Impulse and Momentum
Chapter 9 IMPULSE AND MOMENTUM
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
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