Solutions to Momentum
Classwork and Homework
Warped Warm-Up
Picture if you will...a frog sitting on a frictionless lily­pad. This frog has been a bad frog, experimenting with steroids
and human growth hormone to improve his tongue speed. On this day he shoots his tongue out with such an incredible
velocity that it rips loose and keeps going. If the tongue's velocity is 140 m/s and it's mass is .2 kg, and the frog recoils
at 24 m/s. What is the frog's mass?
What do you ask for when you are all out of mentum?_______________
Before
After
V=?
20 kg
8 kg
8 kg
2.6 m/s
20 kg
Ʃpi = Ʃpf
0 = (8 kg)V + (20 kg)(2.6 m/s)
V = ­ 6.5 m/s
Before
6 slugs
After
m
1.4 ft/s
6 slugs
m
44 ft/s
Ʃpi = Ʃpf
0 = (6 slug)(­1.4 ft/s) + m(44 ft/s)
m = .19 slugs
3. A 76 kg astronaut floats near her capsule, unable to reach it as she has run out of fuel for her jet pack. In
desperation she throws a 0.04 kg wrench at ­18 m/s oppositely to the direction she wishes to go. What is her newly
acquired speed?
Ʃpi = Ʃpf
0 = (.04 kg)(­18 m/s) + (76 kg)V
V = .0095 m/s
Follow­up Q's
c) Is this collision elastic or in­elastic?
d) Is energy conserved in this collision? If not, where did it go?
2­D Mo or not 2­D Mo...
#1
6 m/s
V=?
L
5 kg
A
4 m/s
8 kg
T
3 kg
E
R
Pythagorize it!
Trig it!
θ=?
#2
V=?
L
300 m/s
A
.03kg
5 m/s
T
4kg
E
R
Solution is just like #1.
Answer...V = 5.4 m/s, θ = 65o
θ=?
#3
L
6 ft/s
V'=?
A
2 slug
V=?
T
33o
3.5 slug
E
1.5 slug
R
#4
V=?
L
6 m/s
2kg
θ=?
A
3kg
2kg
T
E
R
3kg
36o
3 m/s
Pythagorize and Trig it...
#5
V=6.8 m/s
.55 kg
L
8.8 m/s
A
5.7 m/s
.55 kg
65o
T
V=?
E
1.4 kg
θ=?
R
1.4 kg
It's done like #4...
answer is V = 4.02 m/s and θ = 54.6o
WHO SAID 2­D MO?
2­D Moe
1. 62.5 m/s @ 35.8o
2. 59o Below Horizontal
3. 25.7 m/s @ 46o
4. 5.86 m/s @ 31.4o
You sank
my battleship.
#1
40 m/s
30kg
35o
30kg
25o
90 m/s
50kg
Horizontal:
50kg
80 m/s
Σpxi= Σpxf
θ
V
(50)(80) ‑ (30)(90cos25 o) = (50)Vx ‑ (30)(40cos35 o)
Vx = 50.7 m/s
Vertical:
Σpyi= Σpyf
0‑(30)(90sin25 o) = ‑(50)Vy + (30)(40sin35 o)
Vy = 36.6 m/s
Pythagorize it and trig it... V = 62.5 m/s and θ =35.8 o
#2
Horizontal:
Σpxi= Σpxf
(.3)(730) + 0 = (.3)(530) + (1.1)V x
Vx = 54.5 ft/s
Vertical:
Σpyi= Σpyf
0‑(1.1)(91) = 0 + (1.1)V y
Vy = ­91 ft/s
Use tangent function ... θ =59 o below horizon
4 m/s
#3
60 kg
g
5 m/s
60 k
18o
4kg
4kg
θ
V
Horizontal:
Σpxi= Σpxf
(60)(5) + 0 = (60)(4cos18 o) + (4)Vx
Vx = 17.9 m/s
Vertical:
Σpyi= Σpyf
0 + 0 = (60)(4sin18 o) + (4)Vy
Vy = ­18.5 m/s
Pythagorize it and trig it...V = 25.7 m/s and θ =46 o
3 kg
#4
4 m/s
3 kg
30o
20o
9 m/s
5 kg
8 m/s
5 kg
Horizontal:
θ
V
Σpxi= Σpxf
(8)(5) ‑ (3)(9cos20 o) = (5)Vx ‑ (3)(4cos30o)
Vx = 5.0 m/s
Vertical:
Σpyi= Σpyf
0 ‑(3)(9sin20 o) = (5)Vy +(3)(4sin30 o)
Vy = ­3.05 m/s
Pythagorize it and trig it...V = 5.86 m/s and θ =31.4 o
Don't Be a
Dummy!
#1
#2
#3
#4
#5
#6
Deja Space Toast Deluxe
Two slices of space toast collide elastically in outer space. Neglect gravity and friction.
a.) Find the unknown angle and velocity.
b.) If the slices are in contact for .1 sec during the collision, what is the average vertical force exerted on
the 3 kg slice during the collision. (Hint: Impulse Momentum Theorem)
4
m
/s
L
g
3k
30o
A
3
kg
T
9
m
/s
20o
E
5 kg
8 m/s
R
5k
g
θ=?
V=
?
ANSWER: a) 5.8 m/s @31o below horizon b) 152 N
V
200 m/s
10 m/s
1 slug
θ
20o
25 N
The disembodied head of Sir Isaac Newton (weight = 25 N on Erf) is shot into outer space by Wayne's punkin
chunker and collides inelastically with a giant space slug (mass = 1 slug). a. What is the final speed and direction of
Issac's disembodied head with the slug stuck to it with slug mucous? b. What impulse did Issac's head receive during
the collision?
ANSWER: a) 22 m/s @28o below horizon
b) ­453 Ns or kgm/s
Deja Space Toast Deluxe
Two slices of space toast collide elastically in outer space. Neglect gravity and friction.
a.) Find the unknown angle and velocity.
b.) If the slices are in contact for .1 sec during the collision, what is the average vertical force exerted on
the 3 kg slice during the collision. (Hint: Impulse Momentum Theorem)
4
m
/s
L
g
3k
30o
A
3
kg
T
9
m
/s
20o
E
5 kg
8 m/s
R
5k
g
θ=?
V=
?
V
200 m/s
10 m/s
1 slug
θ
20o
25 N
The disembodied head of Sir Isaac Newton (weight = 25 N on Erf) is shot into outer space by Wayne's punkin
chunker and collides inelastically with a giant space slug (mass = 1 slug). a. What is the final speed and direction of
Issac's disembodied head with the slug stuck to it with slug mucous? b. What impulse did Issac's head receive during
the collision?
Vy
Vx
(.06ft/s)sin40o
(.03ft/s)cos25o
(.06ft/s)cos40o
Horizontal:
Σpxi= Σpxf
­(.03ft/s)sin25o
(.003)(.06cos40o) + 0 = (.003)(.03cos25o) + (.002)Vx
Vx = .028 ft/s
Vertical:
Σpyi= Σpyf
(.003)(.06sin40o) + 0 = (.002)Vy ‑ (.003)(.03sin25o)
Vy = .077 ft/s
Pythagorize it and trig it... V = .082 ft/s and θ = 70 o
* Conversion... 5.8 slugs = 84.3 kg
a)
Σpxi= Σpxf
(84.3 kg)(6 m/s) + 0 = (84.3 kg)Vf + (1.6 kg)(13.5 m/s)
Vf = 5.74 m/s
b)
I = Δp = pf ­ pi = (1.6 kg)(13.5 m/s) ­ 0 = 21.6 Ns
c) Mr. B receives an equal but opposite Impulse of ­21.6 Ns
d)
e) The pin receives an equal but opposite average force of +1440N