Chapter on the lookout for IMPULSE AND MOMENTUM

ACCIDENT PROBLEMS

A tennis ball and racquet collision: a microscopic look at

COLLISION: POWER VS TIME GRAPH

A huge force exerted during a tiny interval of your time is called a great impulsive push.

LINEAR MOMENTUM

The product with the particle's mass and velocity is called the linear energy p sama dengan mv Like a vector quantity, the momentum can be showed in terms of their components: px= mvx py= mvy

ALTERNATIVE FORM OF NEWTON'S SECOND LEGISLATION

F = ma = m(dv/dt) sama dengan d(mv)/dt sama dengan dp/dt Therefore , F sama dengan dp/dt i actually. e. the force can be viewed the rate with the change of momentum This can be a much better statement than our earlier version F = mum Why?

The version F = dp/dt allows for the possibility that not only the speed, but as well the mass can change! Model: rocket stuffed with fuel is loosing their mass as it burns the fuel.

IMPULSE

F= dp/dt is a gear equation

tf

It can be transformed ∆p times = g fx − pix sama dengan into an important form.

∫ F (t )dt

x ti

Instinct = L x sama dengan ∫ Fx (t )dt

ti

tf

Area within the Fx (t) curve betwn ti and tf

IMPULSE

Graphic rendering of impulse: Jx is the area underneath the force graph. Jx sama dengan Favg∆t

IMPULSE-MOMENTUM THEOREM

An impulse shipped to a compound changes it is momentum. ∆Px = Jx For one-dimensional motion: pf = professional indemnity + Jx Do not need to find out all the details with the force function Fx(t), the particular integral of the force - the area under the force competition is needed to find pfx.

A RUBBER BALL BOUNCING OFF THE BEATEN TRACK Interaction is very complex, although impulse is we need to know to find pfx

A 10 g rubber ball and a ten g clay ball happen to be thrown at a wall with similar speeds. The rubber ball bounces, the clay ball sticks. Which usually ball exerts a larger behavioral instinct on the wall structure?

1 . The clay ball exerts a greater impulse as it sticks. installment payments on your The rubberized ball exerts a larger impulse because it bounces. 3. That they exert the same impulses since they have the same momenta. 4. Neither applies an impulse on the wall structure because the wall membrane doesn't move.

The cart's change of momentum is…

a) -30 kg m/s b) -20 kg m/s c) -10 kg m/s d) 0 kg m/s e) twelve kg m/s f) 20 kg m/s g) 31 kg m/s

HITTING A BASEBALL

Before and after the collision

The connection force involving the bat plus the ball (simplified)

HITTING A BASEBALL

What is the maximum force Fmax and the average power that the softball bat exerts on your ball? ∆px = mvfx– mvix= (0. 15kg)(40m/s - (-20m/s)) = on the lookout for kg m/s Jx = area sama dengan ½ times Fmax by (0. 006s) = (Fmax) (0. 003s) According to impulse-momentum theorem, ∴ Fmax = on the lookout for kg m/s / 0. 003s =3000N Favg sama dengan Jx/∆t = ∆px/∆t sama dengan 9 kilogram m/s as well as 0. 006s =1500N

A 100g rubberized ball is usually dropped from a ht of 2m onto a hard floor. The figure shows the power that the floors exerts on your ball. How large does the ball bounce? v12y = v02y − two g∆y = 0 − 2 g∆y v1 sumado a = − 2 g∆y = − 2(9. 8m / s i9000 2 )(−2m) = −6m / t J y = you / 2(300 N )(0. 008s ) = 1 ) 2 Natursekt p2 sumado a = p1 y & J sumado a = (−0. 626kg m / s) + 1 ) 2 Ns = 0. 574kg m / s i9000 v2 con = p2 y m = zero. 574kg m / h = your five. 74m / s zero. 1kg

two 2 v32y = zero = v2 y − 2 g∆y = v2 y − 2 gy3

(5. 74m / s) 2 = 1 . 68m y3 sama dengan 2 2(9. 8m as well as s )

CONSERVATION OF MOMENTUM If the net force F can be zero => > The impulse J is actually zero => > Then, the momentum will not change

PRESERVATION OF ENERGY AT COLLISIONS (elastic)

Two-particle collision

m1(vix )1+m2(vix)2=m1(vfx )1+m2(vfx )2

d ( s x )1 d ( px ) 2 = (Fx )2 on 1 = (Fx )1 on 2 sama dengan в€’(Fx )2 on 1 dt dt d ( p times )1 d ( px ) two d (( p times )1 & ( s x ) 2 ) + = dt dt dt = (Fx )2 on you + в€’ (Fx )2 on one particular = zero

(

)

∴ ( px )1 + ( px ) 2 sama dengan const a n capital t ( s fx )1 + ( p foreign exchange ) 2 = ( pix )1 + ( pix ) 2

two equal-mass teach cars colliding and coupling

m1(vfx )1+ m2(vfx)2 = m1(vix )1+ m2(vix )2 mvf +mvf = 2mvf = mvi + 0 vf =1/2 vi

REGULATION OF CONSERVATION OF IMPETUS

The total energy P of your isolated system is a constant. Communications within the program do not replace the system's total momentum. Mathematically, for the isolated program undergoing the collision Pf = Pi

A 10g...

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