Newtons 3rd law and momentum

• Newton worked in the 1600s. He talked about momentum before he talked about force but, maybe because momentum is hard to conceptualise, we learn Newton’s Laws as statements about force.

• Momentum is always conserved. Because momentum is always conserved, pairs of forces must be equal and opposite.

• Let’s look at momentum…

Momentum and collisions

• We use momentum to solve collision problems in isolated systems.

• An isolated system has no external forces eg games of pool, frictionless surface problems

momentum = mass x velocity p = mv(kgms-1) =(kg) x (ms-1)

• Momentum is a vector. Use vector diagrams if story not one dimensional!!

• Calculate momentum two ways:

1) Actual momentum at one time

2) Change in momentum between start and end times

1 Momentum now

• On icy winter roads a 1500kg car is travelling at 21ms-1. Calculate the momentum.

• P = mv

= 1500 x 21

= 31 500 kgms-1

2 Changing momentum

• Initial momentum of A plus initial momentum of B

• mAviA + mBviB

• Equals final momentum of A + final momentum of B

• mAvfA + mBvfB

A B

Question:On icy winter roads a 1500kg car travelling at 21m/s collides with a 1800kg van going 15m/s in the opposite direction. The two vehicles lock together (1D collision) and move off with a

new speed v. Find v.

Answer:• Draw a diagram• Find the initial momentum

of each and add (considering direction)

• Find the combined mass and multiply by new v

• Equate and solve

• Ptruck = 1800 x 15

• Pcar = -1500 x 21

• Ptotal = 4 500 kgms-1

After• Mass = 3300kg• 4 500 = 3300 x v• V = 4500/3300• = 1.4 ms-1 (in direction

of car)

21m/s15m/s1800kg 1500kg

2D change in momentum

• Change in momentum in 2 or 3D needs vectors.

change = final vector – initial vector

Question:• An ice hockey puck of

mass 0.8kg moving at 3.5ms-1 hits the side of a second puck initially at rest. The mass of the second puck is 0.70kg.

• After the collision the 0.8kg puck moves off at 2.8ms-1 at right angles to its original direction.

• Find the velocity of the second puck immediately after the collision

0.7

0.8

3.5m/s

0.8

0.7

2.8m/s

Answer:Change in momentum of 0.8 puck equal and opposite tochange in momentum of 0.7 puck.0.8 puck

Final – initial = Final + opposite

2.24kgm/s

2.8kgm/s

3.6kgm/s on an angle of 37o

So 0.7 puck has equal and opposite change

• Initially at rest so momentum = 0

• final – initial = 3.6 – 0

• v = p /m• = 3.6 / 0.7• = 5.1 m/s on 37o

3.6kgm/s on an angle of 37o

NEWTON’S THIRD LAW :

“ACTION AND REACTION ARE ALWAYS EQUAL AND OPPOSITE”

“IF A BODY A EXERTS A FORCE ON BODY B, THEN B EXERTS AN EQUAL AND OPPOSITELY DIRECTED FORCE

ON A”

• Momentum are calculated at one time or over a change in time.

• Forces re calculated over a change in time. Mathematically, this is in the acceleration number.

I’LL PULL HIM

devishlyclever

WELL, ACTION FORCE AND REACTION FORCE ARE ALWAYS

EQUAL AND OPPOSITE!!

WELL ACTION AND REACTION ARE ALWAYS EQUAL AND OPPOSITE!!

Action force and reaction force are always equal and opposite,

SO WHY DOES THE GIRL MOVE FASTER?

NEWTON’S THIRD LAW PAIRS

• THEY ARE EQUAL IN MAGNITUDE

• THEY ARE OPPOSITE IN DIRECTION

• THEY ACT ON DIFFERENT BODIES

The 2 forces act along the same line

SIMILARITIES

The 2 forces act for the same length of time

The 2 forces are the same size

Both forces are of the same type

DIFFERENCES

The 2 forces act on different bodies

The 2 forces are in opposite directions

NEWTON’S THIRD LAW PAIRS

THE CLUB EXERTS A FORCE F ON THE BALL

FF

THE BALL EXERTS AN EQUAL AND OPPOSITE FORCE F ON THE CLUB