Newtons 3 rd 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 = kgms -1

2 Changing momentum Initial momentum of A plus initial momentum of B m A v iA + m B v iB Equals final momentum of A + final momentum of B m A v fA + m B v fB 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 P truck = 1800 x 15 P car = x 21 P total = kgms -1 After Mass = 3300kg = 3300 x v V = 4500/3300 = 1.4 ms -1 (in direction of car) 21m/s15m/s 1800kg1500kg

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 m/s m/s

Answer: Change in momentum of 0.8 puck equal and opposite to change 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 37 o

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 37 o 3.6kgm/s on an angle of 37 o

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”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 devishly clever 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