# Impulse and Momentum and Collisions and Stuff We will now quantify exactly how it hits the fan.

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Impulse and Momentum and Collisions and Stuff We will now quantify exactly how it hits the fan.

Momentum Momentum = product of mass X velocity Symbol for momentum is “p” (don’t ask me why) – So, p = mv Momentum is also a vector and points in the direction of velocity Units for mass = kg, units for velocity = m/s So, units for momentum = kgm/s – Sadly, there’s no other, more convenient unit for it Plural is ‘momenta’

A closer look at momentum Think of momentum as being how easy/hard it is to get something to stop. Remember, p = mv So if mass or velocity is small, it’s likely that the object has a small momentum, unless the other of the two is very large What would be the momentum of an object at rest?

5 Scenarios Let’s look at the momenta of…. A car at rest Mass = 1000 kg, velocity = 0 m/s P = mv = (1000kg)(0m/s) = 0 kg m/s A bumblebee in flight (low mass, low velocity) Mass = 0.05 kg, velocity = 2 m/s P = mv = (0.05kg)(2m/s) = 0.1 kg m/s Projectile from BB gun (low mass, high velocity) Mass = 0.01 kg, velocity = 150 m/s P = (0.01 kg)(150 m/s) = 1.5 kg m/s

5 Scenarios continued A giant land tortoise (large mass, low velocity) Mass = 300 kg (I looked it up!), velocity = 0.5 m/s P = (300kg)(0.5m/s) = 150 kg m/s Tony Stewart at the Brickyard (large mass, high velocity) – Mass = 1000 kg, velocity = 90 m/s – P = (1000kg)(90 m/s) = 90,000 kg m/s

Understanding momentum Which is easier to stop… A slow moving baseball or a fast moving baseball? A chihuahua racing down the hall towards you or Mr. Barbini racing down the hall towards you? A speeding Volkswagen beetle or a speeding dump truck?

I need a volunteer Rank these momenta from lowest to highest – A fast flying bee – The earth in orbit around the sun – A parked garbage truck – A slowly flying bee – Your grandmother driving down the street in her 1959 Edsel – An oil tanker sailing the seven seas. – Tony Hawk grinding on a rail

Impulse Impulse = Product of Force and how long that force acts. – Impulse = F∆t Impulse is a vector. It acts in the same direction as the force Units of force = Newtons, units of time = seconds. So, units for impulse = NewtonSeconds

Okay, but what does that really mean? Let’s say you have a force of 10 N. You apply this force to a hockey puck on (frictionless) ice for 2 seconds. Impulse = (10N)(2 sec) = 20 Newton seconds Now, let’s say you apply this same force for 4 seconds So the impulse now = 40 Newton seconds

Impulse-momentum theorem Remember the work-energy theorem? This is kind of like that. Impulse = change in momentum In general, the mass of the object(s) in questions stays constant ∆p = p F – p I = mv F – mv I So F∆t = mv F – mv I

Well, Duh… Let’s think about 2 situations: Situation 1: pushing someone in a rolling chair – If you push on them with a certain force for two different times, the longer time will result in a greater change in momentum – So if the chair was at rest to begin with, it will end up going faster at the end for the longer time the force was applied

Less Duh Situation 2: crashing your 1,000 kg car Let’s say you are driving along at 45 m/s (about 96 mph) and you crash into one of two objects: a solid wall or a series of water-filled plastic bins (like they have on the highway) If your car goes from 45 m/s to 0 m/s, you have a change in momentum of: P = mv, so momentum = (1000kg)(45m/s) = 45,000 kgm/s

The car crash continued Recall that impulse = change in momentum, so Ft = mv = 45,000 kgm/s So, Force X time = 45,000 kgm/s Now, would you rather have that force spread out over a long time, or over a short time?

Compare stopping times Change in momentum is 45,000 kgm/s For a stopping time of 5 sec (slamming on your brakes): (F)(5 sec) = 45,000 kgm/s – So F = 9,000 N For a stopping time of 1.5 sec (smashing into the plastic water bins): (F)(1.5 sec) = 45,000 kgm/s – So F = 30,000 N For a stopping time of 0.2 sec (smashing into the concrete wall): (F)(0.2 sec) = 45,000 kgm/s – So F = 225,000 N

Let’s look at those numbers more closely… Remember we have a 1,000 kg car and F = ma If F = 9000N, then accel = 9 m/s (just under acceleration due to gravity) If F = 30,000N, then accel = 30 m/s (just over 3X acceleration due to gravity) If F = 225,000N, then accel = 225 m/s (about 23X the acceleration due to gravity) So, which is the most survivable?

Let’s look at some examples Hitting a baseball (mass = 0.14 kg) Let’s say that a baseball is travelling to the left (negative velocity) at -38m/s. After being hit, it moves to the right at 58 m/s. What is the impulse applied to the ball? If the contact time was 1.6X10 -3 sec, what was the force? Impulse = ∆p = p F – p I = mv F – mv I Impulse = (0.14kg)(58m/s) – (0.14kg)(-38m/s) Impulse = 13.4 kg m/s Impulse = F∆t, so F = impulse/∆t Force = (13.4 kg m/s)/(1.6X10 -3 sec) = 8400 N

Conservation of Momentum In general, momentum is conserved This means that the momentum at the beginning is the same as the momentum at the end I.e. momentum is not created or destroyed Remember, symbol for momentum is ‘p’ So P I = ‘initial momentum’ and P F = ‘final momentum’ Let’s look at some examples

Conservation of Momentum: Example 1 Cannon and cannonball What is momentum before cannon is fired (P I )? Zero So what does final momentum (P F ) have to be if momentum is conserved? Zero http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section3.rhtml

Conservation of Momentum: Example 1 Let’s say that the cannon has a mass of 1000 kg and Cannonball has a mass of 10 kg So total P I = P I of ball + P I of cannon P I of ball = (10kg)(0m/s) = 0 kg m/s P I of cannon = (1000kg)(0m/s) = 0 kg m/s

Conservation of Momentum: Example 1 Now let’s say that the cannonball moves to the right at 75 m/s. How fast does the cannon move to the left to conserve momentum? P I = P F So P F must equal zero

Conservation of Momentum: Example 1 P F must equal zero Total P F = P F of ball + P F of cannon Remember, moving to the right = positive velocity Moving to the left = negative velocity

Conservation of Momentum: Example 1 P F = 0, so (10kg)(75 m/s) + (1000kg)(V cannon ) = 0 So -V cannon = (10kg)(75m/s) / (1000kg) V cannon = -0.75 m/s If ball moves to the right, cannon moves to the left, so V cannon should be negative, which it is

Conservation of Momentum: Example 2 Newton’s cradle (or “Executive ball clicker” or, more crassly, “Newton’s Balls”) How does it work? If one ball is used, how many come up on the other side? If two are used…?

Now, Dr. Mason, remind us again when conservation of momentum holds… Momentum is conserved when there are NO EXTERNAL FORCES ACTING ON THE SYSTEM. If any net external force acts, momentum will NOT be conserved. Let’s look at a contrived example involving pool balls.

Billiards Imagine the scenario where one pool ball is stationary on a pool table. A second pool ball smacks into it. – The ‘system’ is the two pool balls. – Are there any net external forces? Now, imagine that just before they hit, a hole opens up underneath them. – The ‘system’ is the two pool balls. – Now are there any net external forces? Okay, this time, no sudden trapdoor. Same two pool balls collide. – The ‘system’ this time is just one pool ball. – Now are there any net external forces?

Freight trains example A freight train is being assembled out of two boxcars. Car 1 has a mass of 65,000 kg and is moving to the right at V 01 = 0.8m/s. Car 2 has a mass of 92,000 kg and is also moving to the right at V 02 = 1.3 m/s. Car 2 collides with car 1 and couples with it. What is the speed of the train after the coupling? Momentum is conserved. (m 1 + m 2 )V f = m 1 V 01 + m 2 V 02 V f = 1.1 m/s

Ballistic Pendulum A ballistic pendulum is a device that is used to determine the muzzle velocity of a gun. http://session.masteringphysics.com/problemAsset/1010989/26/1010989A.jpg

Collisions in 2-D Conservation of momentum holds in 2 dimensions as well As one may expect, you can treat the X- and Y- components independently – So, P f = P i – And P fX = P iX – And P fY = P iY

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