LEARNING TARGET: I will work with my team to solve basic elastic an inelastic collisions during today's class.

WORDS O' THE DAY:

• momentum (p=mv)
• impulse (∆p or F∆T)
• elastic collisions (sometimes called 'perfectly elastic' collisions. We'll sometimes add the 'perfect' part to emphasize the fact that no kinetic energy is lost during that collision. Like many of our high school physics models, that is a highly simplistic view of collisions that doesn't really happen in nature:
• • m_1iv_1i + m_2iv_2i = m_1fv_1f + m_2fv_2f

 

• Inelastic Collisions can also be classified as 'perfect' or just plain inelastic. Since we aren't going to study just plain inelastic collisions we'll assume that all inelastic collisions stick together (which makes them 'perfect'!)

CALENDAR:

FORMULAE OBJECTUS:

p=mv

J = ∆p

J = F∆t

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

WORK O' THE DAY:

Imagine a fully loaded cement truck (m = 35,555 kg) traveling @ 55.000 mph smacks head on into a parked (and unoccupied) 2017 Scion FRS (m=1275 kg).

• After the collision what is the mass of the cement truck-Scion FRS conglomeration if the two objects stick together with absolutely no loss of mass?

• Now determine the final velocity cement truck-Scion FRS conglomeration (Work with your team to derive an appropriate equation FIRST (this is KEY!))

• Now let's test to see if it is in fact a perfectly inelastic collision by doing... what?

═══════════════════════════

The concept of Impulse (J) is a tough one for many first year physics students. Impulse is really nothing more than the change in momentum an object experiences as a result of a force being applied to that object over a specific amount of time:

Key Points:

• A large amount of force applied over a short amount of time to an object of a certain mass moving at a certain velocity will have the same impact as much smaller amount of force applied over a much longer period of time.

• That means impulse is best viewed as a range of values not a specific number:

  • If you are driving a car down I-5 at 60. mph ( m/s) your brakes have to apply a rather massive amount of force to stop your car in a short time (let's say 3.0 seconds).

  • However, if you are driving a car down I-5 at 60. mph ( m/s) your brakes have to apply a much less amount of force to stop your car in a longer period of time (let's say 6.0 seconds).
• The key point is that your car is ending up at zero velocity in each case. The force and time change but the impulse is the same.

Let's do the math. Go look up the approximate mass of your favorite car and imagine you are driving it as shown above. Calculate the force applied in each case (note that the impulse stays the same!)

 

It sounds simple enough but it has major and profound impact on the physics of collisions not to mention rocket science and other really cool parts of physics!

Let's review an example since baseball season is about to get started!

• The very fastest Major League Baseball pitches can throw up around 101 mph (45.1 m/s) -- I think the actual record for the fastest pitch ever clocked on radar was just over 106 mph. The mass of an official MLB baseball is about .145 kg (there is a range of masses actually: from .14175 kg to .14883 kg -- but I digress)

• What is the the momentum (p) of such a fastball just AFTER it leaves the pitcher's hand? FULL WOLGEMUTHIAN!

• Check out the picture below. How does that relate to our conversations of momentum and ESPCIALLY of impulse?

 

 

 

 

Now let's say that you can impart 750 N of force onto the SAME ball traveling the SAME velocity using the bat. How long must the bat be in contact with the ball to completely stop the ball's forward momentum.

Now Let's explore the relationship between momentum, force & time

• Take that exact same situation, and let's say you didn't *quite* take a good whack at the ball and you were only able to impart 501. N of force on the ball. How long must the ball be in contact with the bat to reverse the ball's momentum?
• Now let's chart F, t and p

F	∆t	∆p = J
→750 N	→.0087s	→6.54 Ns
→501 N	→.013 s	→6.54 Ns
→	→	→6.54 Ns

 

• If the bat is in contact with that ball for 0.0087 seconds (!), how much force did that bat make with the ball? FULL WOLGEMUTHIAN!

Consider the following scenario:

Would you rather be hit in the head with a bag of feathers or a 2 x 4" (board). (Each has the same mass and each is swung at the same velocity)

 

Let's discuss in detail

═══════════════════════════

• Now let's switch gears a bit: One of the big-time-overarching laws of physics is known as the Law of Conservation of Momentum and it is particularly helpful in analyzing collisions:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

 

Take a few moments and quiz each other on that... let's get it firmly implanted into your respective graymatter!

The Law of Conservation of Momentum is *HUGE*

Let's discuss.

If time permits we'll START on these (but don't worry about working on them this weekend)

Practice Problems 1- 10: Pick at least 4 of those and then work on setting up the infamous BLUE FONT problem #10