Newtons 3 rd law and momentum

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Newtons 3 rd law and momentum. S I R I S A A C N E WTON (1647 - 1727). 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.

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### 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:
• Actual momentum at one time
• 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

Initial momentum of A plus initial momentum of B

mAviA + mBviB

Equals final momentum of A + final momentum of B

mAvfA + mBvfB

2 Changing momentum

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.

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)

1800kg

15m/s

21m/s

1500kg

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

change = final vector – initial vector

Use diagrams!

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

2.8m/s

0.8

0.8

0.7

0.7

3.5m/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

2.8kgm/s

3.6kgm/s on an angle of 37o

2.24kgm/s

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

So 0.7 puck has equal and opposite change

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”

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

devishly

clever

I’LL PULL HIM

WELL, ACTION FORCE AND REACTION FORCE ARE ALWAYS EQUAL AND OPPOSITE!!

WELL ACTION AND REACTION 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

NEWTON’S THIRD LAW PAIRS

SIMILARITIES

DIFFERENCES

The 2 forces act for the same length of time

The 2 forces act on different bodies

The 2 forces are in opposite directions

The 2 forces are the same size

The 2 forces act along the same line

Both forces are of the same type

F

F

THE CLUB EXERTS A FORCE F ON THE BALL

THE BALL EXERTS A N EQUAL AND OPPOSITE FORCE F ON THE CLUB