LEARNING TARGET: I will identify the part of momentum calculations that give me the MOST indigestion during today's class

CALENDAR:

• Lab Reports are Due looking Good and Standing Tall, YESTERday (Tuesday, February 14th ❤)
• Impulse and Momentum Test is Friday, February 17th

FORMULAE OBJECTUS:

KE = 1/2mv²

m_1iv_1i + m_2iv_2i = m_1fv_1f + m_2fv_2f (Law of Conservation of Momentum)

p=mv

J = impulse (∆p or F∆T or m∆v or p_f - p_i)

WORDS O' THE DAY:

• momentum (p=mv)
• impulse (∆p or F∆T)
• kinetic energy (energy of motion)
• Energy (Joules) (J)

• Law of Conservation of Momentum (Momentum of the objects BEFORE a collision must equal the momentum of the objects AFTER a collision)
• Elastic collision (Kinetic Energy is conserved AND momentum is conserved)
• Perfectly Inelastic collision (Energy is lost to heat, sound and other types of energy so KE is NOT conserved although momentum IS conserved)

ASTRONOMY OBSERVATIONS: We are GO for tonight! Let's take a look HERE

WORK O' THE DAY:

Problem 15

Problem 18

1) A .456 kg ball moving to the east with a speed of 11.34 m/s collides head-on with a .754 kg ball moving to the west at a speed of 7.53 m/s. The velocity of the first ball is 8.50 m/s AFTER the collision.

What is the velocity of the second ball after the elastic collision if both balls bounce back in directly opposite directions?

My solution is here (did you do a sketch?)

1b) Show that the KE of this system is conserved (that means calculate the entire system KE before AND after the collision!)

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2a) Imagine a 1250 kg sports car going 25.51 m/s east crashing head-on into a fully loaded semi (36,000 kg) hauling iron rails to a construction site with a velocity of 19.43 m/s west. The sports car ends up *sticking* to the semi rig and much energy is lost in the collision.

Why is this an example of a *perfectly* inelastic collusion?

Why is kinetic energy NOT conserved?

How does that change the math in our Law of Conservation of Momentum equation?

Please work with your group to write the LawConMom equation to match that scenario.

 

 

 

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Did you come up with something like this:

initial conditions:

v_ic = 25.51 m/s

Note: Negative velocity is KEY here... why?

v_it = -19.43 m/s

Note: The final velocity of the truck and the final velocity of the car will be the same... why?

v_ft= v_fc = ?

 

Note: The mass of the car and truck individually don't change, however they will be stuck together so we'll combine those in a moment

m_ic = m_fc = 1250 kg

m_it = m_ft = 36,000 kg

initial equation:

m_1iv_1i + m_2iv_2i = m_1fv_1f + m_2fv_2f

rewrite to show actual objects (I'd start this way)

m_civ_ci + m_tiv_ti = m_cfv_cf + m_tfv_tf

now show the masses stuck together *after* the collision with the same velocity!:

m_civ_ci + m_tiv_ti = (m_cf + m_tf)v_f

 

Now please calculate the mass of the car-truck AFTER the collision. Be SURE to isolate first, and then substitute and then solve.

ANSWER:

m_civ_ci + m_tiv_ti = (m_cf + m_tf)v_f

isolate:

(m_civ_ci + m_tiv_ti ) / (m_cf + m_tf) = v_f

substitute:

[(1250 kg)(25.51 m/s) + (36000 kg)(-19.43 m/s)] / (1250kg + 36000kg) = v_f

DON'T FORGET TO ASSIGN ONE VELOCITY POSITIVE AND ONE VELOCITY NEGATIVE IF THEY ARE MOVING IN OPPOSITE DIRECTIONS!!!!

solve:

= -17.9 m/s

or 17.9 m/s west

2b) How much force did the sports car experience if its velocity went to zero in .125 seconds

initial conditions:

v_ic = 25.51 m/s

v_fc = 0.00 m/s

∆t = .125 seconds

 

equation:

Why is this just a wee bit nastier now?

F∆t = ∆p

F = ∆p/∆t

What goes next?

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Need more practice? - Take a look at additional Physics Classroom Problems HERE

or Go to your text on page(s):

• 265 for a worked solution
• 266 problem 44 (only). I will have a solution for you momentarily

Check out the graphic on pag 264 Case #1 and Case #2 (ignore the other two cases please)

 

Intro to Air Bag Research Project is HERE - We'll tend to this AFTER the break