Phys 221 SI :
Momentum
Supplemental Instruction
Iowa State University
Leader:
Course:
Instructor:
Date:
Michael
Phys 221
Prell
02/26/2012
Useful Equations
dp
⃗ NET = ⃗ NET
F
dt
⃗
⃗
✔ Impulse : J =Δ ⃗p =∫ F NET dt
✔ New Vocabulary :
◦ Conservation of Linear Momentum : Δ p=0 whenever the net force on the
system is zero.
◦ Elastic Collision : Kinetic energy is conserved during the collision.
◦ Inelastic Collision : Kinetic energy is NOT conserved during the collision.
◦ Completely Inelastic Collision : The objects stick together upon colliding. (i.e.
leave the collision with the same velocity vector)
✔ Momentum :
⃗p =m ⃗v ;
Problems :
1. (*) A raft is floating on some still water. If a person nearby decided to roll a ball across
the raft (eventually falling off the other side) what (if anything) would happen to the raft?
Is momentum conserved? Is kinetic energy conserved? Why? (Friction is NOT
negligible)
• Momentum is conserved, because there is no net external force (normal vs gravity etc
cancels out).
• Kinetic energy is not conserved because friction is present. However if the ball is
rolling and not sliding then kinetic energy is conserved (static friction is not NONconservative)
• The raft would be pushed in the opposite direction of the ball by friction.
2. (**) An astronaut of mass 100 kg can throw a baseball at 35 m/s. While repairing a piece
of the space-station he begins to drift away from the station at 1 m/s. If he is carrying a
wrench of mass 1 kg, can he make it back to the station?
• In order to make it back to the station he must not only throw the wrench with enough
momentum to stop himself, but he must also reverse direction.
• The astronaut has momentum p i=mA v A . If the astronaut throws the wrench away
from the space-station the final momentum will be p f =mA v f +mW v W ≈mW v W . If
our goal is that this vf be just barely negative, we can approximate it by zero.
• Because we are in space and there are no external forces, momentum is conserved
and p i= p f → m A v A=m W v W , so the speed at which he must throw the wrench
m
100 kg
is v W = A v A=
(1 m/ s)=100 m/s
mW
1 kg
• The astronaut cannot throw nearly that fast – and so he is screwed.
Supplemental Instruction
1060 Hixson-Lied Student Success Center v 294-6624 v www.si.iastate.edu
3. (***) Billiard balls collide elastically. We shoot a ball into a group of three others – one
travels in the direction of the initial ball, the other two move apart at 30o from the initial
(in opposite directions). What is the final velocity of the initial ball? Assume that the
three balls that were hit have equal velocities v.
• The balls are all of equal mass.
• Because they collide elastically, kinetic energy is conserved : v 2i =v 2f +3 v 2
• The initial momentum is ⃗p =〈m v i ,0 〉 (note I am assuming that the positive x
direction is the direction of the first ball)
p⃗ f =〈 m v+2 m v cos θ+m v f , m v sin θ−m v sin θ〉
•
• The y component here is useless, but the x gives us v i =v+2 v cos θ+v f
v −v f
•
v= i
1+2 cos θ
2
v i −v f
3
2
2
v −v f )2
v i =v f +3
•
→ ( v i −v f ) ( v i+v f )=
2 ( i
1+2 cos θ
(1+2 cos θ)
3
3
3
v i +v f =
(v i−v f ) → v f 1+
•
=
−1 v i
2
2
(1+2 cos θ)
(1+2 cos θ)2
(1+2 cos θ)
(1+2 cos θ)2+3
3−(1+2 cos θ)2
3
3
•
1+
=
−1=
(1+2 cos θ) 2
(1+2 cos θ) 2
(1+2 cos θ)2
(1+2 cos θ) 2
3−(1+2 cos θ) 2 (1+2 cos θ)2
•
vf=
vi
(1+2 cos θ)2 3+(1+2 cos θ) 2
• This is messy, but because we know the angle is 30o, (1+2 cos( 30o ))2 =7.464
3−7.464
vf=
v ≈− 0.4266 v i
•
3+7.464 i
• Note the sign tells us that it reverses direction.
(
)
(
)(
)
4. (**) While watching a movie you notice that the shotgun used by the protagonist will
propel enemies hit by the shot backwards. If the average sprinting speed for a human is 8
m/s and the average body mass is 100 kg, how fast must your 28 g slugs (the heaviest
available) be traveling to accomplish such a feat? (Also, for extra credit : why do the
antagonists not have equally impressive guns?)
• Again, momentum is conserved, so the initial momentum p i=mv run−mB v B is the
momentum of the enemy and the bullet (which are in opposite directions). The final
momentum is the bullet + enemy system (which is a completely elastic collision) is
backwards, but we'll be conservative and calculate to the point that the enemy is
completely stopped p f =0
m
100 kg
m v run =m B v B → v B=
v =
(8 m/s )≈28,500 m/s=3.8 Mach
•
mB run 0.028 kg
• For the record a shotgun (in reality) has a muzzle velocity on the order of 100-300
m/s.
5. (***) You want to calculate the muzzle velocity of your personal guns, so you build the
following apparatus : A block of mass M sits on a rough surface of coefficient of friction
μk. A bullet (of mass m) is fired into the block. If you measure the distance the block
travels to be Δx, what is the initial velocity of the bullet?
• If the bullet is fired point-blank, then the force due to gravity barely acts on the bullet
and we can assume that momentum is conserved.
• We know that momentum is conserved – so p i=mv i , p f =( m+M ) v f
m+M
vf
p i= p f → v i =
•
m
• If the block is decelerated by a constant friction, then the acceleration of the block
f
N
and bullet is a= =μ =μ g . Recalling our kinematics knowledge,
m
m
v f =√ 2 a Δ x=√ 2 μ g Δ x
m+M
• Thus, v i =
√ 2μ g Δ x
m