section 6 – 1 momentum and impulse
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Section 6 – 1 Momentum and Impulse. Momentum (p). A vector quantity defined as the product of an object’s mass and velocity. Describes an object’s motion. Why “p”? Pulse (Date: 14th century) from Latin pulsus, literally, beating, from pellere to drive, push, beat. - PowerPoint PPT PresentationTRANSCRIPT
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Section 6 – 1Momentum and
Impulse
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Momentum (p)A vector quantity defined as the product of an object’s mass and velocity.
Describes an object’s motion.
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• Why “p”?
–Pulse (Date: 14th century)
•from Latin pulsus, literally, beating, from pellere to drive, push, beat
http://www.madsci.org/posts/archives/dec99/945106537.Ph.r.html
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Momentum = mass x velocity
p = mv
Units: kg-m/s
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Conceptualizing momentum
Question –
Which has more momentum; a semi-truck or a Mini Cooper cruising the road at 10 mph?
Answer –
The semi-truck has more mass. Since the velocities are the same, the semi has more momentum.
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Conceptualizing momentum
Question –
Which has more momentum; a parked semi-truck or a Mini Cooper moving at 10 mph?
Answer –
The velocity of the semi is 0 mph. That means its momentum is 0 and this time the Mini Cooper has more momentum.
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Conceptualizing momentumQuestion –
Which has more momentum, a train moving at 1 mph or a bullet moving at 2000 mph?
Answer –
The mass of a train is very large, while the mass of a bullet is relatively small. Despite the large speed of the bullet, the train has more momentum.
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Ex 1: Gary is driving a 2500 kg
vehicle, what is his momentum if his
velocity is 24 m/s?
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G: m = 2500 kg, v = 24 m/s
U: p =?
E: p = mv
S: p =
S: p =
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Ex 2: Ryan throws a 1.5 kg football, giving it a momentum of 23.5 kg-
m/s. What is the velocity of the
football?
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G: m = 1.5 kg, p = 23.5 kg-m/s
U: v = ?
E: p = mv or v = p/m
S: v =
S: v =
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A change in momentum
p
(p = mv)
Takes force and time.
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Momentums do not always stay the same. When a force is applied to a moving object, the momentum changes.
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Impulse (J)The product of the force and
the time over which it acts on an object, for a constant
external force.
J = Ft
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The change in the momentum is also
called the impulse.
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Impulse – Momentum Theorem
Impulse causes a change in
momentum
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J = p or
Ft = mvf - mvi
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Ex 3: What is the impulse on a football
when Greg kicks it, if he imparts a force of 70 N
over 0.25 seconds? Also, what is the change
in momentum?
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G: F = 70 N, t = 0.25 s U: J = ?E: J = FtS: J =S: J =
p = Ft =
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Ex 4: How long does it take a force of 100 N
acting on a 50-kg rocket to increase its speed from 100 m/s to 150
m/s?
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G: F = 100 N, vf = 150 m/s, vi = 100m/s, m = 50 kg
U: t = ?
E: t = m(vf - vi)/F
S:t =
S: t =
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Stopping times and distances depend
upon impulse-momentum
theorem.
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Ex 5: Crystal is driving a 2250 kg car west, she slows down from 20 m/s to 5 m/s. How long does it take the car to
stop if the force is 8450 N to the east? How far does the
car travel during this deceleration?
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G: m = 2250 kg F = 8450 N east = 8450 N
vi = 20 m/s west = - 20 m/s
vf = 5 m/s west = - 5 m/sU: t =?
E: F t = J = p = m (vf - vi)
t = m (vf - vi) / F
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S:t =
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St =
S: t =
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B)U: x = ?
E: x = ½(vi + vf )tS:x =
S: x = - ___ m or x = ___ m west
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Frictional forces will be disregarded
in most of the problems unless otherwise stated.
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Section 6-2Conservation of
Momentum
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Law of Conservation of Momentum
The total momentum of all objects interacting with one another, in an isolated system remains constant regardless of the nature of the forces between them.
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What this means:Any momentum lost by
one object in the system is gained by one or more of the other objects in the
system.
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total initial momentum
total final momentum=
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ffii pppp ,2,1,2,1
total initial momentum
total final momentum=
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ffii vmvmvmvm ,22,11,22,11
ffii pppp ,2,1,2,1
total initial momentum
total final momentum=
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For objects that collide:
The momentum of the individual object(s) does not remain constant, but
the total momentum does.
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Momentum is conserved when objects push away from each
other.Ex 1: Jumping, Initially there is
no momentum, but after you jump, the momentum of the you and the earth are equal and opposite.
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Ex 2: 2 skateboarders pushing away from each other. Initially neither has momentum. After pushing off one another they both have the same momentum, but in opposite direction.
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Which skateboarder has the higher
velocity?The one with the
smaller mass.
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Ex 6: John, whose mass is 76 kg, is initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock with an a velocity of 2.5 m/s to the right. What is the final velocity of the boat?
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G: mjohn=76 kg, mboat= 45kg, vjohn,
i = vboat,i = 0 m/s, vjohn,f = 2.5 m/s
U: vboat = ?E: Momentum is conserved.
PJ,i + pb,i = pJ,f + pb,f
mjvJ,i + mbvb,i = mjvJ,f + mbvb,f
0 + 0 = mjvJ,f + mbvb,f
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vb,f = (mJvJ,f) /-mb
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vb,f = (mJvJ,f) /-mb
S: vb,f =
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vb,f = (mJvJ,f) /-mb
S: vb,f =
S: vb,f =
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vb,f = (mJvJ,f) /-mb
S: vb,f =
S: vb,f = - ____ m/s
or ___ m/s to the left
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Newton’s 3rd Law leads to a
conservation of momentum.
Open books to HPg. 219-220
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Forces in real collisions are not
constant. They vary throughout the
collision, but are still opposite and equal.
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Section 6 – 3 Elastic and
Inelastic Collisions
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Perfectly Inelastic Collisions
A collision in which two objects stick together and move with a common velocity.
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fii vmmvmvm 21,22,11
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Ex 7: A toy engine having a mass of 5.0 kg and a speed of 3 m/s, east,
collides with a 4 kg train at rest. On colliding the two engines lock and remain
together.
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(a) What is the velocity of the
entangled engines after the collision?
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G: m1 = 5.0 kg, m2 =4.0kg v1 = 3 m/s, & v2 = 0 m/s
U: v1+2 = ?
E: p1,i + p2,i = pf
m1v1 + m2v2 = m(1+2)vf
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vf =[m1v1 +m2v2]/m(1+2)
vf=
vf= , east
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KE is not constant in inelastic collisions.
Some of the KE is converted into sound energy and internal energy as the objects are deformed.
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This KE can be calculated from the equation in Ch 5.
KE = KEf – KEi
KE =1/2mvf2 - 1/2mvi
2
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Ex 8: Two clay balls collide head on in a perfectly inelastic collision. The 1st ball with a mass of 0.5 kg and an initial velocity of 4 m/s to the right. The 2nd ball with a mass of 0.25 kg and an initial speed of 3 m/s to the left. a)What’s the final speed of the new ball after the collision? b)What’s the decrease in KE during the collision?
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a)G:m1,i= 0.5 kg, v1,i = 4 m/s, m2,i = 0.25 kg, v2,i= -3 m/s
U: vf = ?E: m1v1 + m2v2 = (m1+m2)vf vf, =[m1v1 + m2v2]/(m1+m2)
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vf=
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Vf=
v1+2,f = ____ m/s,
to the right
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b)U: KE = ?
E: KE = KEf – KEi We need to find the combined final KE and both initial KE’s. Using the KE = ½ mv2.
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KEi = ½ mv1,i2 +½ mv2,i
2
KEi =
KEi =
KEf = ½(m1+m2)vf2
KEf =
KEf =
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S: KE = KEf – KEi
=
S: KE = - ____ J
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Elastic CollisionsA collision in which the total momentum and the total KE
remains constant. Also, the objects separate and
return to their original shapes.
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In the real world, most collisions are neither elastic nor perfectly inelastic.
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Total momentum and total KE remain constant through
an elastic collision.
m1v1,I + m2v2,I
= m1v1,f + m2v2,f
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½m1v1,i2 + ½m2v2,i
2
=
½m1v1,f2 + ½m2v2,f
2
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Ex 9: An 0.015 kg marble moving to the right, at 0.225 m/s has an elastic
collision with a 0.03 kg moving to the left at 0.18 m/s. After the collision the
smaller marble moves to the left at 0.315 m/s. Disregard friction. A) What
is the velocity of the 0.03 kg marble after the collision? B) Verify answer by
confirming KE is conserved.
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G:m1=0.015 kg,m2=0.03 kg
v1,i= 0.225m/s
v2,i = - 0.18 m/s
v1,f= - 0.315m/s
U: v2,f = ?
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E:
m1v1,i + m2v2,i=m1v1,f +m2v2,f
m1v1,i =[m1v1,f+m2v2,f-m2v2,i]
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E:
m1v1,i + m2v2,i=m1v1,f +m2v2,f
m2v2,f =[m1v1,i+m2v2,i-m1v1,f]
v2,f = [m1v1,i+m2v2,i-m1v1,f]
m2
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S: v2,f = [(0.015 x 0.225) + (0.03 x – 0.18) –
(0.015 x – 0.315)] / 0.03
S: v2,f = ____ m/s (right)
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KEi = ½m1v1,i2 + ½m2v2,i
2
KEi = ½(0.015)(0.225)2
+ ½(0.03)(-0.18)2
KEi = ____ J
KEf = ½m1v1,f2 + ½m2v2,f
2
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KEf = ½m1v1,f2 + ½m2v2,f
2
KEf = ½(0.015)(0.315)2
+ ½(0.03)(0.09)2
KEf = _____J
Since KEi = KEf,, KE is conserved.