Momentum and Collisions

Momentum and CollisionsMomentum and Collisions

Dr. Robert MacKay

Clark College, Physics

Introduction Introduction

Review Newtons laws of motion Define Momentum Define Impulse Conservation of Momentum Collisions Explosions Elastic Collisions

Introduction Introduction

Newtons 3 laws of motion 1. Law of inertia 2. Net Force = mass x acceleration ( F = M A ) 3. Action Reaction

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

Newton’s 2nd LawNewton’s 2nd Law

Net Force = Mass x Acceleration

F = M A

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



Ball 1

Ball 2

F1,2 F2,1

F1,2 = - F2,1

Momentum , pMomentum , p

Momentum = mass x velocity

is a Vector

has units of kg m/s

Momentum , p (a vector)Momentum , p (a vector)

Momentum = mass x velocity p = m v p = ?

8.0 kg 6.0 m/s


Momentum = mass x velocity p = m v p = 160.0 kg m/s

8.0 kg V= ?


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

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

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

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

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

Newton’s 2nd Law Newton’s 2nd Law Net Force = Mass x Acceleration F = M a or F = ∆p/ ∆t

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

If M=1500 kg and t=0.4 sec,Find p and Favg

30°

50°

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)

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)



Ball 1

Ball 2

F1,2 F2,1

F1,2 = - F2,1

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

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.

““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

““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

ExplosionsExplosions

If Vred=8.0 m/sVblue=?

““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



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


This is a perfectly inelastic collisionThis is a perfectly inelastic collision


M=100.0 kg M=100.0 kgafterbefore

V=?V=8.0 m/s

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) VBefore and After Collision

1/2(m+M)V2=(m+M)ghAfter collision but

Before and After moving up

2-D Collisions2-D Collisions

X axis m1V10 = m1v1cos(50) + m2v2cos(40)

Y axis 0 = m1v1sin(50) - m2v2sin(40)

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.

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.

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

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


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



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



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



&v1,f m1 m2 v1

m1 m2 v2,f

2m1v1

m1 m2

v1

m1 m1m2 m2

v1,f v2,f



&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



&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


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

Elastic CollisionsElastic Collisions

m

m

Mv1,f=?


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

Elastic CollisionsElastic Collisions

m

mv1,f=?


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



v1

mm

m m

if m1 = m2 = m, thenv1,f = 0.0 & v2,f = v1

KE 1

2mv2

m2v2

2m

p2

2m

90°



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°

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

Center of massCenter of mass

X cm m1x1 m2x2 m3x3 . ... ... mnxn

m1 m2 m3... ... .mn

The average position of the mass

CM=? M1=6.0 kg and M2=8.0 kgCM=? M1=6.0 kg and M2=8.0 kgYcm=Ycm=Xcm=Xcm=

CM Center of GravityCM Center of Gravity

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) ?

(M+m)v=M(v+v)+m(v-ve)M v= mve

Thrust=ve(dM/dt)

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

The End

Momentum and Collisions

Documents

momentum f t

p impulse

p t newtons

mass x acceleration

vector p

v2 f t

vt f t

mass x velocity p