PHY2053
Summer 2012
Exam 1 Solutions
1.
The period is proportional to the square root of its length
T∝ L
Forming a ratio:
T2
=
T1
L2
L1
We want T2 = 1.5 T1. Substituting
1.5T1
=
T1
L2
L1
L2
= 1 .5
L1
L2
= (1.5)2 = 2.25
L1
miles
miles
1 hour
1 minute 5280 feet
feet
= 60
×
×
×
= 88
hour
hour 60 minutes 60 seconds
1 mile
second
2.
60
3.
The definition of average velocity is
vx , av =
∆x
∆t
where ∆x is the total distance traveled. For the first part of the trip,
∆x1 = v1∆t = 70
miles 3
× hour = 52.5 miles .
hour 4
For the second part,
∆x2 = v2 ∆t = 50
miles 1
× hour = 25 miles
hour 2
Overall,
1
vx , av =
4.
∆x 52.5 miles + 25 miles
=
= 62 miles/hour
∆t 0.75 hour + 0.5 hour
Use
v fx − vix = 2a∆x
2
2
v fx − vix
2
a=
5.
2∆x
2
(15 m/s) 2 − (9 m/s) 2
=
= 3.6 m/s 2
2(20 m)
First find the time it takes for the cage to hit the ground.
∆y = viy ∆t + 12 a y (∆t )2
= 0 − 12 g (∆t )2
− 2∆y
− 2(−25 m)
=
= 2.26 s
g
9.8 m/s 2
∆t =
Princess needs to run 3 m in that time.
∆x = vix ∆t + 12 a x (∆t ) 2
= 0 + 12 a x (∆t )2
ax =
6.
2∆x
2(3 m)
=
= 1.18 m/s2
2
(∆t )
(2.26 s) 2
Solve for the initial speed.
∆y = viy ∆t + 12 a y (∆t ) 2
= viy ∆t − 12 g (∆t ) 2
∆y g∆t − 37 m (9.8 m/s2 )(3.7 s)
viy =
+
=
+
= 8.13 m/s
∆t
2
3 .7 s
2
7.
Find the components of each vector. For the tee shot
Ax = A sin 30o = (150 yards) sin 30o = 75 yards
Ay = A cos 30o = (150 yards) cos 30o = 130 yards
For the second shot
2
Bx = − B sin 45o = −(120 yards) sin 45o = −84.9 yards
By = B cos 45o = (120 yards) cos 45o = 84.9 yards
r
B
r
C
r
A
Add like components
C x = Ax + Bx = (75 yards) + (−84.9 yards) = −9.9 yards
C y = Ay + By = (130 yards) + (84.9 yards) = 215 yards
The calculator gives
 215 
o
 = −87
−
9
.
9


θ = tan −1 
Since the x-component is negative, we add 180o to get 93o or 3o W of N.
8.
We have
r r r
A+B+C=0
Taking components
Ax + Bx + Cx = 0
Ay + By + C y = 0
Solve for Cx and Cy
C x = − Ax − Bx
C y = − Ay − By
r
r
We need the components of A and B :
Ax = A cos θ A = (50 N ) cos 30o = 43.3 N
Ay = A sin θ A = (50 N) sin 30o = 25.0 N
3
Bx = B cos θ B = (70 N ) cos 90o = 0
By = B sin θ B = (70 N) sin 90o = 70.0 N
r
Finding the components of C ,
C x = − Ax − Bx = −(43.3 N ) − 0 = −43.3 N
C y = − Ay − By = −(25.0 N) − (70.0 N ) = −95.0 N
r
The magnitude of C
C = C x + C y = (−43.3 N )2 + (−95.0 N ) 2 = 104 N
2
2
From the calculator
C 
 − 95.0 N 
 = 65o
tan −1  y  = tan −1 
C
−
43
.
3
N


 x
But since Cx is negative, add 180o to get 245o.
9.
The relative velocity equation for this situation is
r
r
r
v TC = v TG + v GC
The velocity of the ground relative to the car is the negative of the car relative to
the ground, and
r
r
r
v TC = v TG − v CG
The x-component equation
(vTC ) x = (vTG ) x − (vCG ) x = −vTG sin 35o − 0 = −(85 km/hr) sin 35o = −48.8 km/hr
The y-component equation
(vTC ) y = (vTG ) y − (vCG ) y = vTG cos 35o − vCG = (85 km/hr) sin 35o − (110 km/hr)
= 40.4 km/hr
The velocity is
vTC = [(vTC ) x ]2 + [(vTC ) y ]2 = [−48.8 km/hr]2 + [40.4 km/hr]2 = 63 km/hr
4
10.
If a projectile is shot horizontally, viy = 0. All objects thrown horizontally will
take the same amount of time to reach the ground. None of these choices will
make stone stay in the air any longer.
11.
The motion in the x direction gives the initial velocity
∆x = vix ∆t
vix =
∆x 105 m
=
= 25 m/s
∆t
4 .2 s
vix = vi cos 25o
vix
25.0 m/s
=
= 27.6 m/s
o
cos 25
cos 25o
vi =
The x-component of the velocity is unchanged since ax = 0. The y-component of
the velocity is found
v fy − viy = a y ∆t
v fy = viy − g∆t = vi sin 25o − g∆t = (27.6 m/s) sin 25o − (9.8 m/s 2 )(4.2 s)
= −29.5 m/s
The final speed
v f = v fx + v fy = (25.0 m/s) 2 + (−29.5 m/s) 2 = 38.6 m/s
2
2
12.
The forces act on different systems. The chair starts to move because my force is
greater than friction.
13.
The free-body diagram is
r
FP
r
FW
Newton’s second law gives
∑F
x
= max
FP − FW = ma
a=
FP − FW 600 N − 200 N
=
= 4 m/s 2
m
100 kg
5
14.
The free-body diagram is
N
Fext
fk
35o
Mg
The x-component
∑F
x
= Max
Fext cos 35o − f k = Ma
The y-component
∑F
y
= Ma y
N + Fext sin 35o − Mg = 0
There are three unknowns, N, Fext, and fk. We need another equation.
f k = µk N
Solve the x-component equation for fk
f k = Fext cos 35o − Ma
and the y-component equation for N
N = Mg − Fext sin 35o
Now substitute into the definition of µk
Fext cos 35o − Ma = µk ( Mg − Fext sin 35o )
= µk Mg − µ k Fext sin 35o
Fext cos 35o + µk Fext sin 35o = Ma + µ k Mg
6
Fext =
M (a + µk g )
cos 35o + µk sin 35o
(3 kg )[1 m/s2 + (0.4)(9.8 m/s2 )]
=
cos 35o + (0.4) sin 35o
= 14.1 N
15.
The free-body diagram is (assuming the block slides down the incline)
r
N
θ
r
f
θ
r
W
Taking the x-axis along the incline
∑F
x
= ma x
f − W sin θ = 0
f = W sin θ
and the y-axis perpendicular to the incline
∑F
y
= ma y
N − W cos θ = 0
N = W cos θ
Since the block is sliding, use the coefficient of kinetic friction
f k = µk N
W sin θ = µ kW cos θ
tan θ = µ k
θ = tan −1 µ k
= tan −1 (0.4)
= 21.8o
7
16.
The free-body diagram for the left block is
r
N
r
T
r
fs
θ
θ
r
W1
Again, take the x-axis along the incline
∑F
x
= m1ax
T − f s − W1 sin θ = 0
f s = T − W1 sin θ
Along the y-axis,
∑F
y
= m1a y
N − W1 cos θ = 0
N = W1 cos θ
The free-body diagram for the other block is
r
T
r
W2
Using Newton’s second law with this free-body diagram
∑F
y
= m2 a y
T − W2 = 0
T = W2
8
The three equations are
f s = T − W1 sin θ
N = W1 cos θ
T = W2
Plug the last equation into the first
f s = W2 − W1 sin θ
N = W1 cos θ
The definition of the coefficient of static friction
f s ≤ µs N
W2 − W1 sin θ 2 ≤ µ sW1 cos θ
Since the blocks have the same mass, W1 = W2 and they cancel
1 − sin θ ≤ µ s cos θ
1 − sin θ
cos θ
1 − sin 37o
µs ≥
cos 37o
µ s ≥ 0.50
µs ≥
17.
The angular speed is defined as
∆θ
∆t
2π rad
1 day
1 hour
1 minute
=
×
×
×
1 day 24 hours 60 minutes 60 seconds
ω=
= 7.27 × 10− 5 rad/s
18.
The free-body diagram is
r
F
Earth
Sun
9
Applying Newton’s second law
∑F
r
= M E ar
F = ME
v2
r
The force is found from Newton’s law of gravity
F=
GM E M S
r2
The velocity can be found from the time it takes to complete one orbit
distance
time
2πr
=
T
2π (1.50 × 1011 m) 1 year
1 day
1 hour
=
×
×
×
1 year
365 days 24 hours 3600 s
v=
= 29,900 m/s
The Sun’s mass can be found from Newton’s second law
v2
F = ME
r
GM E M S
v2
=
M
E
r2
r
2
vr
MS =
G
(29,900 m/s) 2 (1.50 × 1011 m)
=
6.67 × 10−11 N ⋅ m 2 /kg 2
= 2.0 × 1030 kg
19.
After 60 revolutions, the angular speed is 16 rev/s. Convert all to radians
∆θ = 60 rev ×
ω = 16
2π rad
= 120π rad
rev
rev 2π rad
rad
×
= 32π
s
rev
s
10
Find the angular acceleration
ω f 2 − ωi 2 = 2α ∆θ
ω f 2 − ωi 2 (32π rad/s) 2 − (0) 2
=
= 13.4 rad/s 2
α=
2∆θ
2(120π rad/s)
The time is found
∆θ = ωi ∆t + 12 α (∆t ) 2
= 0 + 12 α (∆t ) 2
2∆θ
∆t =
20.
α
=
2(68.2 rev) 2π rad
×
= 8 .0 s
13.4 rad/s 2
rev
The radial acceleration is
ar =
v 2 (0.5 m/s) 2
=
= 0.833 m/s 2
r
0 .3 m
The tangential acceleration is
at = αr = (2.5 rad/s 2 )(0.3 m) = 0.750 m/s2
The net acceleration is
a = ar + at = (0.833 m/s 2 )2 + (0.750 m/s 2 ) 2 = 1.12 m/s 2
2
2
11