Transcript pp\momentum - Dr. Robert MacKay

Momentum and Collisions
Dr. Robert MacKay
Clark College, Physics
Introduction
Review Newtons laws of motion
 Define Momentum
 Define Impulse
 Conservation of Momentum
 Collisions
 Explosions
 Elastic Collisions

Introduction
Newtons 3 laws of motion
 1. Law of inertia
 2. Net Force = mass x acceleration

(F=MA)
 3. Action Reaction

Law of interia (1st Law)

Every object continues in its state of rest, or
of uniform motion in a straight line, unless
it is compelled to change that state by forces
impressed upon it.

acceleration = 0.0 unless the objected is
acted on by an unbalanced force
Newton’s 2nd Law


Net Force = Mass x Acceleration
F = MA
Newton’s Law of Action
Reaction (3rd Law)

You can not touch without being touched
For every action force there is
and equal and oppositely directed reaction force
Newton’s Law of Action
Reaction (3rd Law)
Ball 1
F2,1
Ball 2
F1,2
F1,2 = - F2,1
For every action force there is
and equal and oppositely directed reaction force
Momentum , p

Momentum = mass x velocity

is a Vector

has units of kg m/s
Momentum , p (a vector)
Momentum = mass x velocity
 p=m v
 p=?

8.0 kg
6.0 m/s
Momentum , p
Momentum = mass x velocity
 p=m v
 p = 160.0 kg m/s

8.0 kg
V= ?
Momentum , p
Momentum is a Vector
 p=m v
 p1 = ?
p2 = ?
V= +8.0 m/s
m1=
7.5 kg

V= -6.0 m/s
m2=
10.0 kg
Momentum , p
Momentum is a Vector
 p=m v
 p1 = +60 kg m/s
p2 = - 60 kg m/s

m1=
7.5 kg
V= -6.0 m/s
V= +8.0 m/s
m2=
10.0 kg
Momentum , p
Momentum is a Vector
 p=m v
 p1 = +60 kg m/s
p2 = - 60 kg m/s
 the system momentum is zero.,

m1=
7.5 kg
V= -6.0 m/s
V= +8.0 m/s
m2=
10.0 kg
Momentum , p
Momentum is a Vector
 p=m v
 Total momentum of a system is a vector
sum:
 p1+p2+p3+……..
p3
p2

ptotal
p1
Newton’s 2nd Law
Net Force = Mass x Acceleration

F=M a

F = M (∆V/∆t)
 F ∆t = M ∆V
 F ∆t = M (V1-V2)
 F ∆t = M V1 - M V2

F ∆t = ∆p
Impulse= F∆t
 The Impulse = the change in momentum

Newton’s 2nd Law
Net Force = Mass x Acceleration

F=M a
 or

F = ∆p/ ∆t

Newton’s 2nd Law
Net Force = Mass x Acceleration
 F ∆t = ∆p
Impulse= F ∆t
 The Impulse = the change in momentum

If M=1500 kg and
Dt=0.4 sec,
Find Dp and Favg
30°
50°
Impulse
The Impulse = the change in momentum

F ∆t = ∆p

Impulse
The Impulse = the change in momentum

F ∆t = ∆p

Newton’s Law of Action
Reaction (3rd Law)
Ball 1
F2,1
Ball 2
F1,2
F1,2 = - F2,1
For every action force there is
and equal and oppositely directed reaction force
Newton’s Law of Action
Reaction (3rd Law)
Ball 1
F2,1
Ball 2
F1,2
F1,2 = - F2,1
F1,2∆t = - F2,1 ∆t
∆p2 = - ∆p1
Conservation of momentum
Ball 1
F2,1
Ball 2
F1,2
If there are no external forces acting
on a system (i.e. only internal action
reaction pairs), then the system’s total
momentum is conserved.
“Explosions”
2 objects initially at rest

A 30 kg boy is standing on a stationary 100 kg raft
in the middle of a lake. He then runs and jumps
off the raft with a speed of 8.0 m/s. With what
V=8.0 m/s
speed does the raft recoil?
after
V=?
M=100.0 kg
before
M=100.0 kg
“Explosions”
2 objects initially at rest

A 30 kg boy is standing on a stationary 100 kg raft in the middle of a lake. He then runs
and jumps off the raft with a speed of 8.0 m/s. With what speed does the raft recoil?
after
V=8.0 m/s
V=?
M=100.0 kg
before
M=100.0 kg
p before = p after
0
= 30kg(8.0 m/s) - 100 kg V
100 kg V = 240 kg m/s
V = 2.4 m/s
Explosions
If Vred=8.0 m/s
Vblue=?
“Stick together”
2 objects have same speed after colliding
 A 30 kg boy runs and jumps onto a stationary 100
kg raft with a speed of 8.0 m/s. How fast does he
and the raft move immediately after the collision?
V=8.0 m/s
V=?
M=100.0 kg
before
M=100.0 kg
after
“Stick together”
2 objects have same speed after colliding

A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How
fast does he and the raft move immediately after the collision?
V=8.0 m/s
before
M=100.0 kg
V=?
M=100.0 kg
p before = p after
30kg(8.0 m/s) = 130 kg V
240 kg m/s = 130 kg V
V = 1.85 m/s
after
“Stick together”
2 objects have same speed after colliding
This is a perfectly inelastic collision
V=8.0 m/s
before

M=100.0 kg
V=?
M=100.0 kg
after
A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How
fast does he and the raft move immediately after the collision?
A 20 g bullet lodges in a 300 g Pendulum. The pendulum and bullet
then swing up to a maximum height of 14 cm. What is the initial
speed of the bullet?
mv = (m+M) V
Before and After Collision
1/
2(m+M)V
2=(m+M)gh
After collision but
Before and After moving up
2-D Collisions
X axis m1V10 = m1v1cos(50) + m2v2cos(40)
Y axis
0 = m1v1sin(50) - m2v2sin(40)
2-D Stick together (Inelastic)
 Momentum
Before = Momentum After
 P before=
P after
 For both the x & y components of P.
A 2000 kg truck traveling 50 mi/hr East on
McLoughlin Blvd collides and sticks to a 1000 kg
car traveling 30 mi/hr North on Main St. What is
the final velocity of the wreck? Give both the
magnitude and direction OR X and Y
components.
2-D Stick together (Inelastic)
 Momentum
Before = Momentum After
 P before=
P after
 For both the x & y components of P.
A 2000 kg truck traveling 50 mi/hr East on
McLoughlin Blvd collides and sticks to a 1000 kg
car traveling 30 mi/hr North on Main St. What is
the final velocity of the wreck? Give both the
magnitude and direction OR X and Y
components.
2-D Stick together (Inelastic)
A 2000 kg truck traveling 50 mi/hr East (V1) on McLoughlin Blvd collides and
sticks to a 1000 kg car traveling 30 mi/hr North (V2) on Main St. What is the final
velocity (V) of the wreck? Give both magnitude and direction OR X and Y
components.
P
=P
before
PBx=PAy
2000Kg(50 mi/hr)=3000KgVx
Vx=33.3 mi/h
after
&
PBy=PAy
& 1000kg(30mi/hr)=3000kgVy
& Vy=10 mi/hr
Or
V= 34.8mi/hr = (sqrt(Vx2+Vx2) & q =16.7° = tan-1(Vy/Vx)
V
V1
2000Kg
3000Kg
V2
1000Kg
Elastic Collisions
Bounce off without loss of energy
p before = p after
&
 KE before = KE after

m1
v1
m2
m1
v1,f
m2
v2,f
Elastic Collisions
Bounce off without loss of energy
 p before = p after
&
KE before = KE after
m1v1  m1v1,f  m 2 v2,f


m1 v1  v1, f  m2 v 2,f
1
1
1
2
m1v12  m1v1,f
 m 2 v22,f
2
2
2
2
m1v12  m1v1,f
 m 2 v22,f
2
m1v12  m1v1,f
 m 2 v22, f


m1v1  v1,f v1  v1,f  m 2 v2,f v2, f
v1  v1,f  v2,f
2
m1 v12  v1,f
 m2 v 22,f
v1  v2,f  v1,f
Elastic Collisions
Bounce off without loss of energy
 p before = p after
&
KE before = KE after


m1 v1  v1, f  m2 v 2,f
v1  v2,f  v1,f
m1v1  m1v 2,f  m1v1,f
m1v1  m1v1,f  m 2 v2,f
+
m1v1  m1v1,f  m 2 v2,f
m1v1  m1v 2,f  m1v1,f
2m1v1  m1  m 2 v 2,f
or
v 2,f 
2m1v1
m1  m 2 
Elastic Collisions
Bounce off without loss of energy
 p before = p after
&
KE before = KE after


m1 v1  v1, f  m2 v 2,f
v1  v2,f  v1,f
m 2 v1  m 2 v2,f  m 2 v1,f
m1v1  m1v1,f  m 2 v2,f
or
m1v1  m1v1,f  m 2 v2,f
m 2 v1  m 2 v2,f  m 2 v1,f
m1  m 2 v1  m1  m 2 v1,f
v1,f 
m1  m 2 v1
m1  m 2 
Elastic Collisions
Bounce off without loss of energy
 p before = p after
&
KE before = KE after
v1,f 
m1
v1
m1  m 2 v1
m1  m 2 
m2
&
v 2,f 
m1
2m1v1
m1  m 2 
v1,f
m2
v2,f
Elastic Collisions
Bounce off without loss of energy
 p before = p after
&
KE before = KE after
v1,f
m  m 2 v1
 1
m1  m 2 
&
v 2,f 
2m1v1
m1  m 2 
if m1 = m2 = m, then
v1,f = 0.0 & v2,f = v1
m
m
v1
m
m
v1,f= 0.0
v2,f = v1
Elastic Collisions
Bounce off without loss of energy
 p before = p after
&
KE before = KE after
v1,f
m  m 2 v1
 1
m1  m 2 
&
v 2,f 
if m1 <<< m2 , then m1+m2 ≈m2
v1,f = - v1
& v2,f ≈ 0.0
M
m
2m1v1
m1  m 2 
& m1-m2 ≈ -m2
m
v1
v1,f=- v1
M
v2,f ≈ 0.0
Elastic Collisions
Bounce off without loss of energy
if m1 <<< m2 and
v 2 is NOT 0.0
v1  v2,f  v1,f
Speed of Approach = Speed of separation
(True of all elastic collisions)
M
m
v1
v2
m
v1,f=- (v1 +v2 +v2)
M
v2,f ≈ v2
Elastic Collisions
if m1 <<< m2 and
v 2 is NOT 0.0
Speed of Approach = Speed of separation
(True of all elastic collisions)
A space ship of mass 10,000 kg swings by Jupiter in a psuedo
elastic head-on collision. If the incoming speed of the ship
is 40 km/sec and that of Jupiter is 20 km/sec, with what speed
does the space ship exit the gravitational field of Jupiter?
v2=20 km/s
m
v1 = 40 km/s
M
v1,f=?
m
M
v2,f ≈ ?
if m1 <<<Elastic
m2 and vCollisions
2 is NOT 0.0
Speed of Approach = Speed of seperation
(True of all elastic collisions)
A little boy throws a ball straight at an oncoming truck
with a speed of 20 m/s. If truck’s speed is 40 m/s and the
collision is an elastic head on collision, with what speed does
the ball bounce off the truck?
v2=40 m/s
m
M
v1,f=?
m
v1 = 20 m/s
v2,f ≈ ?
M
Elastic Collisions
Bounce off without loss of energy
 p before = p after
&
KE before = KE after
if m1 = m2 = m, then
v1,f = 0.0 & v2,f = v1
1
m 2 v2 p2
2
KE  mv 

2
2m
2m
m
m
m
v1
m
90°
Elastic Collisions
Bounce off without loss of energy
 p before = p after
&
KE before = KE after
if m1 = m2 = m, then
v1,f = 0.0 & v2,f = v1
p2f
m
m
1
m 2 v2 p2
2
KE  mv 

2
2m
2m
90°
p1
v1
p1 =p1f +p2f
p1f
2
p12 p1f
p22f


2m 2m 2m
p12  p1f2  p22f
m
m
90°
Center of Mass

The average position of the mass

When we use F=ma

We really mean F = m acm

The motion of an object is the combination of
– The translational motion of the CM
– Rotation about the CM
Center of mass
m1x1  m 2 x2  m 3x3 ....... m n xn
X cm 
m1  m 2  m 3 .......m n
The average position of the mass
CM=? M1=6.0 kg and M2=8.0 kg
Ycm=
Xcm=
CM Center of Gravity
8 kg at
(0,3)
4 kg at
(-2,0)
Where must a 10 kg mass be
placed so the center of mass
of the three mass system
is at (0,0) ?
(M+Dm)v=M(v+Dv)+Dm(v-ve)
M Dv= Dmve
Thrust=ve(dM/dt)
t(sec)
M(kg) y(m)
V(m/s)
0
360
0
0
1 357.5 5.2083 10.417
2
355 20.87 20.906
4000
3500
3000
2500
2000
1500
dM=2.5 kg/s and dt =1.0 sec
1000
500
0
A
5
6
7
8
9
t(sec)
0
=A5+dt
=A6+dt
=A7+dt
=A8+dt
B
M(kg)
360
=B5-dM*dt
=B6-dM*dt
=B7-dM*dt
=B8-dM*dt
C
y(m)
0
=C5+(D5+D6)/2*dt
=C6+(D6+D7)/2*dt
=C7+(D7+D8)/2*dt
=C8+(D8+D9)/2*dt
D
0
50
100
V(m/s)
0
=D5+dM*dt/B5*1500
=D6+dM*dt/B6*1500
=D7+dM*dt/B7*1500
=D8+dM*dt/B8*1500

The End