Download 1. a) Give the formula for the linear momentum of an object

First Name:
Third Exam PHYS 1114
1.
Last Name:
5:30–6:20 PM, April 11, 2017
a) Give the formula for the linear momentum of an object of mass m
and velocity v, together with its SI unit.
b) Formulate the principle of conservation of linear momentum in one
sentence.
c) Repeat this in formula form for the case of masses m1 and m2
colliding with initial velocities v1 , v2 and final velocities v10 , v20 .
d) A bullet of mass 10 g is shot into a wood block of mass 1.0 kg,
which can move without friction and which is originally at rest.
Find the original speed of the bullet, if the collision is completely
inelastic and the wood block moves 18 m/s after the hit.
e) If the collision were a one-dimensional elastic collision, which two
equations would you use to solve this problem? (Give equations
only!)
First Name:
Third Exam PHYS 1114
Last Name:
5:30–6:20 PM, April 11, 2017
2. A ball of mass M = 0.300 kg and radius R = 0.200 m rolls down a slope
without slipping, starting from rest at a height of h = 10.0 m above the
bottom of the slope. What is the speed v of the ball at the bottom?
I = 23 M R2
Hollow sphere
I = 25 M R2
Solid sphere
I = M R2
Hollow cylinder
I = 12 M R2
Solid cylinder
a) Give the equation of conservation of mechanical energy in this case,
equating the initial potential energy and the final kinetic energies
in terms of linear speed v and angular speed !.
b) Give the equation relating v and !.
c) Calculate the moment of inertia of the ball for the case of a solid
ball.
d) Calculate the final speed of this ball.
e) Calculate the final linear momentum p and angular momentum L
of the ball.
First Name:
Third Exam PHYS 1114
3.
Last Name:
5:30–6:20 PM, April 11, 2017
a) What is torque in formula form?
b) What does each letter stand for in this equation? What are the
corresponding SI units?
c) A tricycle with a weight of 150 N is at rest on the floor. The two
rear wheels are 0.750 m behind the front wheel, and the tricycle’s
center of gravity is 0.500 m behind the front wheel. Draw a freebody diagram, indicating the relative position of the wheels, the
center of gravity, the weight, and the normal forces on the wheels.
d) Determine the normal force exerted by the floor on the front wheel.
You can take the position of the rear wheels as the axis of rotation
(pivot) in this calculation.
e) Determine the normal forces on each of the two rear wheels.
First Name:
Third Exam PHYS 1114
Last Name:
5:30–6:20 PM, April 11, 2017
4. A man of mass M = 75.0 kg wants to bungee jump using a rubber
bungee cord of length L = 15.0 m with cross area A determined by
radius r = 0.800 cm. For the kind of rubber used the spring constant
is k = 670 N/m. In the following ignore the mass of the cord.
F
L
a) Using
= Y
, calculate Young’s modulus Y , by comparing
A
L
this with Hooke’s law.
b) By how much will the cord be stretched if its top end is connected
to a bridge and the man is attached to the bottom end and not
moving?
c) After the man jumps from the bridge, what is his velocity the
moment the cord is no longer slack and starts to stretch?
d) Give an equation determining L at the lowest point where the
velocity becomes zero, using energy conservation.
e) Calculate this
L numerically.
First Name:
Last Name:
Third Exam PHYS 1114
1.
Answer Sheet
5:30–6:20 PM, April 11, 2017
a) Give the formula for the linear momentum of an object of mass m
and velocity v, together with its SI unit.
p = mv,
N s = kg m/s
b) Formulate the principle of conservation of linear momentum in one
sentence.
In an isolated system, the total momentum is conserved.
c) Repeat this in formula form for the case of masses m1 and m2
colliding with initial velocities v1 , v2 and final velocities v10 , v20 .
m1 v1 + m2 v2 = m1 v10 + m2 v20
d) A bullet of mass 10 g is shot into a wood block of mass 1.0 kg,
which can move without friction and which is originally at rest.
Find the original speed of the bullet, if the collision is completely
inelastic and the wood block moves 18 m/s after the hit.
v10 = v20 = v 0 = 18 m/s,
m1 v1 = (m1 +m2 )v
0
=)
v2 = 0
(m1 + m2 )v 0
v1 =
= 1.8 ⇥ 103 m/s
m1
e) If the collision were a one-dimensional elastic collision, which two
equations would you use to solve this problem? (Give equations
only!)
m1 v1 + m2 v2 = m1 v10 + m2 v20
v20 v10 = v1 v2
Note:
1
2
2 m1 v1
+ 12 m2 v22 = 12 m1 v10 2 + 12 m2 v20 2
is also acceptable as second equation.
First Name:
Third Exam PHYS 1114
Last Name:
Answer Sheet
5:30–6:20 PM, April 11, 2017
2. A ball of mass M = 0.300 kg and radius R = 0.200 m rolls down a slope
without slipping, starting from rest at a height of h = 10.0 m above the
bottom of the slope. What is the speed v of the ball at the bottom?
I = 23 M R2
Hollow sphere
I = 25 M R2
Solid sphere
I = M R2
Hollow cylinder
I = 12 M R2
Solid cylinder
a) Give the equation of conservation of mechanical energy in this case,
equating the initial potential energy and the final kinetic energies
in terms of linear speed v and angular speed !.
M gh = 12 M v 2 + 12 I! 2
b) Give the equation relating v and !.
v = R!
or ! = v/R
c) Calculate the moment of inertia of the ball for the case of a solid
ball.
I = 25 M R2 = 0.00480 kg·m2
d) Calculate the final speed of this ball.
M gh = 12 M v 2 + 12 Iv 2 /R2 = ( 12 M + 12 I/R2 )v 2
s
s
q
M gh
M gh
v=
=
= 10
1
1
1
1
7 gh = 11.8 m/s
2
M
+
I/R
M
+
M
2
2
2
5
e) Calculate the final linear momentum p and angular momentum L
of the ball.
v
p = M v = 3.55 kg·m/s , L = I! = I = 0.284 kg·m2/s
R
First Name:
Last Name:
Third Exam PHYS 1114
3.
Answer Sheet
5:30–6:20 PM, April 11, 2017
a) What is torque in formula form?
⌧ = F? r = F r? = F r sin ✓
b) What does each letter stand for in this equation? What are the
corresponding SI units?
⌧ = torque,
F = force,
r = arm,
⌧ ! N · m = m · N,
F ! N,
✓ = 6 (F, r)
r!m
c) A tricycle with a weight of 150 N is at rest on the floor. The two
rear wheels are 0.750 m behind the front wheel, and the tricycle’s
center of gravity is 0.500 m behind the front wheel. Draw a freebody diagram, indicating the relative position of the wheels, the
center of gravity, the weight, and the normal forces on the wheels.
F3 F2
F1
0.250m
cg
0.500m
0.750m
mg
d) Determine the normal force exerted by the floor on the front wheel.
You can take the position of the rear wheels as the axis of rotation
(pivot) in this calculation.
F1 ⇥ 0.750 m = (150 N) ⇥ 0.250 m
=)
F1 = 50.0 N
e) Determine the normal forces on each of the two rear wheels.
F1 + F2 + F3 = mg
F2 = F3 = 12 (mg
F1 ) = 12 (150 N
=)
50.0 N) = 50.0 N
Also: 2F2 ⇥ 0.750 m = (150 N) ⇥ 0.500 m, etc., with pivot at F1 .
First Name:
Last Name:
Third Exam PHYS 1114
Answer Sheet
5:30–6:20 PM, April 11, 2017
4. A man of mass M = 75.0 kg wants to bungee jump using a rubber
bungee cord of length L = 15.0 m with cross area A determined by
radius r = 0.800 cm. For the kind of rubber used the spring constant
is k = 670 N/m. In the following ignore the mass of the cord.
F
L
= Y
, calculate Young’s modulus Y , by comparing
A
L
this with Hooke’s law.
a) Using
F =k
L=
AY
L
L =) Y =
kL
kL
= 2 = 5.00 ⇥ 107 N/m2
A
⇡r
b) By how much will the cord be stretched if its top end is connected
to a bridge and the man is attached to the bottom end and not
moving?
Mg = k
Leq =)
Leq =
Mg
= 1.0970 m = 1.10 m
k
c) After the man jumps from the bridge, what is his velocity the
moment the cord is no longer slack and starts to stretch?
p
M gL = 12 M v 2 =) v = 2gL = 17.1 m/s
d) Give an equation determining L at the lowest point where the
velocity becomes zero, using energy conservation.
M g(L +
L) = 12 k
L2 =)
L2
2
Mg
L
k
2
Mg
L=0
k
L2 2 Leq L 2 Leq L = 0 (Only first equation needed.)
=)
e) Calculate this
L=
L numerically.
q
2 Leq ± 4 L2eq + 8L Leq
2
(Minus sign is not acceptable.)
=
Leq ±
=)
q
L2eq + 2L Leq
L = 6.94 m