Chapter 7Momentum

In golfing or in baseball – why is the follow through of the swing so important in hitting the LONG BALL?

Momentum = mass x velocity (speed)

momentum (p) = mv

What is Momentum measured in?A) Newtons B) kg m/sC) m/s² D) Joules

By definition:

The momentum of a 6 kg bowling ball going 6 m/s down the alley. How much momentum does the ball have?

A bowler throws the same ball down the alley at the same velocity, but the ball strikes a wall at the end and recoils back towards you at 2 m/s. What is the change in momentum?

Would it hurt more getting hit from the boxer with the gloves

on or the gloves off?

A) Gloves on B) Gloves off

WHY?

Impulse = Change in Momentum

If momentum changes – what does not have to change?

A) Mass B) Velocity C) Time

When things change momentum, most of the time the object is changing itsVelocity, and not its mass. Think about it… while walking down the hallway, if Wanted to change your momentum, would you change your mass or your speed?

So…If you change your velocity, you are ACCELERATING – if you are accelerating

Some FORCE must be happening – it all goes back to Newton’s 2nd law (F = MA)

The greater the force, the greater the acceleration (change in velocity), the greater the momentum.

The amount of time Force is applied is very important to IMPULSE (change in momentum)

Impulse = Force x time intervalChange in velocity = Force x time

Δmv = FΔt(this equation is just a modification of F=ma)

Increasing momentum:Apply the greatest force possible for the longest time possible

Why is a follow through so important in a golf swing?

Δmv = FΔt

If you have a strong follow through, the ball and the golf clubAre in contact with each other longer. That teeny weeny littleBit of extra contact time gives a longer golf shot.

A .2 kg baseball is pitched going 20 m/s and is hit by a batter where the ball is going 50 m/s off the bat. What is the impulse applied to the ball by the bat?

If the bat is in contact with the ball for .016 sec, how much average force did the bat apply to the ball?

.2(20) + .2(50) = 14 kg m/s

14 kg m/s = F (.016 sec)

F = 875 N

Conservation of Momentum

The momentum before = momentum after a collision

mv = mvAn 1500 kg SUV going 10 m/s hits a 5,000 kg bus

going 6 m/s, and lock up together. What is the speed of the locked up SUV and Bus?mvbefore = mvafter

1500kg(10m/s) + 5000kg(6m/s) = mvbefore

(1500kg+5000kg) vsuv/bus = mvafter

45,000 kg m/s = 6500 kg vV = 6.9 m/s

A 60kg skater pushes a shopping car of mass 20 kg. The shopping car takes off moving at 3 m/s, what is the recoil

speed of the skater?

mvbefore = mvafter

0 = (60 kg) v + (20 kg)(3 m/s)

vskater= 60 kg m/s / 60 kg = 1 m/s

Elastic Collisions – total KE before collision is equal to total KE after.Inelastic collision – total KE before collision is not equal to total KE after.

A ) Elastic B) InelasticA ball of clay is thrown against the wall and sticks. A person jumping on a trampoline

In elastic collisions: mvbefore = mvafter

½ mv2before = ½ mv2

after

A ball of mass .250 kg moving at 5 m/s collides head on with a heavier ball (.8kg) that is at rest. What are the velocities of the ball if it is a completely elastic collision?

mvbefore = mvafter

½ mv2before = ½ mv2

after

m1v1 + m2v2 = m1vf1 + m2vf2

vf1 = (m1 – m2) v01 vf2 = ( 2 m1 ) v01

(m1 + m2) m1 + m2

Vf1 = -2.62 m/s vf2 = 2.38 m/s

Ballistic Pendulum

Ballistic pendulums are used to find the speed of bullets. A block of wood (2.5 kg) hangs from a pendulum. A fired bullet (.01 kg) hits the wood with speed v01 and lodges itself in the wood. The pendulum swings upward .65 m as a result of the collision. What is the speed of the bullet before it struck the wood?m1v01 = (m1 + m2) vf

V01 = (m1 + m2) vf

m1

We need vf

With the pendulum, the KE of the bullet/block at the bottom will equal the PE at the top of the swingKEat bottom = Peat top

½ mv2f = mgh for m, it is equal to the bullet+block

Vf = 2gh

V01= 896 m/s

2 Dimensional collisions

Collisions in 2 dimensions is the same as 1D collisions except that you need to find the x and y components of the velocities.

50o

Ɵo

Vf1=

V02 = .54 m/sM2 = .26 kg

V01 = .9m/sM1 = .15 kg

35o

Vf2 = .7 m/s

X component: m1v01 + m2v02 = m1vf1 + m2vf2

.15kg(.9 m/s sin 50o) + (.26 kg) .54m/s = .15(vf1) + .26 kg(.7m/s cos 35o)

solving for vf1 = .63 m/s

Y component: m1v01 + m2v02 = m1vf1 + m2vf2

..15kg(-.9m/s cos 50o) + .26kg(0 m/s) = .15kg(vf1) + .26kg(-.7 sin 35o)

solving for vf1 = .12 m/s

50o

Ɵo

Vf1= .64 m/s

V02 = .54 m/sM2 = .26 kg

V01 = .9m/sM1 = .15 kg

35o

Vf2 = .7 m/s

.63 m/s

.12 m/s

a2+b2 = c2

.632 + .122 = vf12

vf1 = .64 m/s

Ɵ= tan-1(.12/.63)Ɵ = 11o

Center of MassAn objects center of mass is a point that

represents the average location for the total mass of the object.

(the cm does not always have to be within the confines of the object)

m1x1 + m2x2

Center of mass xcm = m1 + m2

Center of mass

Find the center of mass for the planetary system below:

X1= 5 x2= 40M1 = 5 m2 = 100

38.3

Center of Mass Experiment

1. Stand 2 ft (your own feet) away from a wall

2. Place a chair between you and the wall

3. Lean over, with your head against the wall and grab the seat of the chair

4. Lift up the chair and stand up straight without moving your feet.

5. Using your clickerA) You did it B) No you didn’t