Momentum Can be defined as inertia in motion

Momentum Can be defined as inertia in motion (or by Newton: an object’s quantity of motion) Equation: p = m v units: kg m/s So if the object’s at rest, then its p = 0 no matter how massive it is. n Since momentum is based on an object’s velocity, which is a vector quantity, it too is an vector quantity – direction matters again! n

Change in Momentum Δp ≠ p of an object! Δp = pf – pi where each term can have different values. Since Δp most commonly from a Δv, Δp = mvf – mvi Δp = m (vf – vi ) [= mΔv, but don’t use that in math sol’ns!] But it can also be from a Δm Ex: full vs empty salt truck rocket ship burning fuel

True or False? n If an object’s Δp = 0, then its p = 0 as well. False Ex: Any moving object with a constant velocity n If an object’s p = 0, then its Δp = 0 as well. also False, although not as often Only if p = 0 only lasts for an instant during the time when we’re determining Δp Ex: a ball at the top of a free fall climb

The Cause of a Change in Momentum Δp is caused by a net force applied for a period of time This is called impulse (J) equations: J = Δp = ΣFΔt = m (vf – vi) units: N s = kg m/s (kg m/s 2) s = kg m/s Notice how net force can be defined in terms of p? ΣF = Δp/Δt or “net F = rate of Δp” This is actually how Newton originally stated his 2 nd law in his book, Philosophiæ Naturalis Principia Mathematica aka…

Mathematical Principles of Natural Philosophy or often simply referred to as the Principia a work in three books by Isaac Newton first published in Latin in July 1687. Lex II: Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur. Translated by Motte 1729 as: Law II: The alteration of motion is ever proportional to the motive force impress'd; and is made in the direction of the right line in which that force is impress'd. So, according to modern ideas of how Newton was using his terminology, Law II: The change of momentum of a body is proportional to the impulse impressed on the body, and happens along the straight line on which that impulse is impressed.

The Cause of a Change in Momentum During a collision, net force can be assumed to be due to the forces between the interacting objects. Any other forces, like gravity or friction, are typically small compared to those between the colliding objects. So ‘ΣF’ in ΣFΔt, can just be F, & often it’s actually avg F… & Δt – the time it takes the collision to take place – is often very short … watch… Classic F vs Δt graphs for a typical collision

Impending Collisions Notice, since Δp = m (vf – vi) = J = F Δt, If an object is destined to undergo a particular Δp (usually because it is going to crash or collide with something else…) then force and time are inversely proportional to each other for that situation. n Since most often it is a lot of force that causes damage to things, we often want to minimize that force, so we try to extend the amount of time the change in momentum (collision) takes place in because, n The longer the time an object takes to change momentum, the less force will be needed, therefore the less damage to the object. Ex: hard floor vs carpet cushioning a catch landing bent legged vs stiff

More examples: Run-away truck ramps

More examples: Air bags in vehicles BUT sometimes we want to “damage” the object, so a lot of force for a short time is ok Ex: hammering a nail into a wall

The Significance of Bouncing When an object bounces, not only did something have to get the object stopped from its original motion, but then it also needed to get it moving again from rest in the opposite direction n This requires a greater Δp or impulse than to simply stop the object from moving – and since a “bounce off” often takes no more time than just stopping, it follows that a larger force is present Examples n Pelton paddle wheel n Rubber bumpers on cars? ? (like those on Grande Prix) n karate chop that doesn’t work n

The Law of Conservation of Momentum The momentum of any isolated system remains constant. n system refers to the objects you’ve chosen to be included - if they interact, it’s with INTERNAL forces n isolated – a system that has no net force being applied from objects outside the defined system called EXTERNAL forces 2 Ways Con of p is Easily Seen: 1. if one object loses p, then another one must gain = p Ex: billiard balls during a game of pool Newton’s Cradle air track

Figure 7 -5 Connecting Train Cars What can we say about the CG of the system… before? After?

Figure 7 -34 Bumper Car Rear End Collision CG of the system… before? After? v’A = 3. 62 m/s v’B = 4. 42 m/s Math is not doable in head, but note: n Slower one gains speed, n Faster one loses speed, n v. A - v. B = - (v’A - v’B ) 0. 8 m/s = - (-0. 8 m/s) Note… always true for 1 D elastic

The Law of Conservation of Momentum [The momentum of any isolated system remains constant. 2 Ways Con of p is Easily Seen: ] 2. if one object starts moving one way, then another will move the opposite way with = p Ex: 2 students face off on skateboards

Example of Conservation of Momentum: Rocket Propulsion Recall by 3 rd law, Fr on g = -Fg on r Now if we ·Δt Then we get Δpg bc r = -Δpr bc g… con of p! What can we say about the CG of the system… before? After?

Example of Conservation of Momentum: Recoil of a Fired Gun CG of the system… before? After?

Figure 7 -3 Momentum is conserved in a collision of two balls, labeled A and B. Where is the CG of the system… before? After?

Forces on the balls during the collision of Fig. 7 -3. FBon. A FAon. B These are the A/R forces from Newton’s 3 rd Law n To find the net internal force of an isolated system, they will be added together! n And will, therefore, cancel… n So the ΣFint = 0 for any isolated system

Conservation of Momentum When it comes to defining your system, you get to pick the objects included in the system n If they’re not in, and they apply a force, it’s an external force, so then it’s no longer an isolated system, so we can’t expect p to be conserved. So then Δpsys ≠ 0, and Impulse (J) = Δpsys = Fext Δt = m (vf – vi) Ex: 1. push on car from outside of it 2. drop ball – it accelerates to ground 3. any interaction /problem from 7. 1 & 7. 3 These are NOT exceptions to the law of conservation of momentum, we just aren’t satisfying the requirements of the law! n

Conservation of Momentum [When it comes to defining your system, you get to pick the objects included in the system] n If they’re in and they apply a force, it’s an internal force, so Δpsys = 0 for the system as a whole, even though objects inside may be changing their individual p’s Ex: 1. 2 students face off on skateboards 2. push on car from inside it 3. push car from outside, but include earth 4. watch ball drop, but include earth Does the Earth really gain = & - momentum? Yes – but too small to measure No – since it probably isn’t even a net force

The math for the law of conservation of p: for an isolated system: pisys = pfsys [pi 1 + pi 2 + … = pf 1 + pf 2 + …] m 1 vi 1 + m 2 vi 2 = m 1 vf 1 + m 2 vf 2 m 1 = • Back to vectors big time!! Use directions! m 2 = • ID the given carefully & use sketch with dir key vi 1 = vi 2 = • Instead of subscripts 1 & 2, use letters that vf 1 = represent specific objects in the problem vf 2 = • If something starts from or goes to rest, then that entire term = 0 • If the objects are stuck together, then they have = v’s so you can pull it out as a common factor: use (m 1+ m 2) vi , not m 1 vi 1 + m 2 vi 2

Types of Interactions n Elastic Collisions – kinetic energy is conserved before and after the collision But not during the collision: KE PEe, until “minimum separation distance” is reached So then there’s no heat loss, no permanent deformation, no sound created! Ex: atomic & molecular collisions But, in the macro world, only occurs as an ideal… since really there’s always some energy lost to heat, so true elastic collisions don’t really exist – but in Physicsland, we have close approximations: Ex: billiard balls in a game of pool steel balls on Newton’s cradle spring bumpers on gliders on air track

Types of Interactions n Inelastic Interactions – where KE is not conserved n Inelastic Collisions: KEi > KEf n Ex: minor car accident, bouncing ball, any hit in sports, fired bullet passes straight thru object n Inelastic Explosions: KEi < KEf n Ex: firecracker, bomb (spring bug toy- not isolated) Completely Inelastic Collision – where not only is KE not conserved, but the maximum lost during the collision is how much stays lost even after it’s over. When the objects entangle / stick together as a result of the collision – maximum permanent deformation n Ex: serious car accident, fired bullet lodges into object, train cars hooking up (gliders w/ magnets) Note: p is conserved throughout all types of interactions! n

Figure 7 -13 Elastic and Inelastic Collisions (e) If completely inelastic, they’d have stuck together – maybe still moving, maybe not

The Math of Different Types of Interactions n Any Isolated Interaction (elastic or inelastic) If only 1 unknown – only need 1 eq’n: Use Law of Con of p: pisys = pfsys Elastic Collisions If 2 unknowns – need 2 eq’ns: Use Law of Con of p: pisys = pfsys Use Con of KE: Ki = Kf n Head-on Use Law of Con of p: pisys = pfsys Use: vi 1 – vi 2 = vf 2 – vf 1 (see derivation p. 176) n

The Math of Different Types of Interactions Inelastic Collisions – classic ex: ballistic pendulum Must watch because often outside forces act at least for a portion of the entire situation, and if so, then Δpsys ≠ 0 p is not conserved in that portion n For the collison portion, it’s isolated, so Use Law of Con of p: pisys = pfsys n For the stages where there are external forces, then try to conserve energy In general, use L of C of E: Emechi + WNC = Emechf If external forces are conservative forces, then WNC = 0 If external forces are nonconservative forces, then usually Emechf = 0 n

Ballistic Pendulum Set Up Stage I: isolated system bullet moving toward stationary block Stage II: bullet imbeds in block, moving together still isolated, but total inelastic collision Stage III: gravity transforms all KE into PEg no longer isolated, but only conservative forces acting But what if block was not a pendulum, but attached to a spring that either extended or compressed after catching the bullet? Or slid to rest along a surface after catching the bullet?

Momentum Conservation in 2 or 3 Dimensions Individual momentum vectors of the various objects must be used then combine the momentum vectors by vector addition recall tip to tail or parallelogram method? then apply rt Δ trig or law of sine/cosine to solve? and still, pisys = pfsys What can we say about the CG of the system… before? After? n Try sample problem on handout…