1. Objectives Recall and apply Newton’s three laws of motion Explain Force, Momentum, and conservation of momentum Carry out calculations based on these principles Key Words Newton’s laws, force, momentum, conservation Mechanics 6 Newton’s Laws IFP Monday 10 th November 2015

2. Sir Isaac Newton (1643 – 1727) Came up with three laws. Write down any you already know on whiteboards

3. Objectives Recall and apply Newton’s three laws of motion Explain Force, Momentum, and conservation of momentum Carry out calculations based on these principles Key Words Newton’s laws, force, momentum, conservation Newton’s FIRST law In the absence of any resultant force acting on an object, it remains either at rest or continues to move at a constant velocity So - No force ↔ no change of motion Also called law of inertia resultant means net force

4. Objectives Recall and apply Newton’s three laws of motion Explain Force, Momentum, and conservation of momentum Carry out calculations based on these principles Key Words Newton’s laws, force, momentum, conservation Demo Hovercraft Air track

5. Objectives Recall and apply Newton’s three laws of motion Explain Force, Momentum, and conservation of momentum Carry out calculations based on these principles Key Words Newton’s laws, force, momentum, conservation Newton’s SECOND law Force = mass x acceleration Unit of force is one kg ms -2 - known as one newton F is the resultant force (if several forces act on an object) Also note that Weight: W = m g A free-falling object is unsupported but not weightless (g is always there)

6. Think about a lift accelerating The direction of net (resultant) force on an object is the same as the direction of acceleration Physics of lifts (lift of mass m, supported by a cable): - Lift at rest: tension T = weight (m g) - Lift accelerating as it moves upwards: T > mg (a >0) - Lift accelerating as it moves downwards: T<mg (a <0 ) - Lift decelerating as it moves upwards: T< mg (a < 0) - Lift decelerating as it moves downwards: T > mg (a>0)

7. Newton’s THIRD law For every action (force) there is an equal and opposite reaction. Then why don’t they add up to zero? Answer: because they are exerted on different objects! Example: box on a table 2 forces on the box: (1) weight and (2) normal force from table Reaction to (1): force that the box exerts on the earth! Reaction to (2): force that the box exerts on the table! Therefore you cannot add them.

8. Try the questions on the sheet in assigned groups

9. Objectives Recall and apply Newton’s three laws of motion Explain Force, Momentum, and conservation of momentum Carry out calculations based on these principles Key Words Newton’s laws, force, momentum, conservation Momentum Momentum = mass x velocity p = m v Vector quantity, units: kg ms -1 F= m a (for an object with fixed mass m) So F = m (v t – v 0 )/t = (m v t – m v 0 )/t Which is Rate of change of momentum. This is a more general statement of Newton’s second law.

10. Conservation of momentum Principle of conservation of momentum: When two or more bodies interact, then the total momentum is conserved if no external forces act on the bodies. This can be derived from Newton’s third law (action/ reaction exerted on different bodies) It is a very important principle!

11. Example – two balls colliding By law of conservation of momentum, momentum before collision is equal to momentum after collision. So try to solve for v 2

12. Example – two balls colliding m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 v 2 = (m 1 u 1 + m 2 u 2 – m 1 v 1 )/m 2 (Rearrange, to solve for v 2 ) v 2 = (10 x 20 + 8 x 5 – 10 x 10)/ 8 ms -1 = 17.5 ms -1

13. Conservation example 2 A car crashes into a wall at 25 ms -1 and is brought to rest in 0.1s. Calculate (i)the average force exerted on a 75 kg test dummy by the seatbelt. (ii)the dummy’s average acceleration Assume that the seatbelt does all the stopping of the dummy (e.g. ignore friction from the seats or normal forces from airbags)

14. Solution Dummy’s initial and final momentum: p 0 = m v 0 = 75 kg x 25 ms -1 = 1875 kg ms -1 p f = m v f = 75 kg x 0 = 0 Therefore change of momentum : Δp=p f -p 0 =-1875 kg ms -1 Impulse = Δp = Force x time F = Δp/Δt = - 18,750 N (opposite direction to the initial velocity)

15. Solution dummy’s average acceleration: a = f/m = -18,750/75 ms -2 = -250 ms -2 ≈ -25.5 g Similarly: a = Δv/Δt = (0 – 25) ms -1 /0.1 s = -250 m/s2

16. Momentum questions

17. Solutions