COLLISION PROBLEMS

A tennis ball and racket collision: a microscopic view

COLLISION: FORCE VS TIME GRAPH

A large force exerted during a small interval of time is called an impulsive force.

LINEAR MOMENTUM

The product of the particle’s mass and velocity is called the linear momentum p = mv As a vector quantity, the momentum can be represented in terms of its components: px= mvx py= mvy

ALTERNATIVE FORM OF NEWTON’S SECOND LAW

F = ma = m(dv/dt) = d(mv)/dt = dp/dt Therefore, F = dp/dt i.e. the force can be viewed as the rate of the change of momentum This is a much stronger statement than our previous version F = ma Why?

The version F = dp/dt allows for the possibility that not only the velocity, but also the mass can change! Example: rocket filled with fuel is loosing its mass as it burns the fuel.

IMPULSE

F= dp/dt is a differential equation

tf

It can be converted ∆p x = p fx − pix = into an integral form.

∫ F (t )dt

x ti

Impulse = J x = ∫ Fx (t )dt

ti

tf

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

IMPULSE

Graphic representation of impulse: Jx is the area under the force graph. Jx = Favg∆t

IMPULSE-MOMENTUM THEOREM

An impulse delivered to a particle changes its momentum. ∆Px = Jx For one-dimensional motion: pf = pi + Jx Do not need to know all the details of the force function Fx(t), only the integral of the force - the area under the force curve is needed to find pfx.

A RUBBER BALL BOUNCING OFF THE WALL Interaction is very complex, but impulse is all we need to know to find pfx

A 10 g rubber ball and a 10 g clay ball are thrown at a wall with equal speeds. The rubber ball bounces, the clay ball sticks. Which ball exerts a larger impulse on the wall?

1. The clay ball exerts a larger impulse because it sticks. 2. The rubber ball exerts a larger impulse because it bounces. 3. They exert equal impulses because they have equal momenta. 4. Neither exerts an impulse on the wall because the wall 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) 10 kg m/s f) 20 kg m/s g) 30 kg m/s

HITTING A BASEBALL

Before and after the collision

The interaction force between the bat and the ball (simplified)

HITTING A BASEBALL

What is the maximum force Fmax and the average force that the bat exerts on the ball? ∆px = mvfx– mvix= (0.15kg)(40m/s - (-20m/s)) = 9 kg m/s Jx = area = ½ x Fmax x (0.006s) = (Fmax) (0.003s) According to impulse-momentum theorem, ∴ Fmax = 9 kg m/s / 0.003s =3000N Favg = Jx/∆t = ∆px/∆t = 9 kg m/s / 0.006s =1500N

A 100g rubber ball is dropped from a ht of 2m onto a hard floor. The figure shows the force that the floor exerts on the ball. How high does the ball bounce? v12y = v02y − 2 g∆y = 0 − 2 g∆y v1 y = − 2 g∆y = − 2(9.8m / s 2 )(−2m) = −6m / s J y = 1 / 2(300 N )(0.008s ) = 1.2 Ns p2 y = p1 y + J y = (−0.626kg m / s) + 1.2 Ns = 0.574kg m / s v2 y = p2 y m = 0.574kg m / s = 5.74m / s 0.1kg

2 2 v32y = 0 = v2 y − 2 g∆y = v2 y − 2 gy3

(5.74m / s) 2 = 1.68m y3 = 2 2(9.8m / s )

CONSERVATION OF MOMENTUM When the net force F is zero =>> The impulse J is zero =>> Then, the momentum does not change

CONSERVATION OF MOMENTUM AT COLLISIONS (elastic)

Two-particle collision

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

d ( p x )1 d ( px ) 2 = (Fx )2 on 1 = (Fx )1 on 2 = −(Fx )2 on 1 dt dt d ( p x )1 d ( px ) 2 d (( p x )1 + ( p x ) 2 ) + = dt dt dt = (Fx )2 on 1 + − (Fx )2 on 1 = 0

(

)

∴ ( px )1 + ( px ) 2 = const a n t ( p fx )1 + ( p fx ) 2 = ( pix )1 + ( pix ) 2

2 equal-mass train 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

LAW OF CONSERVATION OF MOMENTUM

The total momentum P of an isolated system is a constant. Interactions within the system do not change the system’s total momentum. Mathematically, for the isolated system undergoing the collision Pf = Pi

A 10g...