MomentumMomentum
Momentum can be defined as "mass in motion."
Momentum = mass * velocity
p = m * v
kg*m/s
Units are
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/sof forward momentum
A 84 kg running back moving sideways at 6 m/s will have
(84 kg) (6m/s)
= 504 kgm/sof sideways momentum
pD = 1170 kgm/s
pB 504 kgm/s
What happens when the defender tackles the running back?
Conservation of MomentumConservation of Momentum
If there are no outside forces acting on an object its momentum will stay thesame.
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 collisionwill be the same just after the collision
Total Momentum just before collision =
Total Momentum just after collision
p +
pDi + pBi =
pDf + pBf
mDvDi + mBvBi =
mDvDf + mBvBf
p -
If the collision occurs in a straight line you must assign velocitypositive or negative
If the two objects stick together after they collide there is only one final velocity
mDvDi + mBvBi =
mDvf + mBvf =
(mD + mB) vf
Conservation of MomentumConservation of Momentum
A 130 kg defender moving forwards at 9m/s tackles a 84 kg running back moving towardshim at 6 m/s. If a sticky (perfectly inelastic) collision takes place what velocity do theplayers move off after the collision?
p +
p -
If the collision is at an angle you must add the initial momentum vectors tip to tail and use theparallelogram rule to find the total momentum
mDvDi + mBvBi =
(mD + mB) vf
(130kg)(+9 m/s) + (84kg)(-6m/s) =
(130kg + 84kg) vf
+666 kgm/s =
(214kg) vf
+3.12 m/s = vf
pDi
pBi
pTotal I = pTotal f
Conservation of MomentumConservation of Momentum
A 130 kg defender moving forwards at 9m/s tackles a 84 kg running back moving sidewaysat 6 m/s. If a sticky (perfectly inelastic) collision takes place what velocity (speed anddirection) do the players move off after the collision?
pDi = 1170 kgm/s
pBi = 504 kgm/s
pTotal I = pTotal f
pTotal f = (pBi2 + pDi2)
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
= 23.30
Using scale diagram:1cm = 100 kgm/sand measure the anglewith a protractor
ExplosionsExplosions
In an explosion internal forces are responsible for the object breaking apart. Because thepieces impart equal and opposite forces on each other (Newton’s third law) these internalforces cannot provide a net change in momentum so momentum must be conserved inexplosions
Consider a falling firecracker that explodes into two pieces. The momenta of thefragments combine by vector rules to equal the original momentum of the fallingfirecracker.
Note: If an object is at rest and then explodes the total momentum of all the piecesjust after the explosion must all add up to zero!
ImpulseImpulse
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 ofmotion
FNET = ma =
FNET t = m v
m (v / t)
therefore
= (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 balland bat are in contact for 0.002s determine the average force on the ball
Fav t = m v =
mball = 0.2 kg vbi = 40 m/s vBf = -50 m/s t = 0.002s Fav = ?
p +
p -
m (vBf - vBi)
therefore
Fav = m (vBf - vBi) / t
Fav = (0.2 kg) (-50 m/s - 40 m/s) / (0.002s)
= - 9000N
the area under the f-t curverepresents the change inmomentum or impulse appliedto the object
INET = Fav t = Area
ImpulseImpulse
There are two ways of changing the momentum of anobject.
It can be changed with a large force exerted over ashort period of time
or with a smaller force exerted over a longer period of time
Extending the time for which aforce acts is especially seen insports -
pads, mats, gloves, following through
and vehicle safety -
shocks, padded dash board, air-bags,seat belts, crush zones, safety barriersaround bridge columns
The strings on a tennis racket bendand the ball compresses. This gives alonger time for the rebounding forcesto act, thus increasing the velocity ofthe ball.
Note: the force does not increase byhaving strings. This depends on theswing.
The front end of a modern car isdesigned to cave in in order to extendthe time of the collision. This reducesthe collision force experienced by theoccupants of the car.
Impulse and Conservation of MomentumImpulse and Conservation of Momentum
According to Newton’s third law when two object interact
FA t = - FB t
FA = - FB
Because the two objects interact for the same period of time
This can be written in terms of impulse as….
IA = - IB
Written in terms of changes in momentum….
pA = - pB
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?
mCar vCar = MTruck vTruck
ICar = - ITruck
i) Both vehicles have the same impulse acting on thembut in opposite directions
ii) Both vehicles have the same change in momentum butin opposite directions
iii) The change in velocity of the car is greater than the change in velocity of the truck because of itssmaller inertia (mass). The car’s acceleration is therefore greater than the truck’s acceleration. Thecar’s passengers would therefore feel the acceleration more than the truck driver.
ICar = - ITruck
acar >> aTruck
Deriving Conservation of MomentumDeriving Conservation of Momentum
According to Newton’s third law when two object interact
FA t = - FB t
FA = - FB
Because the two objects interact for the same period of time
This can be written in terms of impulse as….
IA = - IB
Written in terms of changes in momentum….
pA = - pB
Substitute for p….
mA (vAf - vAi) = - mB (vBf - vBi)
mAvAf - mAvAi = - mBvBf + mBvBi
mAvAf + mBvBf = mAvAi + mBvBi
rearrange….
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 totalmass 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
(3.0 kg) vgf + (0.05 kg) (+525 m/s) = 0 kgm/s
vgf = - (0.05 kg) (+525 m/s) / (3.0 kg)
vgf = - 8.75 m/s = - 9 m/s