Transcript Document

Momentum
Momentum can be defined as "mass in motion."
Momentum = mass * velocity
p=m*v
Units are
kg*m/s
Momentum is a vector quantity. It has both magnitude and direction
A 130 kg defender moving forwards at 9 m/s will have (130 kg) (9m/s) = 1170 kgm/s
of forward momentum
pD = 1170 kgm/s
A 84 kg running back moving sideways at 6 m/s will have (84 kg) (6m/s)
pB 504 kgm/s
What happens when the defender tackles the running back?
= 504 kgm/s
of sideways momentum
Conservation of Momentum
If there are no outside forces acting on an object its momentum will stay the
same.
Friction is an outside force that will change the momentum of an object
In a collision between two objects the total momentum just before the collision
will be the same just after the collision
Total Momentum just before collision =
pDi + pBi
mDvDi + mBvBi
Total Momentum just after collision
=
pDf + pBf
=
mDvDf + mBvBf
If the two objects stick together after they collide there is only one final velocity
mDvDi + mBvBi
=
mDvf + mBvf =
(mD + mB) vf
p+
p-
If the collision occurs in a straight line you must assign velocity
positive or negative
Conservation of Momentum
A 130 kg defender moving forwards at 9m/s tackles a 84 kg running back moving towards
him at 6 m/s. If a sticky (perfectly inelastic) collision takes place what velocity do the
players move off after the collision?
mDvDi + mBvBi
p+
=
(130kg)(+9 m/s) + (84kg)(-6m/s) =
p-
+666 kgm/s
=
+3.12 m/s
=
(mD + mB) vf
(130kg + 84kg) vf
(214kg) vf
vf
If the collision is at an angle you must add the initial momentum vectors tip to tail to find the total
momentum or add the individual components
pBi
pDi
pTotal I = pTotal f
Conservation of Momentum
A 130 kg defender moving forwards at 9m/s tackles a 84 kg running back moving sideways
at 6 m/s. If a sticky (perfectly inelastic) collision takes place what velocity (speed and
direction) do the players move off after the collision?
pBi = 504 kgm/s
pDi = 1170 kgm/s
pTotal I = pTotal f
pTotal f =  (pBi2 + pDi2)

 = tan-1 (pBi / pDi)
 = tan-1 (504 / 1170)
 = 23.30
pTotal f =  ((504 kgm/s)2 + (1170kgm/s)2)
pTotal f = 1273.9 kgm/s
(mD + mB) vf = 1273.9 kgm/s
vf = 1273.9 kgm/s / (130kg + 84kg)
vf = 5.95 m/s = 6 m/s
Explosions
In an explosion internal forces are responsible for the object breaking apart. Because the
pieces impart equal and opposite forces on each other (Newton’s third law) these internal
forces cannot provide a net change in momentum so momentum must be conserved in
explosions
Consider a falling firecracker that explodes into two pieces. The momenta of the
fragments combine by vector rules to equal the original momentum of the falling
firecracker.
Note: If an object is at rest and then explodes the total momentum of all the pieces
just after the explosion must all add up to zero!
Impulse
An impulse is a force exerted over a finite period of time
A net impulse will change the momentum of an object in accordance with Newton’s second law of
motion
FNET = ma = m (v / t)
therefore
FNET t = m v =  (mv) = p
A 0.2 kg baseball traveling at 40 m/s is hit and returned at 50 m/s in the opposite direction. If the ball
and bat are in contact for 0.002s determine the average force on the ball
p+
p-
mball = 0.2 kg vbi = 40 m/s
Fav t = m v =
vBf = -50 m/s t = 0.002s Fav = ?
m (vBf - vBi) therefore
Fav = (0.2 kg) (-50 m/s - 40 m/s) / (0.002s)
the area under the f-t curve
represents the change in
momentum or impulse applied
to the object
INET = Fav t = Area
Fav = m (vBf - vBi) / t
= - 9000N
Impulse
There are two ways of changing the momentum of an
object.
It can be changed with a large force exerted over a
short period of time
or with a smaller force exerted over a longer period of time
Extending the time for which a
force acts is especially seen in
sports - pads, mats, gloves, following through
The strings on a tennis racket bend
and the ball compresses. This gives a
longer time for the rebounding forces
to act, thus increasing the velocity of
the ball.
Note: the force does not increase by
having strings. This depends on the
swing.
and vehicle safety - shocks, padded dash board, air-bags,
seat belts, crush zones, safety barriers
around bridge columns
The front end of a modern car is
designed to cave in in order to extend
the time of the collision. This reduces
the collision force experienced by the
occupants of the car.
Impulse and Conservation of Momentum
According to Newton’s third law when two object interact
Because the two objects interact for the same period of time
FA = - FB
FA t = - FB t
This can be written in terms of impulse as….
IA = - IB
Written in terms of changes in momentum….
pA = - pB
acar >> aTruck
ICar = - ITruck
A massive truck and a small car are involved in a collision. Which object has the largest I)
impulse ii) change in momentum iii) acceleration during the collision?
i) Both vehicles have the same impulse acting on them
but in opposite directions
ICar = - ITruck
mCar
vCar
=
MTruck v
Truck
ii) Both vehicles have the same change in momentum but
in opposite directions
iii) The change in velocity of the car is greater than the change in velocity of the truck because of its
smaller inertia (mass). The car’s acceleration is therefore greater than the truck’s acceleration. The
car’s passengers would therefore feel the acceleration more than the truck driver.
Deriving Conservation of Momentum
According to Newton’s third law when two object interact
Because the two objects interact for the same period of time
FA = - FB
FA t = - FB t
This can be written in terms of impulse as….
IA = - IB
Written in terms of changes in momentum….
pA = - pB
mA (vAf - vAi) = - mB (vBf - vBi)
Substitute for p….
mAvAf - mAvAi = - mBvBf + mBvBi
rearrange….
mAvAf + mBvBf = mAvAi + mBvBi
ptotal f = ptotal i
The total momentum after the collision = The total momentum before the collision
If a shotgun of mass 3.0 kg fires shot having a total
mass of 0.05kg with a muzzle velocity of 525 m/s,
what is its recoil velocity?
mg = 3.0 kg
ms = 0.05 kg
ptotal I = 0 kgm/s vsf = 525 m/s vgf = ?
ptotal I = ptotal f = mg vgf + ms vsf = 0 kgm/s
vgf = - (0.05 kg) (+525 m/s) / (3.0 kg)
(3.0 kg) vgf + (0.05 kg) (+525 m/s) = 0 kgm/s
vgf = - 8.75 m/s = - 9 m/s