PHY2053
Summer C 2013
Exam 2 Solutions
1. Since the normal force is perpendicular to the incline and motion is along the incline, the
angle between the force and displacement is 90º.
W
F r cos
F r cos 90
0
The normal force does no work.
2. The acceleration due to the force is
F
ma
a
F
m
4N
2 kg
2 m / s2
The final velocity of the 2-kg block is
vf
vi
a t
vf
vi
a t
0 (2 m/s )(6 s) 12 m/s
The 2-kg block has kinetic energy,
mv 2
1
2
K
1
2
(2 kg)(12 m/s) 2
144 J
The potential energy of the 3-kg block will be 144 J. Its height is
U
mgy
y
U
mg
144 J
(3 kg)(9.8 m/s 2 )
4.9 m
3. The initial mechanical energy of the ball is
ME1
K1 U 1
2
1
2
mv1
1
2
(0.1 kg )(8 m/s ) 2
mgy1
(0.1 kg )(9.8 m/s 2 )(10 m)
13 J
When the ball is 5 m above the ground its potential energy will be
U2
mgy 2
(0.1 kg )(9.8 m/s 2 )(5 m)
1
4.9 J
Since mechanical energy is conserved, ME2 = ME1 = 13 J. The kinetic energy when the
ball is 5 m above the ground is
ME2
K2 U2
K2
ME2 U 2
13 J 4.9 J
8.1J
The velocity is
1
2
K2
mv 2
2
2K 2
m
v2
2(8.1 J)
0.1 kg
12.7 m/s
4. Use the work-energy theorem
W NC
K
U
The initial kinetic energy is
K1
1
2
mv1
2
1
2
(75 kg )(2 m/s ) 2
150 J
The final kinetic energy is K2 = 0 since the bicycle stops. The initial potential energy is
U1
mgy1
(75 kg )(9.8 m/s 2 )(10 m)
7350 J
The final potential energy is U2 = 0 since the bicycle is at the bottom of the hill. The
work done by nonconservative forces is
WNC
K
U
(K 2
K1 ) (U 2 U 1 )
(0 150 J) (0 7350 J)
7500 J
Work is defined as
W
F r cos
Since the frictional forces opposes the displacement,
r
W
F cos
7500 J
(150 N) cos180
= 180º. Solving for r,
50 m
5. When in equilibrium, the force of the spring pulling up is equal to the weight of the mass,
Fs
W1
2
kx m1 g
m1 g
x
k
(1.4 kg )(9.8 m/s 2 )
7.0 cm
1.96 N/cm
When the mass is added, the total displacement from equilibrium is
x2
x1
x
7 cm 10 cm 17 cm
The new mass m2, can be found using the same reasoning as above
kx2
m2 g
m2
kx2
g
(1.96 N/cm)(17 cm)
9.8 m/s 2
3.4 kg
The mass added is
m
m2
m1
3.4 kg 1.4 kg
2.0 kg
6. The spring constant is
F
kx
k
F
x
80 N
0.196 m
408 N/m
As the ball is shot up, only conservative forces act and mechanical energy is conserved.
U1
K1
U2
K2
Initially, the ball is at rest. At the highest point it is a rest again. Therefore, K1 = K2 = 0.
There are two types of potential energy. The elastic potential energy stored into the
spring turns into gravitational potential energy.
U1
1
2
K1
2
kx
U2
K2
0
mgy 0
y
kx2
2mg
(408 N/m)(0.196 m) 2
2(0.20 kg )(9.8 m/s 2 )
4.0 m
7. The car rolls at constant speed. By Newton’s first law, the net force must equal to zero.
The component of the weight along the incline must be cancelled by air resistance and
friction. Call the net resistive force F. The free body diagram is
3
F
N
mg
Taking the x-axis along the incline,
mg sin
Fx
0
F
0
F
mg sin
(1500 kg )(9.8 m/s 2 ) sin 2
513 N
When the car is driven at 20 m/s on level ground, the engine must supply a force equal to
the resistive force of 513 N. The power needed is
P
Fv cos
(513 N)(20 m/s ) cos 0 10,000 W
8. The impulse-momentum theorem states that the change in linear momentum is caused by
the impulse.

F t

p
To work with vectors, we must take components
p2 x
px
Fx t
p1x
Fx t
p2 x
p1x
mv1x
Fx t
Fx t
The initial velocity is to the right, which we take to be the positive direction. The force
acts to the left, which has a negative component,
p2 x
mv1x
Fx t
(2 kg )(5 m/s ) ( 20 N)(4 s)
70 kg m/s
Since the x-component is negative, the momentum is 70 kg·m/s to the left.
4
9. Linear momentum is conserved in an explosion. Since the block is initially at rest, pi = 0.
After the explosion,
v1f =10 m/s
v2f
m1 = 2 kg
m2 = 3 kg
The final momentum is

pf

m1v1 f

m2 v2 f
Taking components
p fx
m1v1 f
m2v2 f
Since m1 moves to the left, its velocity has a negative component. Using conservation of
linear momentum,
pix
0
v2 f
p fx
m1v1 f
m1v1 f
m2
m2v2 f
(2 kg )(10 m/s )
3 kg
6.7 m/s
The positive sign means the 3-kg mass moves to the right.
10. Before the collision,
v1i = 3 m/s
v2i = 2 m/s
m1 = 4 kg
m2
After the collision,
v1f = 1.8 m/s
v2f
m1 = 4 kg
m2
5
In class, we derived the general equations for a one dimensional elastic collision,
v1 f
m1 m2
v1i
m1 m2
2m2
v2 i
m1 m2
v2 f
2m1
v1i
m1 m2
m2 m1
v2 i
m1 m2
Solve the first equation for m2 since we are not given its final speed.
m1 m2
v1i
m1 m2
v1 f
2m2
v2i
m1 m2
(m1 m2 )v1 f
(m1 m2 )v1i
2m2v2i
m1v1 f
m1v1i
2m2v2i
m2 (v1 f
m2v1 f
v1i
m2v1i
2v2i ) m1 (v1i v1 f )
m2
m1
v1i v1 f
v1 f
(4 kg )
v1i
2v2i
3.0 m/s 1.8 m/s
1.8 m/s 3.0 m/s 2(2.0 m/s)
6.0 kg
11. Linear momentum is conserved in the collision,
pix
m1v1i
m2 (v2i
m2v2i
p fx
m1v1 f
m2v2 f
v2 f ) m1 (v1 f
m2
m1
v1i )
v1 f
v1i
v2i
v2 f
(10 kg )
0 5 m/s
3 m/s 2 m/s
10 kg
12. Linear momentum is conserved in the collision. Since the collision is two dimensional,
we must use components. For the x-component (along the east direction),
pix
p fx
mt vt
(mc
mt )v cos 45
6
v
45º
truckcar
vt
truck
vc
car
For the y-component (along the north direction)
piy
p fy
mcvc
(mc
mc vc
mt vt
(mc mt )v sin 45
(mc mt )v cos 45
mc vc
1
mt vt
mt )v sin 45
Divide the two equations,
Solve for vt
mc vc
mt vt
vt
mc vc
mt
(1200 kg )(15 m/s )
12 m/s
1500 kg
13. The center of mass is defined as
xcm
xi mi
x A mA xB mB
mA mB
mi
In this problem, the mass are equal. Let’s call them m. Substituting above gives
7
xcm
x AmA xB mB
mA mB
x A m xB m
m m
m( x A x B )
2m
1
2
( xA
xB )
The velocity is related to the position by x = v t. Substituting above,
xcm
1
2
( xA
vcm t
1
2
(v A t v B t )
vcm
1
2
(v A v B )
xB )
We have vA = 0. The velocity of the center of mass is
1
2
vcm
1
2
(v A vB )
(0 v B )
1
2
vB
Since B moves away from A, the center of mass moves away from A at half the speed of
B.
14. The rotational kinetic energy is given by
K
I
1
2
I
2K
2
2
2(50 J)
(30 rad/s) 2
0.11 kg m 2
15. The figure is
x1 = 2 m
x2 = 1 m
m1 = 3 kg
m2 = 7 kg
The net torque is
1
2
m1gx1 m2 gx2
(3 kg )(9.8 m/s 2 )(2 m) (7 kg )(9.8 m/s 2 )(1m)
Since the torque is negative, it is in the clockwise direction.
16. For a system in equilibrium, we have the net force equal to zero,
Fx
0 and
And the net torque must vanish as well,
8
Fy
0
9.8 N m
0
FW
mg
N
f
Using the rotational equilibrium condition and taking torques at the foot of the ladder,
0
W
FW L cos
mg
0
mg
L
sin
2
0
mg tan
2
(10 kg )(9.8 m/s 2 ) tan 25
2
23 N
FW
The x-component equation gives,
FW
Fx
0
f
0
f
23 N
17. The rotational inertia for a solid sphere is
I
2
5
MR2
2
5
(1.5 kg)(0.30 m)2
Newton’s second law for rotational motion is
I
9
0.054 kg m2
8.1 N m
0.054 kg m 2
I
150 rad/s 2
18. Using conservation of energy
K1 U1
K2 U 2
0 U1
K2 0
The ball starts from rest and ends up at the bottom of the hill.
0 U1
K2 0
1
2
mgy1
mv 2
2
1
2
I
2
2
For a spherical shell,
2
3
I
mR 2
The condition for rolling is v = R. Substituting into the energy equation,
mgy1
1
2
mv 2
2
1
2
mv 2
1
2
5
6
I
2
1
2
( 23 mR 2 )(v2 / R) 2
mv 2
2
1
3
mv 2
mv 2
2
2
2
6(9.8 m/s 2 )(20 m)
5
6 gy1
5
v2
2
1
2
15 m/s
19. The definition of angular momentum is
L
I
I
50 kg m 2 /s
300 rad/s
L
0.166 kg m 2
For a ring, I = MR2. Solving for R,
R
I
M
0.166 kg m 2
0.5 kg
0.58 m
20. Angular momentum is conserved, L1 = L2. Using the definition of angular momentum,
I1
1
I2
10
2
For a solid sphere,
2
MR2
5
I
The angular velocity is the angular displacement divided by the time,
t
When the time interval is equal to the period of the rotation (T) the angular displacement
is 2 radians
2 rad
T
The conservation of angular momentum relation becomes,
I1
2
2
MR1
5
1
I2
2 rad
T1
2
2
2
MR2
5
2 rad
T2
2
T2
T1
R2
R1
T2
2 R1
T1
R1
2
(30 days)(4)
120 days
11