Transcript Momentum and impulse

Momentum and impulse
• The linear momentum - p of an object of
mass m moving with velocity v is the product
of its mass and velocity:
p = mv
SI unit: kg m/s
- Doubling the mass of the velocity of an object
double its momentum
- Doubling both quantities, quadruples its
momentum
Momentum is a vector quantity
( quiz 6.1/161)
• Changing the momentum of an object
require application of a force:
Fnet = ma = m Δv/Δt = Δ(mv)/Δt
The change in an object’s momentum Δp
divided by the elapsed time Δt equals
the constant net force Fnet acting on the
object
If a constant force acts on a object. The
impulse I delivered to the object over a
time interval Δt is given by:
I = F Δt
SI unit: kg m/s (ex 6.2/163)
• Conservation of momentum
• A2 /188
• When a collision occurs in an isolated
system, the total momentum of the
system doesn’t change with the passage
of time
• The total momentum is conserved!
• If we consider an isolated system (no
external forces) of two particles before and
after they collide
F21Δt = m1v1f-m1v1i
F12 Δt = m2v2f-m2v2i
F21= -F12
F21 Δt= -F12 Δt
m1v1f-m1v1i = -(m2v2f-m2v2i)
m1v1i +m2v2i= m1v1f+m2v2f
When no net forces acts
on the system, the total
momentum remains ct.
in time (conservation of
momentum)
• Collisions
• For any type of collision, the momentum of
the system just before the collision= to the
momentum after the collision as long as the
system is isolated
• Elastic collision- both momentum and
kinetic energy are conserved
• Inelastic collision- momentum is conserved
but kinetic energy is not
• Perfect inelastic collision, momentum is
conserved, KE is not, and the 2 objects stick
together after the collision (quiz 6.3/168)
• Perfectly inelastic collision:
m1v1i +m2v2i =(m1+m2)vf
vf=(m1v1i +m2v2i )/(m1+m2)
(quiz 6.6/172)
• Elastic collision:
m1v1i +m2v2i= m1v1f+m2v2f
m1(v1i-v1f) = m2(v2f-v2i)
1/2m1v1i2+1/2 m2v2i2=1/2
m1v1f2=1/2m2v2f2
m1(v1i2-v1f2)=m2(v2f2-v2i2)
m1(v1i-v1f)(v1i+v1f)= m2(v2fv2i)(v2f+v2i)
v1i+v1f =v2f+v2i
v1i-v2i = -(v1f-v2f)
• One dimensional collision; pb. Strategies
- Choose coordinate axis (along direction of
motion)
- Diagram
- Conservation of momentum: write total
momentum before and after
- Conservation of energy: write the
expression (for elastic or perfect inelastic)
- Solve
(ex. Page 182-188)
Glancing collisions
• For a general collision in 3D, the
conservation of momentum principle
implies that total momentum of the system
in each direction is conserved
m1v1ix +m2v2x= m1v1fx+m2v2fx
m1v1i y+m2v2iy= m1v1fy+m2v2fy
Glancing collision: if an object with mass
m1 collides with a mass m2 that is initial at
rest, after collision, obj.1 moves at an
angleθ with horizontal, and obj.2 moves at
an angleΦ with horizontal
x component:
m1v1i + 0 = m1v1f cosθ+m2v2f cosΦ
y component:
0 + 0 = m1v1f sinθ+m2v2f sinΦ
Conservation of energy:
1/2m1v1i2+0 =1/2 m1v1f2+1/2m2v2f2
(if the collision is inelastic, the KE of the
system is not conserved, the 3rd equation
does not apply)
• Two dimensional collision; pb. Strategies
- Choose coordinate axis (use both x, and
y- coordinates)
- Diagram
- Conservation of momentum: write total
momentum before and after
- Conservation of energy: write the
expression (for elastic or perfect inelastic)
- Solve
Rocket Propulsion (page 177)
• Equating the total initial momentum of the
system with the total final momentum:
(M + Δm)v =M(v+ Δv) + Δm(v-ve)
ve –fuel ejected speed
M Δv = - ve Δm
but Δm =- ΔM
M Δv = - ve ΔM
(calculus:) vf-vi =veln(Mi/Mf)
• The thrust of the rocket = the force everted
on the rocket by the ejected exhaust
gases
instantaneous thrust= Ma
= M Δv/ Δt
= |ve ΔM/ Δt|