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### Forces and Dynamics

Momentum and Impulse

Momentum

Whenever there is a collision between moving bodies or whenever an explosion causes bodies to move apart, we can use momentum to analyse the situation.

Momentum is a vector quantity with direction determined by the velocity.

Q Estimate the momentum of a book sliding across the table.

Momentum = mass x velocity

or… p = mv

Note: The units of momentum, p are kgms-1

Newton’s 2nd Law – A better definition

This is often stated as follows…

This leads to

Where F is the average force causing a change in momentum, Δp over a time, Δt.

The rate of change of momentum of a body is directly proportional to the resultant force acting upon it and occurs in the same direction as the force.

F = Δp

Δt

If a force F acts on a body of mass m, changing its velocity from u to v, then…

change of momentum = mv – mu

rate of change of momentum = mv – mu

t

From Newton 2… F mv – mu

t

F m (v - u)

t

but... a = (v – u) / t so F ma

or… F = kma

From the definition of a Newton, k = 1.

so… F = ma

- Change in momentum
- When calculating Δp, it is vital to consider direction.
- E.g.1. A football hits a wall at 10ms-1 and rebounds at 7ms-1. If it has mass 200g and was in contact with the wall for 0.4s determine
- a. The change in momentum of the ball
- b. The force exerted by the wall on the ball.
- Taking towards the wall as positive…
- For ball: Δp = mv – mu
- = (0.2 x -7) - (0.2 x 10)
- = - 3.4 kgms-1
- F = rate of change of momentum = Δp / Δt
- = - 3.4/ 0.4 = - 8.5 N

Change in momentum

- The principle of conservation of momentum
- This applies to all collisions and explosions:
- We consider three types of problem:
- Inelastic collisions: One body collides with another, they stick together and move on together (no bounce!)
- Elastic collisions: One body collides with another, they bounce apart and move separately (only a perfectly elastic collision will also have conservation of KE)
- Explosions: E.g. in a gun, resulting in the bullet and gun moving apart.

The total momentum of a system before a collision is equal to the total momentum of the system after the collision, so long as there are no external forces acting.

A bullet of mass 10g travelling at a speed of 100ms-1 embeds itself into a block of wood of mass 1990g, suspended by a string. Determine the velocity of the block and bullet immediately after the impact.

Before

After

m = 0.01 kg

u1 = 100ms-1

m = 1.99 kg

u2 = 0

m = 2.00 kg

v = ?

After

m1 = 0.01 kg

u1 = 100ms-1

m2 = 1.99 kg

u2 = 0

m3 = 2.00 kg

v = ?

Applying principle of conservation of momentum:

momentum before = momentum after

m1u1 + m2u2 = m3v

(0.01 x 100) + (1.99 x 0) = 2.00 x v

1 + 0 = 2v

v = 0.5ms-1

A railway truck A of mass 2 x 104 kg travelling at 0.5ms-1 collides with another truck B of half its mass moving in the opposite direction with velocity of 0.4ms-1. If the trucks join on collision find the common velocity with which they move on.

An ice skater of mass 60kg throws a ball of mass 1.5kg with a horizontal velocity of 25ms-1. What will be the recoil speed of the skater (assuming a frictionless surface)?

An aeroplane of total mass 50 000kg travels at a velocity of 200ms-1. If a 100kg passenger then walks up to the front of the plane at 2ms-1, what will be… a. the new speed of the plane (to three decimal places)?

b. the change in speed of the plane?

The impulse of a force depends upon the magnitude of the force and the duration of its application:

So…Impulse = Δp

But…

(The units of impulse, like momentum, are kgms-1 or Ns)

Impulse of a force on a body is equal to the total change in momentum of the body.

F = Δp

Δt

Note: In momentum changes, the force usually changes. Thus F is the average force.

Impulse = FΔt

A graph of force against time will have area equal to F x Δt. Therefore…

Impulse of a force on a body is equal to the area under a Force - time graph.

F

t

Δt

F

In reality most F-t graphs are curves showing changing forces over time.

t

F

B

t

E.g.Two cars of equal mass, moving at the same initial velocity, collide with a wall and stop. The forces the cars experienced are represented by the two graphs below.

- If they have the same initial velocity, what can you say about the change in momentum for each car?
- What about the impulse for each car?
- So what can you say about the area under each graph?
- Which one has experienced the smaller average force? How can you tell?
- What could have been different about the design of each car?

This question illustrates how increasing the time of an impact can reduce the average force experienced by an object e.g. during a car crash.

Can you apply Newton 2 to this?!

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