Download 5.1 Impulse and Momentum

Chapter 5 Momentum
Ewen et al. 2005)
Objective:
•Use impulse and momentum in describing motion.

Is a measure of the amount of inertia and
motion and object has.
Another way to think about momentum is
how difficult it is to bring a moving object to
rest.
It is the product of mass and velocity.

Units?


p  mv
 kgm/s and slugft/s

Momentum is a vector quantity.

Find the momentum of an auto with mass
105 slugs traveling 60.0 mi/h.

Find the momentum of an auto with mass
1350 kg traveling 75.0 km/h.

Find the velocity a bullet of mass 1.00 x 10-2
kg would have to have so that it has the same
momentum as a lighter bullet of mass 1.80 x
10-3 kg and velocity of 325 m/s.
Problems 5.1:
Do problems 1-11, odd.

Impulse on an object is the product of the
force and the time interval during which the
force acts on the object.
Impulse  Ft

How are impulse and momentum related?
 Recall that
and
F  ma (Eq. 1)
a
v f  vi
t
(Eq. 2)
 Combining Eq. 1 and Eq. 2
 v f  vi 

F  m
 t 
 Then, distributing m
F
mv f  mvi
t
 Multiply both sides by t
Ft  mv f  mvi
 Thus impulse (Ft) is equal to final minus initial
momentums (mv), i.e. change in momentum.
Impulse  Ft  p  mv f  mvi  mv


After this analysis, we see
that impulse is the
measure of the change in
momentum of an object in
response to an exerted
force.
Some interesting
outcomes
 Given enough time and
distance, any force no matter
how small can bring the
largest object to p = 0 when
given enough time.
 The importance of
“followthough” in sports.

A 17.5-g bullet is fired at a muzzle velocity of
582 m/s from a gun with a mass of 8.00 kg
and a barrel length of 75.0 cm.
a. How long is the bullet in the barrel?
b. What is the force on the bullet while it is in the
barrel?
c. Find the impulse exerted on the bullet while it is
in the barrel.
d. Find the bullet’s momentum as it leaves the
barrel.

When no outside forces are acting on a
system of moving objects, the total
momentum of a system remains constant.
BOTH AT REST
AFTER PUSH
vboy = -0.40 m/s
vman = ?
mman = 75 kg
mboy = 35 kg
ptotal  mboyvboy  mmanvman  0
mboyvboy  mmanvman  0
(35 kg)(-0.40 m/s)  (75 kg) vman  0
vman  0.19 m/s
Ft  p  mv  mv f  mvi

What force is required to slow a 1450 kg care
traveling 115 km/h to 45.0 km/h within 3.00s?
How far does the car travel during its
deceleration?