Chapter 7

Impulse and Momentum• Impulse-Momentum Theorem

• Principle of Conservation of Linear Momentum

• Collisions in One Dimension

• Collisions in Two Dimensions

• Centre of Mass

Impulse and Momentum

Newton’s 2nd Law:

Or

So,

Define the momentum:

Then, (impulse – momentum theorem)

maF

tm

vF

vF

mt

FF

force the of impulse the ist

vp

m

pvF

mt

Impulse = change in momentum

That is, momentum is conserved when the net forceacting on an object is zero.

This applies also to an isolated system of two or moreobjects (no external forces) that may be in contact - thetotal momentum is conserved.

Compare Newton’s first law:

velocity is constant when the net force is zero.

pvF

mt

00 pF

thenIf

Alternative formulation of Newton’s second law

pvF

mt

t :or

pF

The net force acting on an object is equal to the rateof change of momentum of the object.

Example: A 0.14 kg baseball has an initial velocity v0 = -38 m/sas it approaches a bat. The bat applies an average force F that is much larger than the weight of the ball.

After being hit by the bat, the ball travels at speed vf = +58 m/s.

a) The impulse applied to the ball is

m/skg or sN

m/skg

mm

4413

3858140

.

.

)( 0f vvv

b) The bat is in contact with the ball for 1.6 ms.So the average force of the bat on the ball is:

NssN

tm

84001061

44133

.

.vF

Problem 7.13:

A 0.047 kg golf ball strikes a hard, smooth floor at an angle of 30°, and rebounds at the same angle.

What is the impulse applied to the golf ball by the floor?

Problem 7.14:

A dump truck is being filled with sand. The sand falls straight downward from rest from a height of 2.00 m above the truck bed, and the mass of sand that hits the truck per second is 55.0 kg/s. The truck is parked on the platform of a weight scale.

By how much does the scale reading exceed the weight of the truck and sand?

Conservation of Momentum

Two isolated masses collide. Theinitial total momentum is:

With

While the masses are in contact, theyexert equal and opposite forces on eachother (Newton’s third law).

So the impulse acting on m1 is equal in magnitude and opposite indirection to the impulse acting on m2

21 ppp

0222

0111

vp

vp

m

m

2112 FF

Therefore, (change in momentum = impulse)

After the collision:

So, the total momentum after the collision is:

That is, total momentum is conserved:

21 pp

12222

111

ppppvp

ppvp

f2

f1

m

m

p

pp

ppppppp

21

121121

2121 pppp

Problem 7.C13:

An ice boat slides without friction horizontally across the ice. Someone jumps vertically down from a bridge onto the boat.

Does the momentum of the boat change?

As the momentum of the person isdownward, not sideways, the horizontalmomentum of the boat is unchanged.

As the mass of the boat is increased bythe mass of the person, the boat movesmore slowly, so that the momentum is unchanged.

MmMv

v

vMmMv

f

fo

0

Problem 7.16:

A 55-kg swimmer is standing on a stationary 210-kg floating raft. The swimmer then runs off the raft horizontally with a velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.

The freight car on the left catches up with the one on the rightand connects up with it. They travel on with the same speed vf.

Conservation of momentum:

21

2211

212211

mmvmvm

v

vmmvmvm

f

f

21

2211

mmvmvm

vf

Example:

Kinetic Energy:

What happened to the missing energy?

m/svm/s,v

kgmkg,m

3180

0009200065

0201

21

..

,,

m/svf 0919200065000

31920008065000.

..

JvmmKE

JvmvmKE

ff

i

932662

1

985402

1

2

1

221

2022

2011

Missing 5274 J

This was an inelastic collision – some of the kinetic energy was converted to heat.

Problem 7.28:

A projectile (mp = 0.2 kg) is fired at and embeds itself in a target (mT = 2.5 kg). The target (with the projectile in it) flies off after being struck.

What percentage of the projectile’s incident kinetic energy does the target (with the projectile in it) carry off after being struck?

The Ballistic Pendulum How fast was the bullet?

Mass of bullet: m1 = 0.01 kgMass of block: m2 = 2.50 kg

• Elastic collision: the total kinetic energy after collision is equalto the total before collision.

• Inelastic collision: the total kinetic energy is not conserved. Ifobjects stick together after collision, the collision is “perfectly

inelastic” – no bouncing of one object from the other.

Elastic CollisionExample: A ball of mass m1 = 0.25 kg makes a perfectly elastic

collision with a ball of mass m2 = 0.8 kg.

221111

2211

11 0

ff

ff

vmvmvm :conserved is Momentum

vmvm impact after Momentum

vm momentum Initial

2211011 ff vmvmvm :Momentum Conserved

So,

Then use conservation of energy

Combine the twoequations and aftersome algebra:

2

110112 m

vmvmv f

f

222

211

2011 2

1

2

1

2

1ff vmvmvm

21

1012

21

21011

2

mmm

vv

mmmm

vv

f

f

0121

21

0 vvv then

mmif

ff ,

,

21

1012

21

21011

2

mmm

vv

mmmm

vv

f

f

m/svm/sv

then

m/svand

kgm kg,mif

382622

5

80250

21

01

21

.,.

,..

ff

• so, the lighter ball bounces back from the heavier ball

• momentum is conserved

• kinetic energy is conserved

Problem 7.37:An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line.

What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision?

Dec 2003 Final, Q26:

A bullet of mass m is fired with speed v0 into a block of wood of mass M. The bullet comes to rest in the block. The block with the bullet inside slides along a horizontal surface with coefficient of kinetic friction k.

How far does the block slide before it comes to rest?

Collisions in Two Dimensions

2121 ooff

of

pppp

pp

Conservation of momentum:

Satisfied along both axis separately:

yyyyyy

xxxxxx

2121

2121

ooffof

ooffof

pppppp

pppppp

m/skgp

p

pppp

x

x

x

09470

35702605402605015090

3550

1

1

2121

.

cos....sin..

cossin

f

f

ffoo

m/skgp

p

ppp

y

y

y

01760

50901503570260

3550

1

1

211

.

cos..sin..

sincos

f

f

ffo

510

18590

640

096300176009470

111

221

.

.

./

...

0.09470.0176

tan

m/smpv

m/skgp

ff

f

1fpm/skg p y 017601 .f

m/skg p y 094701 .f

x

y

Problem 7.19: A fireworks rocket breaks into two pieces of equalmass.

Find the velocities of the pieces.