momentum and collisions
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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Momentum and CollisionsMomentum and Collisions
Dr. Robert MacKay
Clark College, Physics
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Introduction Introduction
Review Newtons laws of motion Define Momentum Define Impulse Conservation of Momentum Collisions Explosions Elastic Collisions
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Introduction Introduction
Newtons 3 laws of motion 1. Law of inertia 2. Net Force = mass x acceleration ( F = M A ) 3. Action Reaction
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Law of interia (1st Law)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
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Newton’s 2nd LawNewton’s 2nd Law
Net Force = Mass x Acceleration
F = M A
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Newton’s Law of Action Newton’s Law of Action Reaction (3rd Law)Reaction (3rd Law)
You can not touch without being touched
For every action force there is and equal and oppositely directed reaction force
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Newton’s Law of Action Newton’s Law of Action Reaction (3rd Law)Reaction (3rd Law)
For every action force there is and equal and oppositely directed reaction force
Ball 1
Ball 2
F1,2 F2,1
F1,2 = - F2,1
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Momentum , pMomentum , p
Momentum = mass x velocity
is a Vector
has units of kg m/s
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Momentum , p (a vector)Momentum , p (a vector)
Momentum = mass x velocity p = m v p = ?
8.0 kg 6.0 m/s
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Momentum , pMomentum , p
Momentum = mass x velocity p = m v p = 160.0 kg m/s
8.0 kg V= ?
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Momentum , pMomentum , p
Momentum is a Vector p = m v p1 = ? p2 = ?
m2=10.0 kg
V= -6.0 m/s
m1=7.5 kg
V= +8.0 m/s
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Momentum , pMomentum , p Momentum is a Vector p = m v p1 = +60 kg m/s p2 = - 60 kg m/s
m2=10.0 kg
V= -6.0 m/s
m1=7.5 kg
V= +8.0 m/s
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Momentum , pMomentum , p Momentum is a Vector p = m v p1 = +60 kg m/s p2 = - 60 kg m/s the system momentum is zero.,
m2=10.0 kg
V= -6.0 m/s
m1=7.5 kg
V= +8.0 m/s
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Momentum , pMomentum , p Momentum is a Vector p = m v Total momentum of a system is a vector
sum: p1+p2+p3+……..
p1
p2
p3
ptotal
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Newton’s 2nd Law 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
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Newton’s 2nd Law Newton’s 2nd Law Net Force = Mass x Acceleration F = M a or F = ∆p/ ∆t
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Newton’s 2nd Law Newton’s 2nd Law Net Force = Mass x Acceleration F ∆t = ∆p Impulse= F ∆t The Impulse = the change in momentum
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If M=1500 kg and t=0.4 sec,Find p and Favg
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30°
50°
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Impulse Impulse The Impulse = the change in momentum F ∆t = ∆p
Impulse
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5
time(s)
Fo
rce (
N)
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Impulse Impulse The Impulse = the change in momentum F ∆t = ∆p
Impulse
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5
time(s)
Fo
rce (
N)
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Newton’s Law of Action Newton’s Law of Action Reaction (3rd Law)Reaction (3rd Law)
For every action force there is and equal and oppositely directed reaction force
Ball 1
Ball 2
F1,2 F2,1
F1,2 = - F2,1
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Newton’s Law of Action Newton’s Law of Action Reaction (3rd Law)Reaction (3rd Law)Ball 1
Ball 2
F1,2 F2,1
F1,2 = - F2,1 F1,2∆t = - F2,1 ∆t ∆p2 = - ∆p1
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Conservation of momentumConservation of momentumBall 1
Ball 2
F1,2 F2,1
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.
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““Explosions”Explosions”2 objects initially at rest2 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?
M=100.0 kg M=100.0 kg
after
before
V=?
V=8.0 m/s
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““Explosions”Explosions”2 objects initially at rest2 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?
M=100.0 kg M=100.0 kg
after
before
V=?
V=8.0 m/s
p before = p after
0 = 30kg(8.0 m/s) - 100 kg V100 kg V = 240 kg m/s V = 2.4 m/s
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ExplosionsExplosions
If Vred=8.0 m/sVblue=?
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““Stick together”Stick together”2 objects have same speed after colliding2 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?
M=100.0 kg M=100.0 kg
afterbefore
V=?
V=8.0 m/s
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““Stick together”Stick together”2 objects have same speed after colliding2 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?
M=100.0 kg M=100.0 kgafterbefore
V=?V=8.0 m/s
p before = p after
30kg(8.0 m/s) = 130 kg V240 kg m/s = 130 kg V V = 1.85 m/s
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““Stick together”Stick together”2 objects have same speed after colliding2 objects have same speed after colliding
This is a perfectly inelastic collisionThis is a perfectly inelastic collision
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?
M=100.0 kg M=100.0 kgafterbefore
V=?V=8.0 m/s
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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?
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mv = (m+M) VBefore and After Collision
1/2(m+M)V2=(m+M)ghAfter collision but
Before and After moving up
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2-D Collisions2-D Collisions
X axis m1V10 = m1v1cos(50) + m2v2cos(40)
Y axis 0 = m1v1sin(50) - m2v2sin(40)
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2-D Stick together (Inelastic)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.
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2-D Stick together (Inelastic)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.
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2-D Stick together (Inelastic)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.
VV1
V2
Pbefore=Pafter
PBx=PAy & PBy=PAy
2000Kg(50 mi/hr)=3000KgVx & 1000kg(30mi/hr)=3000kgVy
Vx=33.3 mi/h & Vy=10 mi/hrOr
V= 34.8mi/hr = (sqrt(Vx2+Vx
2) & =16.7° = tan-1(Vy/Vx)
2000Kg
1000Kg
3000Kg
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after
& KE before = KE after
v1
m1 m1m2 m2
v1,f v2,f
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after & KE before = KE after
m1v1 m1v1,f m2v2,f
m1 v1 v1, f m2v2,f
1
2m1v1
2 1
2m1v1,f
2 1
2m2v2,f
2
m1v12 m1v1,f
2 m2v2,f2
m1v12 m1v1,f
2 m2v2, f2
m1 v12 v1,f
2 m2v2,f2
m1 v1 v1,f v1 v1,f m2v2,f v2, f
v1 v1,f v2,f
v1 v2,f v1,f
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after & KE before = KE after
v1 v2,f v1,fm1 v1 v1, f m2v2,f
m1v1 m1v1,f m2v2,f
m1v1 m1v1,f m2v2,f
+
m1v1 m1v2,f m1v1,f
m1v1 m1v2,f m1v1,f
2m1v1 m1 m2 v2,f
v2,f 2m1v1
m1 m2 or
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after & KE before = KE after
v1 v2,f v1,fm1 v1 v1, f m2v2,f
m1v1 m1v1,f m2v2,f
m1v1 m1v1,f m2v2,f
-
or
m2v1 m2v2,f m2v1,f
m2v1 m2v2,f m2v1,f
m1 m2 v1 m1 m2 v1,f
v1,f m1 m2 v1
m1 m2
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after & KE before = KE after
&v1,f m1 m2 v1
m1 m2 v2,f
2m1v1
m1 m2
v1
m1 m1m2 m2
v1,f v2,f
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after & KE before = KE after
&v1,f m1 m2 v1
m1 m2 v2,f
2m1v1
m1 m2
v1
mmm m
v1,f= 0.0 v2,f = v1
if m1 = m2 = m, thenv1,f = 0.0 & v2,f = v1
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after & KE before = KE after
&v1,f m1 m2 v1
m1 m2 v2,f
2m1v1
m1 m2
v1
mmM
v1,f=- v1
v2,f ≈ 0.0
if m1 <<< m2 , then m1+m2 ≈m2 & m1-m2 ≈ -m2 v1,f = - v1 & v2,f ≈ 0.0
M
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
v1
mmM
v1,f=- (v1 +v2 +v2) v2,f ≈ v2
if m1 <<< m2 and v 2 is NOT 0.0
M
v1 v2,f v1,f
Speed of Approach = Speed of separation (True of all elastic collisions)
v2
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Elastic CollisionsElastic Collisions
m
m
Mv1,f=?
if m1 <<< m2 and v 2 is NOT 0.0
v2,f ≈ ?
M
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?
v1 = 40 km/s
v2=20 km/s
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Elastic CollisionsElastic Collisions
m
mv1,f=?
if m1 <<< m2 and v 2 is NOT 0.0
v2,f ≈ ?
Speed of Approach = Speed of seperation (True of all elastic collisions)
A little boy throws a ball straight at an oncoming truckwith 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?
v1 = 20 m/s
v2=40 m/s
MM
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after & KE before = KE after
v1
mm
m m
if m1 = m2 = m, thenv1,f = 0.0 & v2,f = v1
KE 1
2mv2
m2v2
2m
p2
2m
90°
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Elastic CollisionsElastic CollisionsBounce off without loss of energyBounce off without loss of energy
p before = p after & KE before = KE after
v1
mm
m m
if m1 = m2 = m, thenv1,f = 0.0 & v2,f = v1
KE 1
2mv2
m2v2
2m
p2
2m
90°
p1 =p1f +p2f
p1
p2fp1f
p12
2m p1f
2
2m p2f
2
2m
p12 p1f
2 p2f2
90°
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Center of MassCenter 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
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Center of massCenter of mass
X cm m1x1 m2x2 m3x3 . ... ... mnxn
m1 m2 m3... ... .mn
The average position of the mass
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CM=? M1=6.0 kg and M2=8.0 kgCM=? M1=6.0 kg and M2=8.0 kgYcm=Ycm=Xcm=Xcm=
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CM Center of GravityCM Center of Gravity
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4 kg at(-2,0)
8 kg at(0,3)
Where must a 10 kg mass be placed so the center of massof the three mass systemis at (0,0) ?
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(M+m)v=M(v+v)+m(v-ve)M v= mve
Thrust=ve(dM/dt)
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t(sec) M(kg) y(m) V(m/s)0 360 0 01 357.5 5.2083 10.4172 355 20.87 20.906
0
500
1000
1500
2000
2500
3000
3500
4000
0 50 100 150
t(sec) M(kg) y(m) V(m/s)0 360 0 0=A5+dt =B5-dM*dt =C5+(D5+D6)/2*dt =D5+dM*dt/B5*1500=A6+dt =B6-dM*dt =C6+(D6+D7)/2*dt =D6+dM*dt/B6*1500=A7+dt =B7-dM*dt =C7+(D7+D8)/2*dt =D7+dM*dt/B7*1500=A8+dt =B8-dM*dt =C8+(D8+D9)/2*dt =D8+dM*dt/B8*1500
dM=2.5 kg/s and dt =1.0 sec
56789
A B C D
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