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Objectives • • Mechanics 6 Newton’s Laws IFP Monday 10th November 2015 • 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 Sir Isaac Newton (1643 – 1727) • Came up with three laws. • Write down any you already know on whiteboards Objectives 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 • • • 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 • So - No force ↔ no change of motion • Also called law of inertia • resultant means net force Newton’s laws, force, momentum, conservation Objectives Demo • Hovercraft • Air track • • • 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 Objectives 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) 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 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) 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. Try the questions on the sheet in assigned groups Objectives Momentum • • Momentum = mass x velocity p=mv • Recall and apply Newton’s three laws of motion Explain Force, Momentum, and conservation of momentum Carry out calculations based on these principles Vector quantity, units: kg ms-1 F= m a (for an object with fixed mass m) So F = m (vt – v0)/t = (m vt – m v0)/t Which is Rate of change of momentum. This is a more general statement of Newton’s second law. Key Words Newton’s laws, force, momentum, conservation 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! Example – two balls colliding By law of conservation of momentum, momentum before collision is equal to momentum after collision. So try to solve for v2 Example – two balls colliding m1 u1 + m2 u2 = m1 v1 + m2 v2 v2 = (m1 u1 + m2 u2 – m1 v1)/m2 (Rearrange, to solve for v2) v2 = (10 x 20 + 8 x 5 – 10 x 10)/ 8 ms-1 = 17.5 ms-1 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) Solution Dummy’s initial and final momentum: p0 = m v0 = 75 kg x 25 ms-1 = 1875 kg ms-1 pf = m vf = 75 kg x 0 = 0 Therefore change of momentum : Δp=pf-p0=-1875 kg ms-1 Impulse = Δp = Force x time F = Δp/Δt = - 18,750 N (opposite direction to the initial velocity) 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 Momentum questions Solutions