1.
Time graphs• The slope of a distance-time graph is equal to the velocity• i.e. y2 - y1 => s2 - s1• x2 - x1 t2 - t1
2.
Examples• A boy travels at constant velocity, covering a distance of 15m in 5 s. He remains at that point for a further 6 s before returning at constant velocity in 10 s. Draw a distance – time graph of the boys motion and use it to calculate his velocity on the outward journey.
3.
Velocity – Time Graphs• The slope of a velocity – time graph is equal to the acceleration• i.e. y2 - y1 => v2 - v1 => v2 – v1• x2 - x1 t2 - t1 t• The area under the graph is equal to the distance traveled
4.
Question• An object starts from rest and accelerates at 2 ms-2. By calculating the distance traveled after each second, sketch the distance – time graph for the first 3 s of its motion.
5.
Momentum• Mass (m) is a measure of the amount of matter in a body• Mass is a scalar quantity with a SI unit of Kilograms (kg)• Momentum (p) is a product of mass and velocity.• Momentum = mass X velocity• Momentum is a vector quantity, unit kgms-1
7.
Example• What is the momentum of a rugby player with a mass of 110kg traveling east at 8ms-1?
8.
Example• What is the momentum of a rugby player with a mass of 110kg traveling east at 8ms-1?
9.
Example• What is the momentum of a rugby player with a mass of 110kg traveling east at 8ms-1?• P = mv
10.
Example• What is the momentum of a rugby player with a mass of 110kg traveling east at 8ms-1?• P = mv• P = (110)(8) = 880 kg m s-1 east
11.
Principle of conservation of momentum• A total momentum before an interaction is equal to the total momentum after, provided no resultant force acts on the system• m1u1 + m2u2 = m1v1 + m2v2
12.
Where this applies: Spacecraft• When a spacecraft is accelerating (speed up, slowing down or changing direction), it must expel gas at high velocity in the opposite direction to the acceleration. When the rocket accelerates in a particular direction, its momentum increases in that direction. In order for the principle of conservation of momentum to hold, the gas must carry an equal amount of momentum in the opposite direction.
13.
Where this applies: Snooker• In the games of snooker, pool or billiards, balls are hit off each other. The principle of conservation of momentum applies to each of these collisions.
14.
Question• Two snooker balls of the same mass, moving in opposite directions, collide head on. The pink ball is moving to the right at 5ms-1 and the blue 3ms-1. The pink is brought to rest by the collision.• (i) What is the velocity of the blue after the collision• (ii) What is the change in momentum of each ball.
15.
Question• An 85kg man collides head on with a 55kg woman while the pair are skating in opposite directions on an ice rink. The two of them get tangled up in the collision and move off together after it. Given the man has a speed of 8ms-1 and the woman 6ms-1.• (i) What speed do the pair move at after the collision?• (ii) If the man and woman had been traveling at right angles to each other before the collision, calculate their speed afterwards.
16.
Force• Force: (F) is that which can cause acceleration• It is a vector quantity with an SI unit of Newton• 1 Newton is the amount of force that will cause a body of mass 1 kilogram to accelerate by 1 meter per second squared.
17.
Newton’s Laws of Motion• Law 1: A body will remain at rest or continue moving at constant velocity unless acted upon by an unbalancing external force.• Law 2: The rate of change of momentum is proportional to the applied force and takes place in the direction of the force• Law 3: For every action there is an equal and opposite reaction, action and reaction do not happen to the same body
22.
Force formula• F = ma• F = mv –mu or Ft = mv-mu
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Force formula• F = ma• F = mv –mu or Ft = mv-mu• t
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Force formula• F = ma• F = mv –mu or Ft = mv-mu• t
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Force formula• F = ma• F = mv –mu or Ft = mv-mu• t• Example: Calculate the force required to give a football of mass 420g and acceleration of 200ms-1
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Force formula• F = ma• F = mv –mu or Ft = mv-mu• t• Example: Calculate the force required to give a football of mass 420g and acceleration of 200ms-1
27.
Force formula• F = ma• F = mv –mu or Ft = mv-mu• t• Example: Calculate the force required to give a football of mass 420g and acceleration of 200ms-1• Question: A golfer strikes a golf ball with a force of 8kN. The ball and club are in contact for 0.5ms. Calculate the change in momentum of the ball.
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