Transcript inelastic collision

The force on an object
may not be constant, but
may vary over time. The
force can be averaged
over the time of
application to find the
impulse.
The impulse of a force is the
product of the average force
and the time interval during
which the force acts.
Impulse = Fave Δt
The unit is the newton•second (N•s)
The linear momentum p
of an object is the product
of the object’s mass m and
the velocity v.
p = mv
The unit is the kilogram•meter/second (kg•m/s)
Remember F = ma, and
a = (vf - v0)/ Δt.
The final term can be substituted
for a in F = ma.
F = m vf - v0 / Δt or
F = mvf - mv0 / Δt
and finally:
F Δt = mvf - mv0
This is the impulsemomentum theorem, where
impulse is equal to the
change in momentum.
F Δt = mvf - mv0
impulse
final
momentum
initial
momentum
While it may be
difficult to find
average force, it is
usually easy to find
mvf and mv0.
Ex. 1 - A baseball (m = 0.14 kg) has
an initial velocity of v0 = -38 m/s as it
approaches a bat. The ball leaves the
bat with a velocity of
vf = +58 m/s. (a) Determine the
impulse applied to the ball by the
bat.
(b) If the time of contact is Δt =
1.6 x 10-3 s, find the average force
exerted on the ball by the bat.
Ex. 2 - During a storm, rain comes
straight down with a velocity of
v0 = -15 m/s and hits the roof of a car
perpendicularly. the mass of the rain
per second that strikes the roof of the
car is 0.060 kg/s. Assuming that the
rain comes to rest upon striking the car
roof, find the average force exerted
by the rain on the car roof.
In an isolated system no net
external forces act on two
colliding objects (The net
external forces are zero). The
sum of the internal forces
(those caused by the collision
itself) is also zero by Newton’s
third law.
Therefore:
(F1+ F2)Δt = (mvf1+ mvf2) - (mv01+ mv02)
becomes:
0 = (mvf1+ mvf2) - (mv01+ mv02)
or:
(mvf1+ mvf2) = (mv01+ mv02)
This is the principle of conservation
of linear momentum.
The total linear momentum of an
isolated system remains constant.
(mvf1+ mvf2) = (mv01+ mv02)
or:
P f = P0
Ex. 5 - A freight train is being
assembled in a switching yard.
Car 1 has a mass of m1 = 65 x103 kg
and moves with a velocity of
v01 = +0.80 m/s. Car 2, with a mass of
m2 = 92 x 103 kg and a velocity of
v02 = =1.2 m/s, overtakes car 1 and
couples to it. Find the common velocity
vf of the two cars after they become
coupled.
Ex. 6 - Starting from rest, two
skaters “push off” against each
other on smooth level ice. One is a
woman (m1 = 54 kg), and one is
a man (m2 = 88 kg).
The woman recoils with a velocity
of vf1 = +2.5 m/s. Find the recoil
velocity vf2 of the man.
Ex. 7 - When a gun fires
a blank, is the recoil
greater than, the same
as, or less than when
the gun fires a
standard bullet?
The total linear
momentum may be
conserved even when the
kinetic energies of the
individual parts of a
system do change. (e. g.
the skaters)
Collisions are often
classified by whether
the total kinetic
energy changes
during the collision.
An elastic collision is one in
which the total kinetic energy of
the system after the collision is
equal to the total kinetic energy
before the collision.
K.E. is conserved in the collision.
An inelastic collision is one in which
the total kinetic energy of the system
is not the same before and after the
collision; if the objects stick together
after colliding, the collision is said to
be completely inelastic.
Kinetic energy is not conserved.
The coupling boxcars is an example
of an inelastic collision.
In inelastic collisions, kinetic
energy is lost in two main
ways. First, it may become
heat due to friction. Second,
kinetic energy is lost when
an object is permanently
deformed by the collision.
Ex. 8 - A ball of mass
m1 = 0.250 kg and velocity
v01 = +5.00 m/s collides
head-on with a ball of mass
m2 = 0.800 kg that is initially at
rest. If the collision is elastic,
what are the velocities of the
balls after the collision?
Ex. 9 - A ballistic pendulum consists of
a block of wood (mass m2 = 2.50 kg)
suspended by a wire.
A bullet (mass m1 = 0.0100 kg) is fired
with a speed v01. Just after the bullet
collides with it, the block (now
containing the bullet) has a speed vf
and then swings to a maximum height
of 0.650 m above the initial position.
Find the speed of the bullet.