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Page 1: Forces & Motion .

Forces & Forces & MotionMotionwww.hse.k12.in.us/staff/tjohnsonwww.hse.k12.in.us/staff/tjohnson

Page 2: Forces & Motion .

Newton’s Third LawNewton’s Third Law Newton’s Third Law of MotionNewton’s Third Law of Motion

When one object exerts a force on a When one object exerts a force on a second object, the second object exerts second object, the second object exerts an equal but opposite force on the first.an equal but opposite force on the first.

Page 3: Forces & Motion .

Newton’s Third LawNewton’s Third Law Problem:Problem:

How can a horse How can a horse pull a cart if the cart pull a cart if the cart is pulling back on is pulling back on the horse with an equal but opposite force? the horse with an equal but opposite force?

NO!!!

Aren’t these “balanced forces” resulting in no Aren’t these “balanced forces” resulting in no acceleration?acceleration?

Page 4: Forces & Motion .

Newton’s Third LawNewton’s Third Law

forces are equal and opposite but act on forces are equal and opposite but act on differentdifferent objects objects

they are not “balanced forces”they are not “balanced forces” the movement of the horse depends on the the movement of the horse depends on the

forces acting forces acting on the horseon the horse

Explanation:Explanation:

Page 5: Forces & Motion .

Newton’s Third LawNewton’s Third Law

Action-Reaction PairsAction-Reaction Pairs

The hammer exerts a The hammer exerts a force on the nail to the force on the nail to the right.right.

The nail exerts an equal The nail exerts an equal but opposite force on but opposite force on the hammer to the left.the hammer to the left.

Page 6: Forces & Motion .

Newton’s Third LawNewton’s Third Law

Action-Reaction PairsAction-Reaction Pairs The rocket exerts a The rocket exerts a

downward force on the downward force on the exhaust gases.exhaust gases.

The gases exert an equal The gases exert an equal but opposite upward force but opposite upward force on the rocket.on the rocket.

FG

FR

Page 7: Forces & Motion .

Newton’s Third LawNewton’s Third Law

Action-Reaction PairsAction-Reaction Pairs Both objects accelerate.Both objects accelerate.

The amount of acceleration depends on the The amount of acceleration depends on the mass of the object.mass of the object.

a Fm

Small mass Small mass more acceleration more acceleration

Large mass Large mass less acceleration less acceleration

Page 8: Forces & Motion .

MomentumMomentum

MomentumMomentum quantity of motionquantity of motion

p = mvp: momentum (kg ·m/s)m: mass (kg)v: velocity (m/s)m

p

v

Page 9: Forces & Motion .

MomentumMomentum Find the momentum of a bumper car if it has a Find the momentum of a bumper car if it has a

total mass of 280 kg and a velocity of 3.2 m/s. total mass of 280 kg and a velocity of 3.2 m/s.

GIVEN:

p = ?

m = 280 kg

v = 3.2 m/s

WORK:

p = mv

p = (280 kg)(3.2 m/s)

p = 896 kg·m/s

m

p

v

Page 10: Forces & Motion .

MomentumMomentum The momentum of a second bumper car is 675 The momentum of a second bumper car is 675

kg·m/s. What is its velocity if its total mass is kg·m/s. What is its velocity if its total mass is 300 kg? 300 kg?

GIVEN:

p = 675 kg·m/s

m = 300 kg

v = ?

WORK:

v = p ÷ m

v = (675 kg·m/s)÷(300 kg)

v = 2.25 m/s

m

p

v

Page 11: Forces & Motion .

Conservation of MomentumConservation of Momentum

Law of Conservation of MomentumLaw of Conservation of Momentum The total momentum in a group of objects The total momentum in a group of objects

doesn’t change unless outside forces act doesn’t change unless outside forces act on the objects.on the objects.

pbefore = pafter

Page 12: Forces & Motion .

Conservation of MomentumConservation of Momentum A 5-kg cart traveling at 4.2 m/s strikes a A 5-kg cart traveling at 4.2 m/s strikes a

stationary 2-kg cart and they connect. Find their stationary 2-kg cart and they connect. Find their speed after the collision. speed after the collision.

BEFORECart 1:m = 5 kgv = 4.2 m/s

Cart 2 :m = 2 kgv = 0 m/s

AFTERCart 1 + 2:m = 7 kgv = ?

p = 21 kg·m/s

p = 0

pbefore = 21 kg·m/s pafter = 21 kg·m/s

m

p

vv = p ÷ mv = (21 kg·m/s) ÷ (7 kg)v = 3 m/s

Page 13: Forces & Motion .

Conservation of MomentumConservation of Momentum A 50-kg clown is shot out of a 250-kg cannon at a A 50-kg clown is shot out of a 250-kg cannon at a

speed of 20 m/s. What is the recoil speed of the speed of 20 m/s. What is the recoil speed of the cannon? cannon?

BEFOREClown:m = 50 kgv = 0 m/s

Cannon:m = 250 kgv = 0 m/s

AFTERClown:m = 50 kgv = 20 m/s

Cannon:m = 250 kgv = ? m/s

p = 0

p = 0

pbefore = 0

p = 1000 kg·m/s

pafter = 0

p = -1000 kg·m/s

Page 14: Forces & Motion .

Conservation of MomentumConservation of Momentum So…now we can solve for velocity. So…now we can solve for velocity.

GIVEN:

p = -1000 kg·m/s

m = 250 kg

v = ?

WORK:

v = p ÷ m

v = (-1000 kg·m/s)÷(250 kg)

v = - 4 m/s (4 m/s backwards)

m

p

v