CHAPTER 6
•Momentum- the product of the mass and velocity of
an object. It is equal to
•In general the momentum of an object can be
conceptually thought of as the tendency for an object to
continue to move in its direction of travel. As such, it is
a natural consequence of Newton's first law.
•Momentum is a conserved quantity, meaning that the
total momentum of any closed system (one not affected
by external forces) cannot be changed.
•The amount of momentum that an object has depends on two
physical quantities: the mass and the velocity of the moving object
in the frame of reference. In physics, the symbol for momentum is
usually denoted by a small p (bolded because it is a vector), so
this can be written:
•where p is the momentum, m is the mass, and v the velocity.
The velocity of an object is given by its speed and its direction.
Because momentum depends on velocity, it too has a magnitude
and a direction and is a vector quantity. For example the
momentum of a 5-kg bowling ball would have to be described by
the statement that it was moving westward at 2 m/s. It is
insufficient to say that the ball has 10 kg m/s of momentum
because momentum is not fully described unless its direction is
given.
Practice Problem
• A 2250 kg pickup truck has a velocity of 25
m/s to the east. What is the momentum of
the truck?
m= 2250 kg v= 25 m/s to the east
p=?
• p=mv=(2250 kg)(25 m/s)= 56,000 kgm/s
to the east
• Practice 6 a 1-3
•A change in momentum takes force and time
•According to Newton's second law, force is equal to the change
in momentum with respect to time
We see F=ma which comes from the original form
F= Dp/Dt = the change in momentum/change in time
So with algebra we rearrange the equation to be FDt=Dp this is
the impulse momentum theorem
The left side is known as the impulse. It is the force applied to a
object in a given time interval that changes the momentum of
the object.
•The common equation relating to force can be applied if the
mass of the object is constant. Luckily, that scenario is
extremely common.
If a system is in equilibrium, then the change in momentum with
respect to time is equal to 0
Practice Problem
• A 1400 kg car is moving west with a
velocity of 15 m/s collides with a pole and
stops in .30 s. Find the magnitude of the
force exerted on the car during collision?
• FDt = Dp = mvf – mvi
• F= mvf – mvi /Dt = (1400kg)(0) –
(1400kg)(15m/s) / .30 s =70,000 N to the
east
• Problems 6 b 1-4
• Stopping times and distances also depend on the
impulse-momentum theorem
• Something that has more mass than it originally did
must undergo a greater change in momentum in
order to stop than it would with less mass.
• Also a change in momentum over a longer time
requires less force
• Coming to a red light the fast your going the more
time your going to need to slow down if you apply
the brakes way in advanced you apply them easier
and it takes less force. If you did it almost right at
the light you have to slam on your brakes and
apply a bigger force to stop.
Practice problem
• A 2250 kg car traveling west slows down
uniformly from 20 m/s to 5 m/s. How long does it
take the car to decelerate if the force on the car
is 8450 N to the east? How far does the car go
during the deceleration?
• FDt = Dp = mvf – m
 Dt = mvf – mvi / F
• = (2250 kg)(-5m/s) –
(2250kg)(20m/s)/8450kgm^2/s^2 = 4s
• d= ½(vi + vf)Dt =1/2(-20 m/s-5 m/s)(4s)
= -50 m = 50 m west
Problems 6c 1-3