PHY2053 Summer 2012
Exam 2
Solutions
1.
The free-body diagram for the block is
N
F2
37o
F1
fk
mg
Using Newton’s second law for the x-components
∑F
x
= ma x
F1 + F2 cos 37o − f k = 0
f k = F1 + F2 cos 37 o = (10 N) + (15 N) cos 37o = 22 N
The work done by kinetic friction
W = f k ∆r cos θ = (22 N)(6 m) cos180o = −130 N
2.
Mechanical energy is conserved
U1 + K1 = U 2 + K 2
mgy1 + 12 mv1 = mgy2 + 12 mv2
2
gy1 + 12 v1 = gy2 + 12 v2
2
2
2
v2 = v1 + 2 g ( y1 − y2 ) = (3 m/s) 2 + 2(9.8 m/s 2 )(60 m − 30 m) = 24 m/s
2
3.
Mechanical energy is not conserved since the sled stops moving.
Wnc = ∆K + ∆U
= ( K f − K i ) + (U f − U i )
= (0 − 0) + (0 − mgyi )
= −mgyi = −(45 kg + 15 kg )(9.8 m/s 2 )(15 m) = −8800 J
Friction does the dissipative work
Wnc = f k ∆r cos θ
fk =
4.
Wnc
− 8800 J
=
= 200 N
o
∆r cos180
(44 m)(−1)
The force information gives the force constant for the spring
F = kx
k=
80 N
F
=
= 400 N/m
x 0.20 m
Mechanical energy is conserved as the ball exits the gun
U 1 + K1 = U 2 + K 2
1
2
kx1 + 12 mv1 = 12 kx2 + 12 mv2
2
1
2
2
2
kx1 + 0 = 0 + 12 mv2
2
v2 = x1
5.
2
2
k
400 N/m
= (0.20 m)
= 30 m/s
m
0.018 kg
The work done by the engine increases the car’s kinetic energy.
W = ∆K
= K f − Ki
= 12 mv f − 12 mvi
2
2
= 12 (1000 kg )(40 m/s) 2 − 12 (1000 kg )(10 m/s) 2 = 7.5 × 105 J
The power output by the engine rate of the work done by the engine
P=
6.
W 7.5 × 105 J
=
= 7.5 × 10 4 W
t
10 s
The force is found from the impulse-momentum theorem. Before the collision
6 m/s
After the collision
2 m/s
The impulse-momentum theorem is
r r
∆p = Fav ∆t
Since this is a vector equation, we must take components. Since the motion is only along
the x-axis, only the x-component is needed.
∆p x = Fav , x ∆t
The change in momentum is
∆p x = mv fx − mvix = (1.5 kg )(2 m/s) − (1.5 kg )(−6 m/s) = 12 kg ⋅ m/s
The average force is
Fav , x =
7.
∆p x 12 kg ⋅ m/s
=
= 60 N
∆t
0.20 s
Linear momentum is conserved since the explosion is an internal force. Before the
explosion
40 m/s
pi = mvi = (15 kg)(40 m/s) = 600 kg ⋅ m/s
After the explosion
40 m/s
v1f
m2
m1
p f = m1v1 f + m2v2 f
Since linear momentum is conserved,
pi = p f
pi = m1v1 f + m2v2 f
v1 f =
8.
pi − m2v2 f
m1
=
(600 kg ⋅ m/s) − (5 kg )(60 kg ⋅ m/s)
= 30 m/s
10 kg
Use the equations for a one dimensional elastic collision derived in lecture
5 m/s
2 kg
2 m/s
3 kg
 2m1 
 m − m1 
v1i +  2
v2i
v2 f = 
m
+
m
m
+
m
 1
2 
 1
2 
 2(2 kg ) 
 3 kg − 2 kg 
(5 m/s) + 
(2 m/s) = 4.4 m/s
= 
 2 kg + 3 kg 
 2 kg + 3 kg 
9.
Momentum is conserved in the collision. Before the collision
v1 =15 m/s
m1
=1200 kg
v2 =10 m/s
m2 =
1500
kg
For the x-component
pix = m1v1 = (1200 kg )(15 m/s) = 1.8 × 10 4 kg ⋅ m/s
For the y-component
piy = m2v2 = (1500 kg)(10 m/s) = 1.5 × 104 kg ⋅ m/s
After the collision
v
θ
m =2700 kg
For the x-component
p fx = mv cos θ
And the y-component
p fy = mv sin θ
Using the conservation of linear momentum
pix = p fx
1.8 × 10 4 kg ⋅ m/s = mv cos θ
And
piy = p fy
1.5 × 10 4 kg ⋅ m/s = mv sin θ
There are two equations
1.8 × 10 4 kg ⋅ m/s = mv cos θ
1.5 × 10 4 kg ⋅ m/s = mv sin θ
To solve for v directly, square the equations and add them together.
(mv cos θ ) 2 + (mv sin θ ) 2 = (1.8 × 10 4 kg ⋅ m/s) 2 + (1.5 × 10 4 kg ⋅ m/s) 2
(mv) 2 (cos 2 θ + sin 2 θ ) = 5.49 × 108 kg 2 ⋅ m 2 /s 2
v=
10.
5.49 × 108 kg 2 ⋅ m 2 /s 2
5.49 × 108 kg 2 ⋅ m 2 /s 2
=
= 8.7 m/s
m2
(2700 kg ) 2
The object looks like
The definition of the center of mass is
xcm =
m1 x1 + m2 x2
m1 + m2
The 1 subscript refers to the rod and the 2 subscript refers to the additional mass. Solving
for m2:
xcm =
m1 + m2
m1 + m2
xcm (m1 + m2 ) = m1 x1 + m2 x2
m1 xcm + m2 xcm = m1 x1 + m2 x2
m2 ( xcm − x2 ) = m1 ( x1 − xcm )
m2 = m1
( x1 − xcm )
( xcm − x2 )
Measuring the locations from the left end of the rod, the location of the rod is x1 = 1 m,
the location of the added mass is x2 = 0, and the location of the center of mass is xcm =
0.75 m. So
m2 = m1
11.
( x1 − xcm )
(1 m − 0.75 m)
= (3 kg )
= 1.0 kg
( xcm − x2 )
(0.75 m − 0)
The object consists of two parts. The rotational inertia can be decomposed into
I = I disk + I mass
The rotational inertia of the disk is
I disk = 12 mdisk R 2 = 12 (3 kg )(0.6 m) 2 = 0.54 kg ⋅ m 2
The rotational inertia of the extra mass is
I disk = mr 2 = (1 kg )(0.6 m) 2 = 0.36 kg ⋅ m 2
Finally,
I = I disk + I mass = 0.54 kg ⋅ m 2 + 0.36 kg ⋅ m 2 = 0.90 kg ⋅ m 2
12.
3m
4m
960 N
The forces on the beam are
θ
T
W
The condition for equilibrium is
∑τ = 0
Taking torques about the left end (the hinge)
∑τ = 0
τW + τ T = 0
The torque due to the weight is clockwise. Its value is
τ W = −Wr⊥W = −(960 N)(4 m) = 3840 N ⋅ m
To find the torque due to the tension, we need the angle θ
3m
 = 37o
4m
θ = tan −1 
The torque is counterclockwise,
τ T = +Tr⊥T = T ((4 m) sin 37 o ) = (2.4 m)T
The tension can be found
τW + τ T = 0
− 3840 N ⋅ m + (2.4 m)T = 0
T=
13.
3840 N ⋅ m
= 1600 N
2 .4 m
The rotational inertia of the hoop is
I = MR 2 = (100 kg )(2 m) 2 = 400 kg ⋅ m 2
Its angular acceleration
α=
∆ω (0 − 20 rad/s)
=
= −0.10 rad/s 2
∆t
200 s
We don’t care about the direction of the acceleration. Drop the minus sign. The torque is
∑τ = Iα
τ = Iα = (400 kg ⋅ m 2 )(0.10 rad/s 2 ) = 40 N ⋅ m
14.
The fastest object reaches the bottom first. Use energy to find the fastest. Take position
1 at the top of the ramp and position 2 at the bottom of the ramp.
K1 + U1 = K 2 + U 2
0 + mgy = ( 12 mv 2 + 12 Iω 2 ) + 0
The rotational inertia for the shapes can be summarized (like our text does) by
I = β mR 2
For the sphere β = 2/5, the cylinder β = 1/2, and the ring β = 1. Also use ω = v/R in the
energy relation:
v
mgy = mv + β mR  
R
gy = 12 v 2 (1 + β )
2
1
2
v=
2
2
1
2
2 gy
(1 + β )
The largest β will be the slowest. The order will be sphere, cylinder, and ring.
15.
Angular momentum will be conserved.
Li = L f
I iωi = I f ω f
The time for one rotation (T) is related to the angular speed (ω)
ωT = 2π
2π
ω=
T
Substituting
I iωi = I f ω f
Ii
2π
2π
= If
Ti
Tf
T f = Ti
16.
No.
17.
At the depth of 2 m
1I 
= (1.8 s) 2 i  = 0.90 s
Ii
 Ii 
If
P2 = P1 + ρgd = (1.01 × 105 Pa ) + (1000 kg/m 3 )(9.8 m/s 2 )(2 m) = 1.21 × 105 Pa
Double that number and find the depth
P2 = P1 + ρgd
d=
18.
P2 − P1 2(1.21 × 105 Pa) − 1.01 × 105 Pa
=
= 14 m
ρg
(1000 kg/m 3 )(9.8 m/s2 )
From the density and the mass the volume is found
ρ=
V=
m
V
m
ρ
=
15 kg
= 5.0 × 10− 3 m3
3
3000 kg/m
Use Archimedes’ principle to find the buoyant force
FB = ρ f gV = (1000 kg/m 3 )(9.8 m/s 2 )(5.0 × 10−3 m3 ) = 49 N
19.
Call the position at the bottom of the pipe 1 and the position at the top of the pipe 2.
Applying Bernoulli’s equation
P1 + ρgy1 + 12 ρv1 = P2 + ρgy2 + 12 ρv2
2
2
The pipe’s diameter does not change so
A1 = A2
By the continuity equation
A1v1 = A2v2
v1 = v2
Since the end of the pipe is exposed to the atmosphere P2 = Patm. Heights are measured
from the lowest point so y1 =0. Making these substitutions
P1 + ρgy1 + 12 ρv1 = P2 + ρgy2 + 12 ρv2
2
2
P1 + 0 + 12 ρv1 = Patm + ρgy2 + 12 ρv1
2
2
P1 = Patm + ρgy2 = 1.01 × 105 Pa + (1000 kg/m 3 )(9.8 m/s2 )(6 m) = 1.60 × 105 Pa
20.
Poiseuille’s law is
∆V π ∆P / L 4
=
r
∆t 8 η
 (8)(1.0 × 10−3 Pa ⋅ s 
 ∆V  8η
−2
3
 = 150 Pa/m
∆P / L = 
 4 = (2.30 × 10 m /s)
4
 ∆t  πr
 π (0.025 m)
