Download Momentum and Collision Notes

Momentum & Impulse
AP Physics
Impulse and Momentum
 Newton’s 2nd Law of motion can be rewritten by
using the definition of acceleration as the change in
velocity over the change in time.
F  ma
v 
F  m 
t 
Impulse and Momentum
 If the change in time is multiplied out of the
denominator, we are left with the following:
Ft  mv
Ft  mv
 The product of force and change in time is
called the impulse (symbol is J).
 Impulse is a vector quantity and is
measured in Newton-seconds (Ns).
Impulse
 If a car hits a haystack or the same car hits a wall,
momentum is decreased by same impulse – the
same products of force and time.
 However, impact force is greater into the wall than
it is into the haystack as the haystack extends
impact time, lessening the impact force.
 Impact time is the time during which momentum is
brought to zero.

  mv
 The product of the mass and the velocity is
called the momentum (symbol  -“rho”) of
an object.
 Momentum is also a vector quantity and is
measured in kgm/s.
 Note that the units for impulse and
momentum appear different, but they are
actually the same unit when simplified.
Momentum
 Momentum can be increased with an
increase in either mass or in velocity or
both.
 Ex: a rolling bowling ball has greater momentum
than a tennis ball rolling at the same speed
because its mass is greater
 Ex: a racecar going forward at 120 mi/hr has
greater momentum than the same size car going
90 mi/hr due to its greater velocity
 If an object is not moving (no matter how big
it is), the momentum is equal to zero.

Ft  mv f  mvi
 The impulse-momentum theorem states
that the impulse on an object is equal to
the object’s final momentum minus the
object’s initial momentum.
 Can also be written as:
Ft   f  i
Example 1
Example 2
A car of mass m, traveling at speed v, stops in time t
when maximum braking force is applied. Assuming
thebraking force is independent of mass, what time
would be required to stop a car of mass 2m traveling
at speed v?
(A) ½ t
(B) t
(C) √2 t
(D) 2t
Conservation of Momentum
 A system is the environment and all of the
objects examined in a problem.
 A closed system is a system in which no mass
is gained or lost.
 An isolated system is a system in which the
net external force is zero… no forces acting
outside of the system have an effect inside
of it.
Conservation of Momentum
 The law of conservation of
momentum states that the sum of
momentum of any closed, isolated
system does not change… or that the
sum of the momentum of the objects
in that system is constant.
Conservation of Momentum
 Mathematically, we can view this as a
BEFORE and AFTER situation.
 For any two objects A and B:
Ai  Bi  Af  Bf
Types of Collisions
 If two objects bounce apart when they collide and KE is NOT
conserved, it is called an inelastic collision and can be written:
 If two objects bounce apart when they collide and KE is
conserved, it is called an elastic collision and can be written:
m1v1i  m2v 2i  m1v1 f  m2v 2 f
 If two objects stick together when they collide and KE is NOT
conserved, it is called an perfectly inelastic collision and can
be written:
m1v1i  m2v 2i  (m1  m2 )v f
More on collisions
 Momentum is conserved in all three types of collisions.
 Kinetic energy is only conserved in an elastic collision.
 In inelastic and perfectly inelastic collisions, some of
the KE converts into internal elastic PE, sound or heat.
Collision
Momentum
KE
End Result
elastic
conserved
conserved
objects bounce
inelastic
conserved
not conserved
objects bounce
perfectly inelastic
conserved
not conserved
objects stick
Example 1
 Tubby and his twin brother Chubby have a
combined mass of 200.0kg and are zooming
along in a 100.0kg amusement park bumper
car at 10.0m/s. They bump Melinda’s car,
which is sitting still. Melinda has a mass of
25.0kg. After the elastic collision, the twins
continue ahead with a speed of 4.12m/s.
How fast is Melinda’s car bumped across the
floor?
Example 1 Picture
Before Collision
After Collision
T&C
Mel
T&C
Mel
m1  300.0kg
m2  125.0kg
m1  300.0kg
v1i  10.0m /s
v 2i  0m /s
v1 f  4.12m /s
m2  125.0kg
v2 f  ?



Example 1 Answer
m1v1i  m2v 2i  m1v1 f  m2v 2 f
m1v1i  m2v 2i  m1v1 f  m2v 2 f
m1v1i  m2v 2i  m1v1 f
m2
 v2 f
Example 1 Answer
[m1v1i  m2v 2i  m1v1 f ]
m2
 v2 f
[(300.0 10.0)  (125.0  0)  (300.0  4.12)]
 v2 f
125.0
14.1m /s  v 2 f
Example 2
 If an 800.kg sports car slows to 13.0m/s to
check out an accident scene and the
1200.kg pick-up truck behind him continues
traveling at 25.0m/s, with what velocity
will the two move if they lock bumpers
after a rear-end collision?
Example 2 Picture
Before Collision
After Collision
m1 1200.kg
m2  800.kg
v1i  25.0m /s
v 2i 13.0m /s


(m1  m2 )  2000.kg
vf  ?
Example 2 Answer
m1v1i  m2v 2i  (m1  m2 )v f
(m1v1i  m2v 2i )
 vf
(m1  m2 )
Example 2 Answer
(m1v1i  m2v 2i )
 vf
(m1  m2 )
[(1200  25.0)  (800 13.0)]
 vf
(1200  800)
20.2 m/s forward  v f