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1
CHAPTER 6
Impulse & Momentum
Pn. Siti Hajar Abd. Rahman
Faculty of Innovative Design and Technology
Universiti Sultan Zainal Abidin,Terengganu
DTS 1113 DYNAMICS
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Which do you think has moremomentum?
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Momentum
Momentum is a measure of how hardit is to stop or turn a moving object.
Momentum is related to both massand velocity.
Momentum is possessed by all moving
objects.
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Calculating Momentum
For one particlep = mv
For a system of multiple particlesP = pi = mivi Momentum is a vector with the same
direction as the velocity vector. The unit of momentum is
kg m/s or Ns
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Sample Problem
Calculate the momentum of a 65-kgsprinter running east at 10 m/s.
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Sample Problem Calculate the momentum of a system
composed of a 65-kg sprinter running east at10 m/s and a 75-kg sprinter running north at9.5 m/s.
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Solutionp = mv= p1 + p2
p1= (65kg)(10 m/s) = 650 kg m/s (east)p2= (75kg)(9.5 m/s) = 712.5 kg m/s (north)
650 kg m/s
712.5 kg m/s
p = (712.52 + 6502)1/2p = 964 kg m/sat tan-1(712.5/650)
= 47.6onorth of east
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Change in momentum
Like any change, change in momentumis calculated by looking at final and
initial momentums. p = pf pi p: change in momentum
pf: final momentum pi: initial momentum
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Momentum change Mini-lab
Using only a meter stick, find the momentumchange of each ball when it strikes the deskfrom a height of exactly one meter.
Turn in: Data for each ball. Include in your data section
the masses Calculation of momentum of each ball just before
and after it strikes the desk. Momentum change for each ball upon striking thedesk.
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Wording
dilemma
In which case isthe magnitude ofthe momentumchange greatest?
In which case is
the change in themagnitude of themomentumgreatest?
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Impulse
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Impulse (J)
Impulse is the product of an externalforce and time, which results in a
change in momentum of a particle orsystem.
J = F t
J =DP
Units: N s or kg m/s(same as momentum)
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Impulsive Forces Usually high
magnitude, shortduration.
Suppose the ball
hits the bat at 90mph and leaves thebat at 90 mph,what is themagnitude of the
momentum change? What is the
change in themagnitude of themomentum?
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Impulse (J) on a graph
Area under the curve
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Sample Problem
Suppose a 1.5-kg brick is dropped on a
glass table top from a height of 20 cm.a) What is the magnitude and direction of the
impulse necessary to stop the brick?
b) If the table top doesnt shatter, and stopsthe brick in 0.01 s, what is the averageforce it exerts on the brick?
c) What is the average force that the brick
exerts on the table top during this period?
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Solution a)
Find the velocity of the brick when it strikes the tableusing conservation of energy.
mgh = mv2
v = (2gh)1/2 = (2*9.8 m/s2*0.20 m) 1/2 = 2.0 m/sCalculate the bricks momentum when it strikes the
table.
p = mv = (1.5 kg)(2.0 m/s) = 3.0 kg m/s (down)
The impulse necessary to stop the brick is the impulsenecessary to change to momentum to zero.
J = Dp = pf pi = 0 3.0 kg m/s = -3.0 kg m/s
or 3.0 kg m/s (up)
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Solution b) and c)
b) Find the force using the otherequation for impulse.
J = Ft3.0 N s = F (0.01 s)
F = 300 N (upward in the same direction
as impulse)c)According the Newtons 3rd law, the
brick exerts an average force of 300
N downward on the table.
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Sample Problem
This force acts on a 1.2 kg object moving at 120.0 m/s.The direction of the force is aligned with the velocity.
What is the new velocity of the object?
0.20 0.40 0.60 0.80 t(s)
F(N)
1,000
2,000
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Solution
Find the impulse from the area under the curve.
A = base * height = (.1 s)(2500 N) = 125 Ns
J = 125 N s
Since impulse is equal to change in momentum and it isin the same direction as the existing momentum,the momentum increases by 125 kg m/s.
Dp = 125 kg m/s
Dp = pf - pi = mvf - mvimvf = mvi + Dp
= (1.2 kg)(120 m/s) + 125 kg m/s = 269 kg m/s
vf = (269 kg m /s) / (1.2 kg) = 224 m/s
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Law of Conservation of Momentum
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Law of Conservation of Momentum
If the resultant external force on asystem is zero, then the vector sum
of the momentums of the objects willremain constant.
Pbefore = Pafter
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Sample problem
A 75-kg man sits in the back of a 120-kgcanoe that is at rest in a still pond. If theman begins to move forward in the canoe at
0.50 m/s relative to the shore, whathappens to the canoe?
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Solution
The momentum before the man moves is equalto the momentum after the man moves.
pb = pa0 = mmvm + mcvc0 = (75 kg)(0.50 m/s) + (120 kg)v
v = - (75 kg)(0.50 m/s)/(120 kg)
v = -0.31 m/sThe canoe slips backward in the water at -0.31
m/s
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External versus internal forces
External forces:forces coming fromoutside the system of particles whosemomentum is being considered.
External forces change the momentum of thesystem.
Internal forces:forces arising frominteraction of particles within a system. Internal forces cannot change momentum of
the system.
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Explosions
When an object separates suddenly, as inan explosion, all forces are internal.
Momentum is therefore conserved in anexplosion.
There is also an increase in kinetic energyin an explosion. This comes from a potential
energy decrease due to chemicalcombustion.
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Recoil
Guns and cannons recoil when fired.
This means the gun or cannon must movebackward as it propels the projectileforward.
The recoil is the result of action-reaction
force pairs, and is entirely due to internalforces. As the gases from the gunpowder explosion
expand, they push the projectile forwards
and the gun or cannon backwards.
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Sample problem
Suppose a 5.0-kg projectile launchershoots a 209 gram projectile at 350
m/s. What is the recoil velocity ofthe projectile launcher?
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Solution
Momentum conservation is used to calculaterecoil speed.
pb = pa0 = mpvp + mlvl0 = (0.209 kg)(350 m/s) + (5.0 kg)v
v = - (0.209 kg)(350 m/s)/(5.0 kg)
v = - 14.6 m/s
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Sample Problem An exploding object breaks into three fragments. A 2.0 kg
fragment travels north at 200 m/s. A 4.0 kg fragment travels east
at 100 m/s. The third fragment has mass 3.0 kg. What is themagnitude and direction of its velocity?
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SolutionThe momentum before is zero, so the momentum after is zero.
This is a vector addition problem. Each fragment has a momentummagnitude of 400 kg m/s according to the formula p = mv.
400 kg m/s
400 kg m/s
400 kg m/s
(4002 + 4002)1/2
566 kg m/sdue southwest
v = p/m= 566/3= 189 m/sdue SW
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Collisions
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Collisions
When two moving objects make contactwith each other, they undergo a collision.
Conservation of momentum is used toanalyze all collisions.
Newtons Third Law is also useful. It tellsus that the force exerted by body A on
body B in a collision is equal and opposite tothe force exerted on body B by body A.
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Collisions
During a collision,external forces areignored.
The time frame ofthe collision is veryshort.
The forces are
impulsive forces(high force, shortduration).
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Collision Types
Elastic collisions Also called hard collisions
No deformation occurs, no kinetic energy lost
Inelastic collisions Deformation occurs, kinetic energy is lost
Perfectly Inelastic (stick together)
Objects stick together and become one object Deformation occurs, kinetic energy is lost
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Sample Problem An 80-kg roller skating
grandma collidesinelastically with a 40-kg kid. What is theirvelocity after thecollision?
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Sample Problem A train of mass 4mmoving 5 km/hrcouples with a
flatcar of mass m atrest. What is thevelocity of the carsafter they couple?
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Elastic Collision
After the collision, there are still twoobjects, with two separate velocities
Kinetic energy remains constant before and
after the collision. Therefore, two basic equations must hold
for all elastic collisions
pb = pa (momentum conservation) Kb = Ka (kinetic energy conservation)
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Sample Problem A 500-g cart on an air track strikes a 1,000-g cart at rest. What
are the resulting velocities of the two carts? (Assume the
collision is elastic, and the first cart is moving at 2.0 m/s whenthe collision occurs.)
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Solution
before after m1v1 = m1v1 + m2v2 1.0 = 0.50v
1+ v
2
m1 v12 = m1v12 + m2v22
2.0 = 0.50v12 + v22
Solve simultaneously
v1 = -0.67 m/s v2 = 1.33 m/s
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Center of Mass
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DEFINITION OF CENTER OF MASS
The center of mass is a point thatrepresents the average location forthe total mass of a system
The center of mass will remainconstant during a collision
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CENTER OF MASS
21
2211
mm
xmxmxcm
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Velocity of the center of mass
D
21
2211
mm
vmvm
vcm
21
2211
mm
xmxm
xcm
DDD
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Velocity of the center of mass
In an isolated system,the total linearmomentum does notchange.
Therefore, thevelocity of the centerof mass does notchange.
21
2211
mm
vmvm
vcm