Chapter 6: Momentum and Collisions. Section 6 – 1 Momentum and Impulse.
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Transcript of Chapter 6: Momentum and Collisions. Section 6 – 1 Momentum and Impulse.
Chapter 6:Momentum
and Collisions
Section 6 – 1Momentum 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
http://www.madsci.org/posts/archives/dec99/945106537.Ph.r.html
Momentum = mass x velocity
p = mv
Units: kg-m/s
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.
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.
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.
Ex 1: Gary is driving a 2500 kg
vehicle, what is his momentum if his
velocity is 24 m/s?
G: m =2500 kg, v =24 m/s
U: p =?
E: p = mv
S: p = (2500 kg)(24m/s)
S: p = 60,000 kg-m/s
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?
G: m=1.5 kg, p =23.5 kg-m/s
U: v = ?
E: p = mv or v = p/m
S:v=(23.5 kg-m/s)/(1.5kg)
S: v = 15.7 m/s
A change in momentum
p
(p = mv)
Takes force and time.
Momentums do not always stay the same. When a force is applied to a moving object, the momentum changes.
ImpulseThe product of the force and
the time over which it acts on an object, for a constant
external force.
Impulse = Ft
The change in the momentum is also
called the impulse.
Impulse – Momentum Theorem
Ft = por
Ft = mvf - mvi
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?
G: F = 70 N, t = 0.25 s U: Impulse = ?E: Impulse = FtS: Impulse =(70 N)(0.25 s)S: Impulse = 17.5 kg-m/sFt = p =17.5 kg-m/s
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?
G: F = 100 N, vf = 150 m/s, vi = 100m/s, m = 50 kg
U: t = ?
E: t = m (vf - vi)/F
S:t=50kg(150–100m/s)/100s
S: t = 25 sec
Stopping times and distances depend
upon impulse-momentum
theorem.
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?
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 = p = m (vf - vi)
t = m (vf - vi) / F
S:t=[2250kg(-5– -20m/s)]
8450 N
St=[2250kg(-5– -20m/s)]
8450 N
S: 4.0 s
B)U: x = ?
E: x = ½(vi + vf )tS:x=½(-5+-20)m/s(4 s)
S: x = - 50 m x = 50 m west
Frictional forces will be disregarded
in most of the problems unless otherwise stated.
Section 6-2Conservation of
Momentum
Law of Conservation of Momentum
The total momentum of all objects interacting with one another remains constant regardless of the nature of the forces between them.
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.
total initial momentum
total final momentum=
ffii pppp ,2,1,2,1
total initial momentum
total final momentum=
ffii vmvmvmvm ,22,11,22,11
ffii pppp ,2,1,2,1
total initial momentum
total final momentum=
For objects that collide:
The momentum of the individual object(s) does not remain constant, but
the total momentum does.
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.
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.
Which skateboarder has the higher
velocity?The one with the
smaller mass.
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?
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
0 + 0 = mjvJ,f + mbvb,f
vb,f = (mJvJ,f) /-mb
vb,f = (mJvJ,f) /-mb
S: vb,f = (76kg x 2.5m/s) - 45kg
vb,f = (mJvJ,f) /-mb
S: vb,f = (76kg x 2.5m/s) - 45kg
S: vb,f = - 4.2 m/s
vb,f = (mJvJ,f) /-mb
S: vb,f = (76kg x 2.5m/s) - 45kg
S: vb,f = - 4.2 m/s
or 4.2 m/s to the left
Newton’s 3rd Law leads to a
conservation of momentum.
Open books to page 219-220
Forces in real collisions are not
constant. They vary throughout the
collision, but are still opposite and equal.
Section 6 – 3 Elastic and
Inelastic Collisions
Perfectly Inelastic Collisions
A collision in which two objects stick together and move with a common velocity.
fii vmmvmvm 21,22,11
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.
(a) What is the velocity of the
entangled engines after the collision?
G: m1 = 5.0 kg, m2 =4.0kg v1 = 3 m/s, & v2 = 0 m/s
U: v1+2 = ?
E: p1,i + p2,i = p1+2
m1v1 + m2v2 =m1+2 v1+2
v1+2 =[m1v1 +m2v2]/m1+2
v1+2= [(5 x 3)+(4 x 0)]/9
v1+2= 1.66 m/s, east
KE is not constant in inelastic collisions.
Some of the KE is converted into sound energy and internal energy as the objects are deformed.
This KE can be calculated from the equation in Ch 5.
KE = KEf – KEi
KE =1/2mvf2 - 1/2mvi
2
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?
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
vf=[(0.5x4)+(0.25x-3)] 0.75
vf=[(0.5x4)+(0.25x-3)] 0.75
v1+2,f = 1.67 m/s,
to the right
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.
KEi = ½ mv1,i2 +½ mv2,i
2
KEi =½(0.5)(4)2+½(0.25)(-3)2
KEi = 5.12 J
KEf = ½(m1+m2)vf2
KEf = ½(0.5+0.25)(1.67)2
KEf = 1.05J
S: KE=1.05 J – 5.12 J
S: KE = - 4.07 J
Elastic CollisionsA collision in which the total momentum and the total KE
remains constant. Also, the objects separate and
return to their original shapes.
In the real world, most collisions are neither elastic nor perfectly inelastic.
Total momentum and total KE remain constant through
an elastic collision.
m1v1,I + m2v2,I
= m1v1,f + m2v2,f
½m1v1,i2 + ½m2v2,i
2
=
½m1v1,f2 + ½m2v2,f
2
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.
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 = ?
E:
m1v1,i + m2v2,i=m1v1,f +m2v2,f
m1v1,i =[m1v1,f+m2v2,f-m2v2,i]
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
S: v2,f = [(0.015 x 0.225) + (0.03 x – 0.18) –
(0.015 x – 0.315)] / 0.03
S: v2,f = 0.09 m/s (right)
KEi = ½m1v1,i2 + ½m2v2,i
2
KEi = ½(0.015)(0.225)2
+ ½(0.03)(-0.18)2
KEi = 0.000866 J
KEf = ½m1v1,f2 + ½m2v2,f
2
KEf = ½m1v1,f2 + ½m2v2,f
2
KEf = ½(0.015)(0.315)2
+ ½(0.03)(0.09)2
KEf = 0.000866 J
Since KEi = KEf,, KE is conserved.