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### Momentum

14th September 2009

Learning Outcomes

- You should be able to:
- Calculate the momentum of an object of known mass and velocity.
- State that momentum is conserved in any collision provided no external forces act on the colliding bodies.
- Apply and rearrange the appropriate equations to two bodies that collide in a straight line

Starter

- Why does it take an oil tanker several km to stop, but a bicycle on a few metres?
- What is different about them?

Momentum

- Momentum is used in contact sports.
- In rugby a player with a lot of momentum is very difficult to stop

- The Momentum of a moving object = mass x velocity
- What are the units for each of these?
- kg m/s, kg, m/s. If you can remember the units or mass and velocity you can remember them for momentum.
- Which rugby player has the most momentum?
- Player A is 45kg running at 6 m/s
- Player B is 50kg running at 4 m/s

- A = 45x 6 = 270 kg m/s
- B = 50x 4 = 200 kg m/s Player A will be harder to stop

Stopping power

- Which of these balls is the most difficult to stop and why â€“ what properties make them difficult to stop?
- Tennis ball
- Football
- Rugby ball
- Bowling ball
- Squash ball
- Ping pong ball
- American football

- There isnâ€™t a right answer so argue your point! Why is their no right answer?
- Momentum = mass x velocity

Momentum

- Momentum is hard to define.
- With your knowledge on momentum write your own definition.
- Remember the equation is
- Momentum = mass x velocity

Collisions

- Now we know what momentum is we can look at what happens when moving objects collide

Momentum Rules

If car A is moving along a road and collides with stationary car B what will happen?

- If the two cars are the same mass then the velocity of car A is halved by the impact. The combined mass after the collision is twice the moving mass before the collision so the momentum after the collision is the same as before.
e.g. cars both 500kg. Velocity of car A 20 m/s

Before car Aâ€™s momentum was 500 x 20 = 10000 kg m/s

After the collision the mass of both moving cars combined was 1000kg the velocity was half as fast (10 m/s) so the momentum is 1000kg x 10m/s = 10000 kg m/s

The momentum is conserved!

Momentum Rules

- If car A with mass 300kg travelling at a velocity of 9 m/s crashes into stationary car B of mass 600kg (twice the size) then:
The velocity of car A is reduced to a third. The combined mass after the collision is three times the initial mass so the mass should be conserved

Work through and prove this rule.

Momentum Rules

- If car A with mass 300kg travelling at a velocity of 9 m/s crashes into stationary car B of mass 600kg (twice the size) then:
The velocity of car A is reduced to a third. The combined mass after the collision is three times the initial mass so the mass should be conserved

Work through and prove this rule.

Car A 300kg x 9 m/s has a momentum of 2700 kg m/s

Car A and B after has a combined mass 900kg and a velocity of 3 m/s so a momentum of 900 kg x 3 m/s = 2700 kg m/s

Momentum is conserved

Shunting Effect

- If there are more than 2 objects involved momentum can be â€˜shuntedâ€™.
- E.g. if there is a line of cars and a vehicle crashes into the back of them, each car is shunted into the one in front and the momentum is transferred along to the one at the front.

Demo Newtonâ€™s cradle

- This effect can be see in Newton's cradle

Questions on momentum

- Complete the momentum calculations to check your use of the equation

Answers

- a) 600 kg m/s
b) 24000 kg m/s

c) 100,000,000 kg m/s

d) 24,000,000 kg m/s

- a) 180 kg m/s
b) 180 / 90 = 2 m/s

- a) 24,000 kg m/s
b) 6000 + 2000 = 8000 kg

24,000/ 8000 = 3 m/s

answers

- 3) a) Kinetic energy = 0.5 X mass X speedÂ²
- 0.5 x 6000 x 4Â² = 48,000 J
- b) 0.5 x 8000 x 3Â² = 36,000 J
- C) because they are travelling slower, even though momentum is coserved.

Momentum is conserved in explosions

- Explosions arenâ€™t only chemical explosions you can get a spring explosion. They occur when some kind of stored energy (chemical or elastic) is suddenly transferred into kinetic. E.g. gun being fires
- During an explosion objects move with equal and opposite momentum.
- Homework â€“ explosions calculations sheet due thusday lesson

Momentum has a size and direction

- When two objects push against each other they move in an equal and opposite motion
- Equation
(mass of A x velocity of A) x (mass of B x velocity of B) = 0

- Or
(mass of A x velovity of A) = -(mass of B x velocity of B)

Alternative explanation

- The explosion creates fast moving molecules of gas which stream out of the open end of the rocket.
- Momentum is conserved soâ€¦
- The gain in momentum of the gas molecules must be equalled by the gain in momentum (in the opposite direction) of the rocket

Conservation of momentum

- As the molecules fly out to the left, the rocket moves to the right.

Have a go

- An artillery gun mass 2000 kg fires a shell of mass 20 kg at a velocity of 120 m/s. Calculate the velocity of the gun.
- (Mass of gun x recoil velocity) = -(mass of shell x velocity of shell)
- (2000kg x ? ) = - (20kg x 120 m/s)
- (2000kg x ? ) = - 2400 kg m/s
- ? = -2400 kg m/s = -1.2 m/s
- 2000kg
- (2000kg x 1.2m/s) = - (20kg x 120 m/s)

Explosion questions

- Two roller skaters, a girl of mass 40 kg and a boy of mass 50 kg stand facing each other on flat level ground. When one of the roller skaters pushes the other one away, they move away in opposite directions at velocities of 2.5 m/s and 2.0 m/s.
- i) Which of the two roller skaters,
- the boy or the girl, moves away fastest?
The girl as she has a smaller mass and therefore will have a faster velocity if their momentum is the same

- ii) Calculate the momentum of the girl just after they move apart. 40kg x 2.5m/s =100 kg m/s
- i) What can you say about the momentum of the boy just after they moved away from each other?
- It will be the same but in the opposite direction (equal and opposite momentum)
- ii) What can you say about the total momentum of both the skaters just after they moved away from each other?
- They will equal 0

Explosion Calculations

- A person of mass 70 kg in a stationary rowing boat of mass 120 kg by the side of a canal jumps out of the rowing boat onto the canal bank. The boat recoils at a velocity of 1.4 m/s.
- Calculate:
- The momentum of the boat when it recoils.
120kg x 1.4 m/s = 168kg m/s

- The velocity at which the person jumps out of the boat.(mass of person x velocity of person) = - (mass of boat x velocity of boat)
- ( 70kg x ? ) = (120kg x 1.4 m/s )
- ? =- 168 kg m/s = -2.4 m/s
70

- 168 kg m/s = - 168 kg m/s

Explosion Calculations

- A cannon of mass 1600 kg fired a cannon ball of mass 5.0 kg and recoiled at a velocity of 0.25 m/s.
- Calculate:
- The momentum of the cannon when it recoiled.
1600kg x 0.25 m/s = 400 kg m/s

- The velocity of the cannon ball when it left the cannon. (5kg x ? ) = - (1600kg x 0.25)
? = - 400 kg m/s

5

? = - 80 m/s

- i) Show that the kinetic energy of the cannon immediately after it recoiled was 50 J. [Higher]
- kinetic energy = 0.5 x mass x speedÂ²
- Kinetic energy = 0.5 x 1600kg x 0.25Â²
kinetic energy = 50 J

- Ii) Calculate the kinetic energy of the cannon ball when it left the cannon. [Higher]
Kinetic energy = 0.5 x 5 kg x 80Â²

kinetic energy = 16,000 J

Changing Momentum

- How can we slow down momentum
- Momentum isnâ€™t always useful. When we have a car accident we want to slow down the momentum.
- We have crumple zones at the front and back of a car

- Equation
- Force (N) = change in momentum kg m/s
time taken (s)

During a crash increasing the crash time reduces the force of the crash

Example

- A bullet of mass 0.004 kg moving at a speed of 90m/s is stopped by a bullet proof vest in 0.0003 seconds. Calculate the impact force
- Initial momentum = mass x velocity
- 0.36 kg m/s = 0.004 kg x 90m/s

- Force (N) = change in momentum kg m/s
time taken (s)

1200 N = 0.36 kg m/s

0.0003 s

Calculate if the impact

had been 0.0001s

Answer

3600 N = 0.36 kg m/s

0.0001 s

Use this to explain why it is important police wear bullet-proof vests.

The egg is moving quite slowly when it hits the ground. The momentum is small so the force needed to stop it is quite small. The egg survives.

When the egg is dropped from a greater height it picks up more speed before it reaches the ground. Its momentum is greater. The force needed to change it is greater. What do you notice about the effect of this greater force?

When the egg is dropped from even higher it picks up even more speed and hits the ground with even more momentum. The force needed to change this momentum is much larger. You know what that means for the egg.

The video clips give a good picture of the force acting to change the eggâ€™s momentum â€“ and an endless supply of omelettes for the crew!

The size of the force controls whether the egg breaks or not. Can we use a smaller force instead? Hereâ€™s a game you mustnâ€™t try at home!

When you catch the egg you move your arm backwards to slow the moving egg down more slowly. In effect you use a smaller force but use it for longer. This stops (you hope!) the egg from breaking.

Use the idea of force and momentum to explain these road safety facts. They refer to the speed of a car when it hits a pedestrian.

Speed of Chance Chance

collision (mph) of death of surviving

20 10% 90%

30 50% 50%

40 90% 10%

If an object is moving you must apply a force to stop it moving.

The larger the momentum of the object the greater the force needed.

A large force can be delivered in a short time or a smaller force can act over a longer time.

Too great a force has disadvantages!

Who is willing to try this? moving.

- Egg game!!!

Plenary moving.

- Use all the equations that you've learnt to write a question for your partner.
- Make sure you build up the question using all three equations.

Learning Outcomes moving.

- You should be able to:
- Calculate the momentum of an object of known mass and velocity.
- State that momentum is conserved in any collision provided no external forces act on the colliding bodies.
- State that during an explosion momentum acts in equal and opposite direction
- Apply and rearrange the appropriate equations for momentum, momentum of an explosion and changing momentum.

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