F = ma F = 8 kN m = 0.042 kg (golf ball mass) a = ? F = ma 8 kN = 0.042 kg x a a = 8 kN/0.042 kg = 190,000 m/s^2 Δp = FΔt Δp = 8 kN x 0.0005 s = 4 kg m/s

2. Time graphs
• The slope of a distance-time graph is
equal to the velocity
• i.e. y2 - y1 => s2 - s1
• x2 - x1 t2 - t1

3. 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.

4. 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

5. 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.

6. 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

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?

10. Example
• What is the momentum of a rugby player
with a mass of 110kg traveling east at
8ms-1?
• P = mv

11. 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

12. 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

13. 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.

14. 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.

15. 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.

16. 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.

17. 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.

18. 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

19. • http://teachertech.rice.edu/Participants/
louviere/Newton/
15

20. Force formula

21. Force formula
• F = ma

22. Force formula
• F = ma

23. Force formula
• F = ma
• F = mv –mu or Ft = mv-mu

24. Force formula
• F = ma
• F = mv –mu or Ft = mv-mu
• t

25. Force formula
• F = ma
• F = mv –mu or Ft = mv-mu
• t

26. 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

28. 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.