IMPULS AND MOMENTUM

Name : Rahmawati Yusri

What happened to this picture ??

NIM : RSA1C311001

THE MATERIAL

RELATIONSHIP

APLICATION

SUBDUCTION

PENDULUM

SUBDUCTION TWO DIMENSIONS

ROCKET

IMPULSE AND MOMENTUM RELATIONSHIP

• Outgrow style that employs at object up to subduction happening can be depicted by relationship graph among f by t, with that assumption constant styled.

tt1 t2

F(t)

t

. • One particle gets mass m one that moving with speed v having linear momentum p one that constitute multiple among that particle speed with its mass

p = mv.

F = ma.

Legally Newton II. resultant inspires that is conected with one straight proportionate object with velocity

dt

dp

dt

mvdF

)(

dp=Fdt

If each diintegralkan therefore acquired:

m v

2

1

2

1

.21

t

t

p

p

dtFdppp

Linear Momentum continuity

If resultant external style that employs at system equal to zero, therefore momentum vector total system makes a abode constant

0dt

dp

For particle system

pppp n ........21

SEVERALLY MOMENTUM PRINCIPLE PURPOSE

• Two logs A and b one gets mA's masses and mB, one that linked by one spiral spring and lie upon horizontal table without friction. Our spiral spring strungs strikingly both of log to sideways as on image.

AB

y

xO

The one log gets positive momentum( A moves in aim + x) and bemomentum negative's the other log (B moving deep tenor –x) from hokum our conservation of momentum gets:

Early momentum = final momentum

AABB vmvm 0

AABB vmvm

OrB

A

BA v

m

mv

SUBDUCTION

sebelum selama setelah

1. Subduction Dashes Away perfect

SUBDUCTION TYPES

A subduction is said dash away perfect if jumlahan object kinetic energies that get good subduction before and after subduction with . (Energy continuity law kinetic)

sebelum sesudah

m1

m1m2 m2

v2

v’2

v’1

v1

Gambar 6.4. Tumbukan dua benda

momentun early totaled: paw = m1v1 + m2v2

total early kinetic energy: Ekaw = m1v12 + m2v2

2.

second totaled momentum that object subduction afters is

pak = m1v’1 + m2v’2

totaled kinetic energy after subduction is Ekak = m1v’12 + m2v2’

2.

paw = pak m1v1 + m2v2 = m1v’1 + m2v’2

Ekaw = Ekak m1v1

2 + m2v22 = m1v’1

2 + m2v2’ 2

m1(v1 − v’1) = m2(v’2 − v2),

m1v12 − m1v’1

2 = m2v2’ 2 − m2v2

2

or

m1 (v1 − v’1)( v1 + v’1) = m2(v’2 − v2) (v’2 + v2)

or

Of two box deep equations tingle to be gotten

v1 + v’1 = v’2 + v2 ot 1''

12

12 vv

vv

In common compare evv

vv

12

12 ''

2. Subduction Dashes Away to play favorites

After subduction there is a portion changed mechanical energy as heat energy, sound or the other energy. So after subduction available exempt energy. Mechanical energy continuity law inoperative. On this subduction is distinguished its elasticity price is 0 <e<1

3. Subduction Does Not Dash Away absolutely

After second object subduction clings to become one and moving with afterses same speed subduction both of menyatu's object. E=0's price

BALLISTIC PENDULUM

V’

v

h

Gambar 6.5 Bandul-Balistik untuk menentukan kecepatan peluru

If shot mass is m and pendulum mass be m, with momentum continuity is gotten

')( vMmmv system energy on the turn as shot potential energy with pendulum until up until shot oscillation top pendulum

ghMmvMm )(')(2

1 2 or ghv 2'

If equation in yellow box at merged acquired:

ghm

Mmv 2

SUBDUCTION IN TWO DIMENSIONS

x x

y y

vo

m1 m2

m1

m2

q

j

kelesterian is momentum for each one aim

jq .cos.cos 2211 vmvmvm om Wicked aim x:

Wicked aim y: jq sinsin0 2211 vmvm

222

211

21 2

1

2

1

2

1vmvmvm o If subduction gets elastic character

But if inelastis's subduction io Evmvmvm 2

22211

21 2

1

2

1

2

1

billiard's ball with speed 30 m / s menumbuk serves a ball biliard II. one holds tongue and get masses with. After subduction, moving i. ball deviates 30 o of aim originally. Look for speed each ball and power aim serves a ball II.. (elastic reputed subduction)

One log gets mass m 1 = 2,0 kg moves along

ultrasmooth table surface with runaway speed 10 m / dt. In front log is it first available one log get mass m 2 = 5,0 kg moves

with runaway speed 3,0 m / unidirectional dt with first log. One spiral spring with tetapan k = 1120 n / m glued on

second log as it were is shown on image gets what far that spiral spring termampatkan upon happens subduction?

10 m/dt

m1 m2

3,0 m/dt

Kunci = 0,25 m

Rocket Thruster energy

• Rockets early momentum p 1 =mv • Upon + dt rocket speed increases v + dv. For example

m mass that gushes about satuan time. Rocket mass stays behind m? dt, detached fueled mass? dt.

• If vr relative rocket speed to fuel that gushes. • v ’ =v vr • Eventual momentum is (m? dt) (v + dv) • Fueled momentum that tersembu is v ’ dt

Therefore prevailing: -

mgdt=((m- dt)(v+dv)+v’ dt)-mv

If m huge therefore m dtdv can be ignored So: mdv=vr dt-mgdt\

dm=- dt, so is gotten:

By integrates is gotten:

v=-vrlnm-gt+C If modan vo mass and kec while t=0 therefore

vo=-vrlnmo+C

Dan v=vo-gt+vrln(mo/m)

gdtm

dmvdv r

Neutrino case

• If two object fly to be diqoined by speeds v 1 and v 2 therefore kinetic energy it also separate:

• Q=K 1 + K 2 =1 / 2 m 1 2 +1 / 2 m 2

2

• Second particle momentum has to equal zero so:

• m 1 v 1 = -m 2 v 2

• If both of dikuadratkan's equation and at divides two therefore acquired:

1/2m12v1

2=1/2m22v2

2

m1K1=m2K2

If this equation compounded by equation upon acquired:

QK mmm

21

2

1 QK mmm

21

1

2

I THINK THAT’S ALL MY PRESENTATION…FORGIVE ME FOR MY MISTAKE..

AND SEE YOU IN THE NEXT CONDITION..ASSALAMU’ALAYKUM WA RAHMATULLAH

WA BARAKATUH…