oct. 5, 2001 dr. larry dennis, fsu department of physics1 physics 2053c – fall 2001 chapter 7...
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Oct. 5, 2001Dr. Larry Dennis,
FSU Department of Physics 1
Physics 2053C – Fall 2001
Chapter 7Linear Momentum
Conservation of Momentum
2
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????
3
Momentum
p = mv
1. Momentum = mass * velocity
2. Momentum is a vector.
3. It is parallel to the velocity.
4
Equivalent Formulation of Newton’s Second Law
Δt
pΔF
The rate of change of momentum of a body is equal to the net force applied to it.
5
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.
amΔt
vmΔΔt
vΔmΔt
pΔF
6
Newton’s Second Law & Momentum
When there is a net force: Momentum changes. p = Ft = Impulse
When there is no net force: Momentum remains constant. p = Ft = 0
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. 1. Find the impulse imparted to the ball.
Impulse = Ft = p = mvf – mvo
Impulse = m(vf – vo) = 0.05kg * (45 – 0)m/s
Impulse = 2.25 kg-m/s = 2.25 N-s
8
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 = Ft ( = 2.25 N-s )
Impulse/t = F
F = 2.25 N-s/5.0x10-3 s = 450 N
9
Application to CollisionsWhen p = 0 then:
Momentum Before Collision = Momentum After Collision
Mathematically this means:
M1V1b + M2V2b = M1V1a + M2V2a
10
Types of Collisions
An Elastic Collision – Kinetic Energy does not change.
An Inelastic Collision – Kinetic Energy changes
11
Inelastic Collision: CAPA 4
Momentum Before Collision = Momentum After Collision
MtVtb + MsVsb = MtVta + MsVsa Vtb
Vta = Vsa = 0
Vsb
MtVtb + MsVsb = (Mt+Ms) 0
MtVtb = - MsVsb Vsb = -Vtb*Mt/Ms
12
Inelastic CollisionMomentum Before Collision = Momentum After
Collision
M1V1b + M2V2b = M1V1a + M2V2a V1b = Vb
V1a = V2a = Va
V2b = 0
M1Vb + M20 = (M1+M2)Va
15
Elastic Collision in the CM
U1b U2b U1aU2a
M1U1b + M2U2b = 0 M1U1a + M2U2a = 0
½M1U2
1b + ½M2U2
2b = ½M1U2
1a + ½M2U2
2a
16
Elastic Collision in the CM
M1U1b + M2U2b = 0 M1U1a + M2U2a = 0
½M1U2
1b + ½M2U2
2b = ½M1U2
1a + ½M2U2
2a
Solution – Mirror Image: U1a = -U1b
U2a = -U2b
17
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.
4.23 m/s3.88 m/s
18
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
3.88 m/s
19
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
U1bU2b U1a U2a
20
Elastic Collision in the CM: CAPA #6-9
3. Set U1a = -U1b and U2a = -U2b
4. Calcuate V1a and V2a
A. V1a = U1a + Vcm and V2a = U2a + Vcm
U1a = - U1b = -0.192 m/s
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
U1bU2b U1a U2a
3.85 m/s4.20 m/s
21
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.
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