Monday January 24, 2011

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Monday January 24, 2011. Ch 9.1 Impulse and Momentum. What is Momentum ?. Momentum (p) = mass x velocity A really slow moving truck and an extremely fast roller skate can have the same momentum. 1 kg. 10 m/sec. 1000 kg. .01 m/sec. Question :.

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Monday January 24, 2011

Ch 9.1

Impulse and Momentum

What is Momentum ?
• Momentum (p) = mass x velocity
• A really slow moving truck and an extremely fast roller skate can have the same momentum.

1 kg

10 m/sec

1000 kg

.01 m/sec

Question :
• Under what circumstances would the roller skate and the truck have the same momentum ?
• Answer: If ratio of vs/vt = mt/ms
• A 1000 kg truck moving at 0.01 m/sec has the same momentum as a 1 kg skate moving at 10 m/sec. Both have a momentum of 10 kg m/sec. ( 1000 x .01 = 1 x 10 = 10 )
Impulse and Momentum
• Impulse = change in momentum (mv)
• Most often v changes (accel)
• And m x a = force
• Applying a force over a time interval to an object changes the momentum
• Force x time interval = Impulse
• Impulse = FΔt or

FΔt = mΔv

MOMENTUM
• An object at rest has no momentum, why?
• (the velocity component is zero for an object at rest)
• To INCREASE MOMENTUM, apply the greatest force possible for as long as possible.
• Examples :
• pulling a sling shot
• drawing an arrow in a bow all the way back
• a long cannon for maximum range
• hitting a golf ball or a baseball . (follow through is important for these !)

FORCE

TIME

MOMENTUM
• SOME VOCABULARY :
• impact : the force acting on an object (N)
• impact forces : average force of impact
• impulse-momentum theorem – The impulse on an object is equal to the change in momentum that it causes.

F∆t = p2-p1

MOMENTUM

• Decreasing Momentum
• Which would be safer to hit in a car ?
• Knowing the physics helps us understand why hitting a soft object is better than hitting a hard one.

Ft

mv

mv

Ft

MOMENTUM
• In each case, the momentum is decreased by the same amount or impulse (force x time)
• Hitting the haystack extends the impact time (the time in which the momentum is brought to zero).
• The longer impact time reduces the force of impact and decreases the deceleration.
• Whenever you need to decrease the force of impact, extend the time of impact !
DECREASING FORCE
• If the time of impact is increased by 100 times (from .01 sec to 1 sec), then the force of impact is reduced by 100 times
• EXAMPLES :
• Airbags in cars or safety nets in circuses
• Moving your hand backward as you catch a fast-moving ball with your bare hand or a boxer moving with a punch.
• Flexing your knees when jumping from a higher place to the ground. or elastic cords for bungee jumping
• Using wrestling mats instead of hardwood floors.
• Dropping a glass dish onto a carpet instead of a sidewalk.
EXAMPLES OF DECREASING MOMENTUM

Ft = change in

momentum

• Bruiser Bruno on boxing …
• Increased impact time reduces force of impact
• Bungee jumping provides another example

Ft = change in

momentum

POOF !

CRUNCH !

Questions :
• When a dish falls, will the impulse be less if it lands on a carpet than if it lands on a hard ceramic tile floor ?
• The impulse = same for either surface (same ∆p)
• forceis less for the impulse on the carpet because of the greater time of momentum change. There is a difference between impulse and impact.
• If a boxer is able to increase the impact time by 5 times by “riding” with a punch, by how much will the force of impact be reduced?
• Since the time of impact increases by 5 times, the force of impact will be reduced by 5 times.
Impulse Momentum - Theorem

Impulse is related to the change

in its momentum.

F∆t = p2 – p1

To find the change of the momentum after a collision, solve for P2

VECTORS

Signs (directions) are VERY important when calculating momentum and velocity.

Angular Momentum
• Angular Momentum – quantity of motion that is used with rotating objects about a fixed axis.

Force changes momentum whereas Torque changes angular momentum.

Momentum vs. Angular Momentum
• Momentum is a change in an object’s mass and velocity
• Angular momentum is a change in an objects mass, displacement from the center of rotation, and velocity perpendicular to displacement

Monday January 24, 2010

Ch 9.2

Conservation of Momentum

Conserving Momentum

Condition 1: A system that does not change its mass is considered a closed system. All of these forces are internal.

Condition 2: When the net external force on a closed system is zero it is called an isolated system.

CONSERVATION OF MOMENTUM
• To accelerate an object, a force must be applied.
• The force or impulse on the object must come from outside the object.
• EXAMPLES : Sitting in a car and pushing on the dashboard doesn’t create movement.
• Internal forces like these are balanced and cancel each other.
• If no outside force is present, no change in momentum is possible.
The Law of Conservation of Momentum
• In the absence of an external force, the momentum of a system remains unchanged.
• This means that, when all of the forces are internal (for EXAMPLE: cars colliding or stars exploding) the net momentum of the system before and after the event is the same.
COLLISIONS
• ELASTIC COLLISIONS
• INELASTIC COLLISIONS – objects stick together

Momentum transfer from one

Object to another . Example:

Billiard ball collisions.

Is a Newton’s cradle like the one

Pictured here, an example of an

elastic or inelastic collision?

Problem Solving #1
• A 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is at rest. Find the velocity of the fish immediately after “lunch”.
• net momentum before = net momentum after
• (net mv)before = (net mv)after
• (6 kg)(1 m/sec) + (2 kg)(0 m/sec) = (6 kg + 2 kg)(vafter)
• 6 kg.m/sec = (8 kg)(vafter)
• vafter = 6 kg.m/sec /8 kg
• 8 kg
• vafter = ¾ m/sec

vafter =

Problem Solving #2
• Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 2 m/sec. Find the velocity of the fish immediately after “lunch”.
• net momentum before = net momentum after
• (net mv)before = (net mv)after
• (6 kg)(1 m/sec) + (2 kg)(-2 m/sec) = (6 kg + 2 kg)(vafter)
• 6 kg.m/sec + -4 kg.m/sec = (8 kg)(vafter)
• vafter = 2 kg.m/sec /8 kg
• 8 kg
• vafter = ¼ m/sec

vafter =

Problem Solving #3 & #4
• Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 3 m/sec.
• (net mv)before = (net mv)after
• (6 kg)(1 m/sec) + (2 kg)(-3 m/sec) = (6 kg + 2 kg)(vafter)
• 6 kg.m/sec + -6 kg.m/sec = (8 kg)(vafter)
• vafter = 0 m/sec
• Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 4 m/sec.
• (net mv)before = (net mv)after
• (6 kg)(1 m/sec) + (2 kg)(-4 m/sec) = (6 kg + 2 kg)(vafter)
• 6 kg.m/sec + -8 kg.m/sec = (8 kg)(vafter)
• vafter = -1/4 m/sec
Practice Problems

Page 210; 7-12

Explosion Equation

Momentum before the explosion is equal to momentum after the explosion

MaV = - MbV

Example Problem #1

A cart with a mass of 3 kg is sitting next to a cart with a mass of 2 kg. Between them is a compressed spring. When the spring is released, the 3 kg cart moves at a speed of 5 m/s. How fast did the 2 kg cart move?

System 1: (3kg)(-5m/s) = -15 kg*m/s

System 2: (2 kg)(V) = 2V

Pa = -Pb

So…

-15 = -2V

V = 7.5 m/s

Example Problem #2

An astronaut at rest in space fires a thruster pistol that expels 50g of hot gas at 750 m/s. The combined mass of the pistol and the astronaut is 80 kg. How fast and in what direction is the astronaut moving after firing the pistol?

System 1 = gas (.050 g)(-750) = -37.5 kg*m/s – gas is moving to the left

System 2 = astronaut and pistol (80 kg)(V)

-astronaut is moving to the right.

Pa = - Pb

So… -37.5 = - 80(V)

V = 0.468 m/s

Practice Problems

Page 21413-15

Collisions in Two or Three Dimensions

Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy.

Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities.

Collisions in Two or Three Dimensions

• Problem solving:
• Choose the system.
• Draw diagrams of the initial and final situations, with momentum vectors labeled.
• Choose a coordinate system.

Collisions in Two or Three Dimensions

5. Apply momentum conservation; there will be one equation for each dimension.

6. Solve.

8. Check units and magnitudes of result.

2-Dimensional momentum

The law of conservation of momentum applies to 2-Dimensional momentum in the same way: The momentum before is equal to the momentum after the collision.

- If you define the x-axis to be in the direction of the initial momentum, the y-component of the initial momentum is zero. Therefore the sum of the final y-component must be zero.

PA2y + PA2y = 0

or PA2y = -PA2y

SO… PA2y + PA2y = 0
• They are equal in magnitude but have opposite signs.
• The sum of the horizontal components are also equal.

Pa1 = Pa2X + Pa2X

In order to solve for the final momentum of the system, you add the momentums of both objects like you add vectors.

Pa + Pb = P2

P2 is the resultant vector

Example Problem 1

A 3 kg ball, A, is moving at a speed of 5 m/s. It collides with a stationary ball, B, of the same mass. After the collision, ball A moves off in a direction 30 degrees to the left of its original direction. Ball B moves 90 degrees to the right of ball A’s final direction. How fast are they moving after the collision?

Example 2

A 1500 kg train is moving North at 23 m/s and collides with a 2000 kg train moving east at 15 m/s. They stick together. In what direction and with what speed do they move after the collision?

Practice Problems

Page 216

17 and 18

MOMENTUMVECTORS
• Momentum can be analyzed by using vectors
• The momentum of a car accident is equal to the vector sum of the momentum of each car A & B before the collision.

A

B

MOMENTUMVECTORS(Continued)
• When a firecracker bursts, the vector sum of the momenta of its fragments add up to the momentum of the firecracker just before it exploded.
• The same goes for subatomic elementary particles. The tracks they leave help to determine their relative mass and type.