Transcript Physics 2053C – Fall 2001

Physics 2053C – Fall 2001
Chapter 7
Linear Momentum
Conservation of Momentum
Oct. 5, 2001
Dr. Larry Dennis,
FSU Department of Physics
1
Conservation of Momentum

Recall:

Generalized Work-Energy Theorem
K1 + U1 + Wnc = K2 + U2
Conservation of Energy means the total
energy doesn’t change.

Conservation of Momentum means the
total momentum doesn’t change.
What’s Momentum????
2
Momentum
p = mv
1. Momentum = mass * velocity
2. Momentum is a vector.
3. It is parallel to the velocity.
3
Equivalent Formulation of Newton’s Second Law

F 

Δp
Δt
The rate of change of
momentum of a body is
equal to the net force
applied to it.
4
Equivalent Formulation of Newton’s Second Law
The rate of change of momentum of a body is equal to the
net force applied to it.

F 

Δp
Δt


Δm v
Δt


mΔv
Δt

 ma
5
Newton’s Second Law & Momentum

When there is a net force: Momentum
changes.


p =  Ft = Impulse
When there is no net force: Momentum
remains constant.

p =  Ft = 0
6
CAPA 1 & 2

A golf ball of mass 0.05 kg is hit off the tee at a speed of 45 m/s.
The golf club was in contact with the ball for 5.0x10-3 s.
1.
Find the impulse imparted to the ball.
Impulse = Ft = p = mvf – mvo
Impulse = m(vf – vo) = 0.05kg * (45 – 0)m/s
Impulse = 2.25 kg-m/s = 2.25 N-s
7
CAPA 1 & 2

A golf ball of mass 0.05 kg is hit off the tee at a speed of 45 m/s.
The golf club was in contact with the ball for 5.0x10-3 s.
2.
Find the average force imparted to the golf ball by the club.
Impulse = Ft ( = 2.25 N-s )
Impulse/t = F
F = 2.25 N-s/5.0x10-3 s = 450 N
8
Application to Collisions
When
p =
0 then:
Momentum Before Collision = Momentum After Collision
Mathematically this means:
M1V1b + M2V2b = M1V1a + M2V2a
9
Types of Collisions
An Elastic Collision – Kinetic Energy does not change.
An Inelastic Collision – Kinetic Energy changes
10
Inelastic Collision: CAPA 4
Momentum Before Collision = Momentum After Collision
MtVtb + MsVsb = MtVta + MsVsa
Vtb
Vsb
Vta = Vsa = 0
MtVtb + MsVsb = (Mt+Ms) 0
MtVtb = - MsVsb  Vsb = -Vtb*Mt/Ms
11
Inelastic Collision
Momentum Before Collision = Momentum After Collision
M1V1b + M2V2b = M1V1a + M2V2a
V1b = Vb
V2b = 0
V1a = V2a = Va
M1Vb + M20 = (M1+M2)Va
12
Inelastic Collisions (cont)
Vb
Va
M1Vb + M20 = (M1+M2)Va
M1Vb = (M1+M2)Va
M1/(M1+M2) Vb = Va
13
Center of Mass Velocity
M1V1b + M2V2b  (M1 + M2 )Vcm
V1b
V2b
U = V - Vcm
U1b
U2b
and V = U + Vcm
14
Elastic Collision in the CM
U1b
U2b
M1U1b + M2U2b = 0
½M1U
2
1b
+ ½M2U
U2a
U1a
2
2b
M1U1a + M2U2a = 0
= ½M1U
2
1a
+ ½M2U
2
2a
15
Elastic Collision in the CM
M1U1b + M2U2b = 0
½M1U
2
1b
+ ½M2U
2
2b
M1U1a + M2U2a = 0
= ½M1U
2
1a
+ ½M2U
2
2a
Solution – Mirror Image:
U1a = -U1b
U2a = -U2b
16
3.88 m/s
4.23 m/s
Elastic Collision in the CM: CAPA #6-9
A pair of bumper cars in an amusement park ride collide
elastically as one approaches the other directly from
the rear. One has mass m1 = 467 kg and the other
mass m2 = 567 kg. If the lighter one approaches at v1
= 4.23 m/s and the other one is moving at v2 = 3.88
m/s, calculate:
• the velocity of the lighter car after the collision.
• the velocity of the heavier car after the collision.
• the change in momentum of the lighter car.
• the change in momentum of the heavier car.
17
3.88 m/s
Elastic Collision in the CM: CAPA #6-9
1. Calcuate Vcm = (M1V1b+M2V2b)/(M1+M2)
2. Calcuate U1b and U2b
1. U1b = V1b – Vcm and U2b = V2b – Vcm
3. Set U1a = -U1b and U2a = -U2b
4. Calcuate V1a and V2a
1. V1a = U1a + Vcm and V2a = U2a + Vcm
18
U1b
U2b
U1a
U2a
Elastic Collision in the CM: CAPA #6-9
1. Calcuate Vcm = (M1V1b+M2V2b)/(M1+M2)
2. Calcuate U1b and U2b
A. U1b = V1b – Vcm and U2b = V2b – Vcm
Vcm = (M1V1b+M2V2b)/(M1+M2)
Vcm = (467*4.23 + 567*3.88)/(467+567) = 4.038 m/s
U1b = V1a - Vcm = 4.23 m/s - 4.038 m/s = 0.192 m/s
U2b = V2b - Vcm = 3.88 m/s - 4.038 m/s = -0.158 m/s
19
U1b
U2b
U1a
U2a
Elastic Collision in the CM: CAPA #6-9
3. Set U1a = -U1b and U2a = -U2b
4.
Calcuate V1a and V2a
A.
U1a = - U1b = -0.192 m/s
V1a = U1a + Vcm and V2a = U2a + Vcm
U2a = -U2b = 0.158 m/s
V1a = U1a + Vcm = -0.192 m/s + 4.038 m/s = 3.846 m/s
V2a = U2a + Vcm = 0.158 m/s + 4.038 m/s = 4.196 m/s
4.20 m/s
3.85 m/s
20
Elastic Collision in the CM: CAPA #6-9
Impulse = mVa – mVb (for each of cars 1 and 2)
Note: Impulse on car 1 = - Impulse on car 2.
21
Next Time



Chapter 7 – Conservation of
Momentum.
Quiz on Chapter 7.
Please see me with any questions or
comments.
See you on Monday.
22