Momentum and Momentum and CollisionsCollisions
Linear Momentum
The linear momentum of a particle or an object that can be modeled as a particle of mass m moving with a velocity v is defined to be the product of the mass and velocity: p = m v
We will usually refer to this as “momentum”, omitting the “linear”.
Momentum is a VECTOR. It has components. Don’t forget that.
Linear Momentum
The dimensions of momentum are ML/T The SI units of momentum are kg · m / s Momentum can be expressed in component
form: px = m vx py = m vy pz = m vz
A 3.00-kg particle has a velocity of . (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum.
Newton and Momentum
Newton called the product mv the quantity of motion of the particle
Newton’s Second Law can be used to relate the momentum of a particle to the resultant force acting on it
with constant mass.
d md dm m
dt dt dt
vv pF a
Newton’s Second Law
The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle This is the form in which Newton presented the Second
Law It is a more general form than the one we used previously This form also allows for mass changes
Applications to systems of particles are particularly powerful
Two Particles (1 dimension for now)
other. on the force
a producing particle one are that those
are system on this acting forcesonly the
thenforces external NO are there
21
IFdt
dp
dt
dp
dt
dptotal
Continuing
00 int
21
ernaltotal
external
total
Fdt
dpF
dt
dp
dt
dp
dt
dp
NO EXTERNAL FORCES: ptotal is constant Force on P1 is from particle 2 and is F21
Force on P2 is from particle 1 and is F12
F21=-F12
Summary
The momentum of a SYSTEM of particles that areisolated from external forces remains a constant ofthe motion.
Conservation of Linear MomentumTextbook Statement Whenever two or more particles in an
isolated system interact, the total momentum of the system remains constant The momentum of the system is conserved,
not necessarily the momentum of an individual particle
This also tells us that the total momentum of an isolated system equals its initial momentum
Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M? (b) Find the original elastic potential energy in the spring if M = 0.350 kg.
A particle of mass m moves with momentum p. Show that the kinetic energy of the particle is K = p2/2m. (b) Express the magnitude of the particle’s momentum in terms of its kinetic energy and mass.
An Easy One…
Consider a particle roaming around that is suddenly subjected to some kind of FORCE that looks something like the last slide’s graph.
change will then 0 If
0 ifconstant
:
pF
Fp
Favp
vp
mdt
dm
dt
d
m
NEWTON
Let’s drop the vector notation and stick to one dimension.
f
i
f
i
t
t
if
f
i
t
t
Fdtpp
Fdtdp
Fdtdp
Fdt
dp
Change of Momentum
Impulse
NEW LAW
IMPULSE = CHANGE IN MOMENTUM
NEW LAW
A 3.00-kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle (Fig. P9.9). If the ball is in contact with the wall for 0.200 s, what is the average force exerted on the ball by the wall?
LETS TALK ABOUT
Consider two particles:
1 2
m1 m2
v1v2
V1 V2
1 2
1 2
What goes on during the collision? N3
Force on m2
=F(12)
Force on m1
=F(21)
The Forces
Equal and Opposite
More About Impulse: F-t The Graph Impulse is a vector quantity The magnitude of the
impulse is equal to the area under the force-time curve
Dimensions of impulse are M L / T
Impulse is not a property of the particle, but a measure of the change in momentum of the particle
Impulse
The impulse can also be found by using the time averaged force
I = t
This would give the same impulse as the time-varying force does
F
SKATEBOARD DEMO
Conservation of Momentum, Archer Example
The archer is standing on a frictionless surface (ice)
Approaches: Newton’s Second Law –
no, no information about F or a
Energy approach – no, no information about work or energy
Momentum – yes
Archer Example, 2
Let the system be the archer with bow (particle 1) and the arrow (particle 2)
There are no external forces in the x-direction, so it is isolated in terms of momentum in the x-direction
Total momentum before releasing the arrow is 0 The total momentum after releasing the arrow is p1f
+ p2f = 0
Archer Example, final
The archer will move in the opposite direction of the arrow after the release Agrees with Newton’s Third Law
Because the archer is much more massive than the arrow, his acceleration and velocity will be much smaller than those of the arrow
An estimated force-time curve for a baseball struck by a bat is shown in Figure P9.7. From this curve, determine (a) the impulse delivered to the ball, (b) the average force exerted on the ball, and (c) the peak force exerted on the ball.
Overview: Collisions – Characteristics We use the term collision to represent an event
during which two particles come close to each other and interact by means of forces
The time interval during which the velocity changes from its initial to final values is assumed to be short
The interaction force is assumed to be much greater than any external forces present This means the impulse approximation can be used
Collisions – Example 1
Collisions may be the result of direct contact
The impulsive forces may vary in time in complicated ways This force is internal to
the system Momentum is
conserved
Collisions – Example 2
The collision need not include physical contact between the objects
There are still forces between the particles
This type of collision can be analyzed in the same way as those that include physical contact
Types of Collisions
In an elastic collision, momentum and kinetic energy are conserved Perfectly elastic collisions occur on a microscopic level In macroscopic collisions, only approximately elastic
collisions actually occur In an inelastic collision, kinetic energy is not
conserved although momentum is still conserved If the objects stick together after the collision, it is a
perfectly inelastic collision
Collisions, cont
In an inelastic collision, some kinetic energy is lost, but the objects do not stick together
Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types
Momentum is conserved in Momentum is conserved in allall collisions collisions
Perfectly Inelastic Collisions
Since the objects stick together, they share the same velocity after the collision
m1v1i + m2v2i =
(m1 + m2) vf
A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg). The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?
. Velcro couplers make the carts stick together after colliding. Find the final velocity of the train of three carts. (b) What If? Does your answer require that all the carts collide and stick together at the same time? What if they collide in a different order?
Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false. Newton’s third law tells us that both objects experience forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at 8.00 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 80.0 kg. Including the drivers, the total vehicle masses are 800 kg for the car and 4 000 kg for the truck. If the collision time is 0.120 s, what force does the seatbelt exert on each driver?
Elastic Collisions
Both momentum and kinetic energy are conserved
1 1 2 2
1 1 2 2
2 21 1 2 2
2 21 1 2 2
1 1
2 21 1
2 2
i i
f f
i i
f f
m m
m m
m m
m m
v v
v v
v v
v v
Elastic Collisions, cont Typically, there are two unknowns to solve for and so you need two
equations The kinetic energy equation can be difficult to use With some algebraic manipulation, a different equation can be used
v1i – v2i = v1f + v2f This equation, along with conservation of momentum, can be used
to solve for the two unknowns It can only be used with a one-dimensional, elastic collision
between two objects The solution is shown on pages 262-3 in the textbook. (Lots
of algebra but nothing all that difficult. We will look at a special case.
Let’s look at the case of equal masses.
Second Particle initially at rest
Explains the demo!
Elastic Collisions, final
Example of some special cases m1 = m2 – the particles exchange velocities When a very heavy particle collides head-on with a very
light one initially at rest, the heavy particle continues in motion unaltered and the light particle rebounds with a speed of about twice the initial speed of the heavy particle
When a very light particle collides head-on with a very heavy particle initially at rest, the light particle has its velocity reversed and the heavy particle remains approximately at rest
Collision Example – Ballistic Pendulum Perfectly inelastic collision –
the bullet is embedded in the block of wood
Momentum equation will have two unknowns
Use conservation of energy from the pendulum to find the velocity just after the collision
Then you can find the speed of the bullet