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Momentum and Collisions

Motion under the action of time-varying force F (t )

Newton’s second law


dv d
F  ma  m
 (m v )
dt dt


dp
F 
dt
Definition of (linear) momentum:


P   m jv j


p  mv
F(t)
F
Δt=t2-t1
Momentum of many-particle system
j
Effect of force on the object’s motion is determined by impulse
t2






p2  p1  p  F t  J   F dt
(Impulse-Momentum Theorem)
t1
Ball hits a wall
mv 2 x  mv 1 x
Fx 

t
0.4kg (20  30)m / s

0.01s
 2000 N
The Principle of Conservation of Momentum
The total momentum of an isolated system remains constant (is conserved).
Proof:


dP
  Fi  0 
dt
i


P   m j v j  const
j
Two-particle collision
Example: Recoil of a rifle
Solution: momentum conservation yields
Pfx = P0x = 0 → mBvBx + mRvRx = 0 → vRx = - mBvBx / mR
vRx = - ( 0.005 kg / 3 kg ) 300 m/s = - 0.5 m/s
KB = (1/2)mB vBx2 = (1/2)mB (mRvRx / mB)2 = KR (mR/mB) >> KR
Elastic and Inelastic Collisions
Elastic collision: total kinetic energy = const
(interaction via only conservative forces)
Inelastic collision: total kinetic energy ≠ const
(nonconservative forces are involved in the interaction:
heating, irreversible deformation, explosion, propulsion, etc.)
Example of completely inelastic collision
Important fact: Conservation of the momentum P does
not depend on the conservation of the kinetic energy K !
Example of the inelastic collision: Ice Skaters
Note: Kf > K0 = 0
(Inelastic collision !!!)
Solution: m1vf1 + m2vf2 = 0 → vf2 = - m1 vf1 / m2
vf2 = - 2.5 m/s ( 54 kg / 88 kg ) = - 1.5 m/s
Exam Example 18: The Ballistic Pendulum (example 8.8, problem 8.43)

ac L


T T
A block, with mass M = 1 kg, is suspended by a
massless wire of length L=1m and, after completely
inelastic collision with a bullet with mass m = 5 g,
swings up to a maximum height y = 10 cm.
Find: (a) velocity v of the block with the
bullet immediately after impact;
(b) tension force T immediately after impact;
(c) initial velocity vx of the bullet.
Solution:
(a) Conservation of mechanical energy K+U=const

(m  M ) g
y
Vtop=0

v
1
1
2
2
K 0  U top  ( m  M )v  ( m  M ) gy  ( m  M )v top
 v  2 gy
2
2 

(b) Newton’s second law ( m  M ) a  ( m  M ) g  2T and a  a  v 2 / L
c
2
mM 
v  (m  M ) g 
2y
yields T 
 g 
 
1 

2
L
2
L 


(c) Momentum conservation for the collision
mM
mM
mv x  ( m  M )v  v x 
v
m
m
2 gy
Two-body Elastic Collision
General property of all elastic collisions:
In an elastic collision the relative velocity of two bodies
has the same magnitude before and after the collision.
In 1-D elastic collision the relative velocity of two bodies changes its sign:
vB2x – vA2x = - ( vB1x – vA1x )
Note: This general property is equivalent to conservation of the total kinetic energy.
The Gravitational Slingshot Effect
Example of how to
give an extra boost
to spacecraft
-(vB2x – vA2x) =
= vB1x – vA1x =
= 20 km/s →
KA2=(29.6/10.4)2 KA1
= 8·KA1
Two-Dimensional Collisions
Exam Example 19: Collision of Two Pendulums
Center of Mass

rcm
It is a point that represents the average location for the total mass of a system

rcm 

m
r
 ii
i
m

 v cm 
i
i

m
v
 ii
i
m
i



P

, P   mi vi
M
i
z
i
nd Law:
Center-of-Mass Dynamics
from
Newton’s
2
x



 ( ext )  (int)
 ( ext )
dvi 
dvcm dP
mi
 Fi  M

  ( Fi  Fi )   Fi
dt
dt
dt
i
i
since
ri
 (int)
 Fi  0
(Newton’s third law)
y

 ( ext )
dP
  Fi
dt
i
 ( ext )
 0)
Conservation of the momentum in the isolated system (  Fi
i




P  const  P  M v cm  const  v cm  const
0
m1
x1
M
m2
xcm
x2
x
xcm
m1 x1  m 2 x 2

m1  m 2
i
Exam Example 20: Head-on elastic collision (problems 8.48, 8.50)
Data: m1, m2, v01x, v02x
Find: (a) v1x, v2x after collision;
(b) Δp1x, Δp2x , ΔK1, ΔK2 ;
(c) xcm at t = 1 min after collision
if at a moment of collision xcm(t=0)=0
Solution: In a frame of reference moving
V01x
y’
V02x
m1
m2
X’
with V02x, we have V’01x= V01x- V02x, V’02x = 0, and
conservations of momentum and energy yield
m1V’1x+m2V’2x=m1V’01x → V’2x=(m1/m2)(V’01x-V’1x)
0
m1V’21x+m2V’22x=m1V’201x→ (m1/m2)(V’201x-V’21x)=V’22x = (m1/m2)2(V’01x- V’1x)2 →
X
V’01x+V’1x=(m1/m2)(V’01x–V’1x)→ V’1x=V’01x(m1-m2)/(m1+m2) and V’2x=V’01x2m1/(m1+m2)
(a) returning back to the original laboratory frame, we immediately find:
V1x= V02x+(V01x– V02x) (m1-m2)/(m1+m2) and V2x = V02x +(V01x– V02x)2m1/(m1+m2)
(a) Another solution: In 1-D elastic collision a relative velocity switches direction
V2x-V1x=V01x-V02x. Together with momentum conservation it yields the same answer.
(b) Δp1x=m1(V1x-V01x), Δp2x=m2(V2x-V02x) → Δp1x=-Δp2x (momentum conservation)
ΔK1=K1-K01=(V21x-V201x)m1/2, ΔK2=K2-K02=(V22x-V202x)m2/2→ΔK1=-ΔK2 (E=const)
(c) xcm = (m1x1+m2x2)/(m1+m2) and Vcm = const = (m1V01x+m2V02x)/(m1+m2)
→ xcm(t) = xcm(t=0) + Vcm t = t (m1V01x+m2V02x)/(m1+m2)
Exam Example 21: Head-on completely inelastic collision
(problems 8.82)
Data: m2=2m1, v10=v20=0, R, ignore friction
Find: (a) velocity v of stuck masses immediately after collision.
(b) How high above the bottom will the masses go after colliding?
Solution: (a) Momentum conservation y
m1v1
1
m1v1  (m1  m2 )v  v 
 v1
m1  m2 3
Conservation of energy: (i) for mass m1 on the
m1

v1

v
h
way to the bottom just before the collision
1
1
x
2
m2
m1 gR  m1v1  v1  2 gR  v 
2 gR
2
3
(ii) for the stuck together masses on the way from the bottom to the top
(b)
2
2
1
v
v
R
2
1
(m1  m2 )v  (m1  m2 ) gh  h 


2
2 g 18 g 9
Rocket Propulsion
Momentum conservation in the co-moving with the rocket frame:
m dv = - vex dm → m dv/dt = - vex dm/dt → thrust F = - vex dm/dt
In the gravity-free outer space for constant vex :
v
m
v ex dm
m0
dv
dm
a

  dv    v ex
 v  v 0  v ex ln
dt
m dt
m
m
v0
m0
“Missing” Momentum and Energy in a β-Decay of Nuclei