1
is
1
1
1
1
(mL  mB )v22  k( x2  xe )2  (mL  mB )gy2  (mL  mB )v12  k( xe  xe )2  (mL  mB )gy1
2
2
2
2
Using v2
and (mL mB
k
y2
y1

v1
,
1
1
0 J  (2000 N/ m)( x2  xe )2  0 J  (60 kg)(8.0 m/s)2  0 J  (60 kg)(9.8 m/s2 )(17.1 m)
2
2
Solving this equation yields (x2  xe )  3.46 m.
(b) As long as the ice is slippery enough to be considered frictionless, we know from
conservation of mechanical energy that the speed at the bottom depends only on the
y. Only the ramp’s height h is important, not its shape or angle.
K max 
1 2
mvmax  5.0 J  Umin
2
 vmax  2(5.0 J  Umin )/ m  2(5.0 J  1.0 J)/0.002 kg  63.2 m/s
 (F
)  Fsp  w  0  ky  mg  0
on m y
 y  mg / k  (49 N)/(1450 N/m)  0.0338 m  3.4 cm