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.