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Chapter 6: Momentum and Collisions. 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)

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momentum p
Momentum (p)

A vector quantity defined as the product of an object’s mass and velocity.

Describes an object’s motion.

slide4

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
Momentum = mass x velocity

p = mv

Units: kg-m/s

conceptualizing momentum
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 momentum1
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 momentum2
Conceptualizing momentum

Question –

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.

slide10
Ex 1: Gary is driving a 2500 kg vehicle, what is his momentum if his velocity is 24 m/s?
slide11
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

slide12
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?
slide13
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
A change in momentum

Dp

(Dp = mDv)

Takes force and time.

slide15
Momentums do not always stay the same. When a force is applied to a moving object, the momentum changes.
impulse
Impulse

The product of the force and the time over which it acts on an object, for a constant external force.

Impulse = FDt

impulse momentum theorem
Impulse – Momentum Theorem

FDt = Dp

or

FDt = mvf - mvi

slide19
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?
slide20
G: F = 70 N, Dt = 0.25 s

U: Impulse = ?

E: Impulse = FDt

S: Impulse =(70 N)(0.25 s)

S: Impulse = 17.5 kg-m/s

FDt = Dp =17.5 kg-m/s

slide21
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?
slide22
G: F = 100 N, vf = 150 m/s, vi = 100m/s, m = 50 kg

U: Dt = ?

E: Dt = m (vf - vi)/F

S:Dt=50kg(150–100m/s)/100s

S: Dt = 25 sec

slide24
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?
slide25
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/s

U: Dt =?

E: F Dt = Dp = m (vf - vi)

Dt = m (vf - vi) / F

slide26
S:

Dt=[2250kg(-5– -20m/s)] 8450 N

slide27
S:

Dt=[2250kg(-5– -20m/s)] 8450 N

S: 4.0 s

slide28
B)

U: Dx = ?

E: Dx = ½(vi + vf )Dt

S:Dx=½(-5+-20)m/s(4 s)

S: Dx = - 50 m

Dx = 50 m west

law of conservation 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
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.

slide33

total initial momentum

total final momentum

=

slide34

total initial momentum

total final momentum

=

slide35

total initial momentum

total final momentum

=

for objects that collide
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
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.

slide38
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
Which skateboarder has the higher velocity?

The one with the smaller mass.

slide41
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?
slide42
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

slide44
vb,f = (mJvJ,f) /-mb

S: vb,f = (76kg x 2.5m/s) - 45kg

slide45
vb,f = (mJvJ,f) /-mb

S: vb,f = (76kg x 2.5m/s) - 45kg

S: vb,f = - 4.2 m/s

slide46
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

slide48
Forces in real collisions are not constant. They vary throughout the collision, but are still opposite and equal.
perfectly inelastic collisions
Perfectly Inelastic Collisions

A collision in which two objects stick together and move with a common velocity.

slide52

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.

slide54
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

slide55
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
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 d ke can be calculated from the equation in ch 5
This D KE can be calculated from the equation in Ch 5.

D KE = KEf – KEi

D KE =1/2mvf2 - 1/2mvi2

slide58
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?
slide59
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

slide61
vf=[(0.5x4)+(0.25x-3)] 0.75

v1+2,f = 1.67 m/s,

to the right

slide62
b)

U: D KE = ?

E: D KE = KEf – KEi

We need to find the combined final KE and both initial KE’s. Using the KE = ½ mv2.

slide63
KEi = ½ mv1,i2 +½ mv2,i2

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

slide64
S: D KE=1.05 J – 5.12 J

S: D KE = - 4.07 J

elastic collisions
Elastic Collisions

A collision in which the total momentum and the total KE remains constant.

Also, the objects separate and return to their original shapes.

total momentum and total ke remain constant through an elastic collision
Total momentum and total KE remain constant through an elastic collision.

m1v1,I + m2v2,I

=

m1v1,f + m2v2,f

slide69
½m1v1,i2 + ½m2v2,i2

=

½m1v1,f2 + ½m2v2,f2

slide70

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.

slide71
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 = ?

slide72
E:

m1v1,i + m2v2,i=m1v1,f +m2v2,f

m1v1,i =[m1v1,f+m2v2,f-m2v2,i]

slide73
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

slide74
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)

slide75
KEi = ½m1v1,i2 + ½m2v2,i2

KEi = ½(0.015)(0.225)2

+ ½(0.03)(-0.18)2

KEi = 0.000866 J

KEf = ½m1v1,f2 + ½m2v2,f2

slide76
KEf = ½m1v1,f2 + ½m2v2,f2

KEf = ½(0.015)(0.315)2

+ ½(0.03)(0.09)2

KEf = 0.000866 J

Since KEi = KEf,, KE is conserved.