Conservation of Momentum & Energy in Collisions

Given some information, & using conservation laws, we can determine a LOT about collisions without knowing the collision forces! To analyze ALL collisions: Rule #1 Momentum is ALWAYS (!!!) conserved in a collision!  m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f H OLDS for ALL collisions!

An Ideal Very Special Case: 2 very hard objects (like billiard balls) collide. An “Elastic Collision” To analyze Elastic Collisions: Rule # 1 Still holds!  m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Rule # 2 For Elastic Collisions ONLY (!!) Total Kinetic Energy (KE) is conserved!! (KE) before = (KE) after  (½)m 1 (v 1i ) 2 + (½) m 2 (v 2i ) 2 = (½)m 1 (v 1f ) 2 + (½)m 2 (v 2f ) 2 Note!!

Total Kinetic energy (KE) is conserved for ELASTIC COLLISIONS ONLY!! Inelastic Collisions  Collisions which are AREN’T elastic. Is KE conserved for Inelastic Collisions? NO!!!!!! Is momentum conserved for Inelastic Collisions? YES!! By Rule # 1: Momentum is ALWAYS conserved in a collision!

Types of Collisions Momentum is conserved in all collisions Inelastic collisions: rubber ball and hard ball –Kinetic energy is not conserved –Perfectly inelastic collisions occur when the objects stick together Elastic collisions: billiard ball –both momentum and kinetic energy are conserved Actual collisions –Most collisions fall between elastic and perfectly inelastic collisions

Collisions Summary In an elastic collision, both momentum and kinetic energy are conserved In a non-perfect inelastic collision, momentum is conserved but kinetic energy is not. Moreover, the objects do not stick together In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same Elastic and perfectly inelastic collisions are limiting cases, most actual collisions fall in between these two types Momentum is conserved in all collisions

Special case: Head-on Elastic Collisions These can be analyzed in 1 dimension. Some Types of head on collisions: For 2 masses colliding elastically: Often we know the masses & the initial velocities. Both momentum & kinetic energy are conserved, so we have 2 equations. Doing algebra, we can solve for the 2 unknown final speeds.

Special case: Head-on Elastic Collisions. 1 dimensional collisions: Some possible types: before collision  or after collision or  v A, v B, (v A ), (v B ), are 1 dimensional vectors!

Elastic Collisions in 1 Dimension Special case: Head-on Elastic Collisions. Momentum is conserved (ALWAYS!) P before = P after m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f v 1i, v 2i, v 1f, v 2f are one dimensional vectors! Kinetic Energy is conserved (ELASTIC!) (KE) before = (KE) after (½)m 1 (v 1i ) 2 + (½) m 2 (v 2i ) 2 = (½)m 1 (v 1f ) 2 + (½)m 2 (v 2f ) 2 2 equations, 6 quantities: v 1i,v 2i,v 1f, v 2f, m 1, m 2  Clearly, we must be given 4 out of 6 to solve problems! Solve with CAREFUL algebra!!

m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (1) (½)m 1 (v 1i ) 2 + (½) m 2 (v 2i ) 2 = (½)m 1 (v 1f ) 2 + (½)m 2 (v 2f ) 2 (2) Now, some algebra with (1) & (2), the results of which will help to simplify problem solving: Rewrite (1) as: m 1 (v 1i – v 1f ) = m 2 (v 2f - v 2i ) (a) Rewrite (2) as: m 1 [(v 1i ) 2 - (v 1f ) 2 ] = m B [(v 2f ) 2 - (v 2i ) 2 (b) Divide (b) by (a):  v 1i + v 1f = v 2i + v 2f or v 1i – v 2i = v 2f – v 1f = - (v 1f – v 2f ) (3) Relative velocity before= - Relative velocity after Elastic head-on (1d) collisions only!!

Summary: 1d Elastic collisions: Rather than directly use momentum conservation + KE conservation, often convenient to use: Momentum conservation: m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (1) Along with: v 1i – v 2i = v 2f – v 1f = - (v 1f – v 2f ) (3) (1) & (3) are equivalent to momentum conservation + Kinetic Energy conservation, since (3) was derived from these conservation laws!

Example: Pool (Billiards) m 1 = m 2 = m, v 1i = v, v 2i = 0, v 1f = ?, v 2f = ? Momentum Conservation: mv +m(0)=mv 1f + mv 2f Masses cancel  v = v 1f + v 2f (I) Relative velocity results for elastic head on collision: v - 0 = v 2f - v 1f (II) Solve (I) & (II) simultaneously for v 1f & v 2f :  v 1f = 0, v 2f = v Ball 1: to rest. Ball 2 moves with original velocity of ball 1 Before: Ball 1Ball 2  v v = 0 Before: Ball 1  v Ball 2 v = 0

Example: Unequal Masses, Target at Rest A very common practical situation is for a moving object (m 1 ) to strike a second object (m 2, the “target”) at rest (v 2 = 0). Assume the objects have unequal masses, and that the collision is elastic and occurs along a line (head-on). (a) Derive equations for v 2f & v 1f in terms of the initial velocity v 1i of mass m 1 & the masses m 1 & m 2. (b) Determine the final velocities if the moving object is much more massive than the target (m 1 >> m 2 ). (c) Determine the final velocities if the moving object is much less massive than the target (m 1 << m 2 ).

Inelastic Collisions Inelastic Collisions  Collisions which Do NOT Conserve Kinetic Energy! Some initial kinetic energy is lost to thermal or potential energy. Kinetic energy may also be gained in explosions (there is addition of chemical or nuclear energy). A Completely Inelastic Collision is one in which the objects stick together afterward, so there is only one final velocity.

Total Kinetic energy (KE) is conserved for ELASTIC COLLISIONS ONLY!! Inelastic Collisions  Collisions which are NOT elastic. Is KE conserved for Inelastic Collisions? NO!!!! Is momentum conserved for Inelastic Collisions? YES!! (Rule # 1: Momentum is ALWAYS conserved in a collision!). Special Case: Completely Inelastic Collisions  Inelastic collisions in which the 2 objects collide & stick together. KE IS NOT CONSERVED FOR THESE!!

Example: Railroad cars again Same rail cars as Ex Car A, mass m 1 = 10,000 kg, traveling at speed v 1i = 24 m/s strikes car B (same mass), initially at rest (v 2i = 0). Cars lock together after collision. Ex. 7-3: Find speed v after collision. Ex. 7-3 Solution: v 2i = 0, v 1f = v 2f = v Use Momentum Conservation: m 1 v 1i +m 2 v 2i = (m 1 + m 2 )v  v = [(m 1 v 1i )/(m 1 + m 2 )] = 12 m/s Before Collision After Collision Ex. 7-9: Cars lock together after collision. Find amount of initial KE transformed to thermal or other energy forms: Initially: KE i = (½)m 1 (v 1i ) 2 = 2.88  10 6 J Finally: KE f = (½)(m 1 + m 2 )(v) 2 = 1.44  10 6 J ! (50% loss!)

More about Perfectly Inelastic Collisions When two objects stick together after the collision, they have undergone a perfectly inelastic collision Conservation of momentum Kinetic energy is NOT conserved

An SUV Versus a Compact  An SUV with mass 1.80  10 3 kg is travelling eastbound at m/s, while a compact car with mass 9.00  10 2 kg is travelling westbound at m/s. The cars collide head-on, becoming entangled. (a) Find the speed of the entangled cars after the collision. (b) Find the change in the velocity of each car. (c) Find the change in the kinetic energy of the system consisting of both cars.

(a) Find the speed of the entangled cars after the collision. An SUV Versus a Compact

(b) Find the change in the velocity of each car. An SUV Versus a Compact

(c) Find the change in the kinetic energy of the system consisting of both cars. An SUV Versus a Compact

More About Elastic Collisions Both momentum and kinetic energy are conserved Typically have two unknowns Momentum is a vector quantity –Direction is important –Be sure to have the correct signs Solve the equations simultaneously

Elastic Collisions A simpler equation can be used in place of the KE equation

Summary of Types of Collisions In an elastic collision, both momentum and kinetic energy are conserved In an inelastic collision, momentum is conserved but kinetic energy is not In a perfectly inelastic collision, momentum is conserved, kinetic energy is not, and the two objects stick together after the collision, so their final velocities are the same

Example: Ballistic pendulum The ballistic pendulum is a device used to measure speeds of projectiles, such as a bullet. A projectile, mass m, is fired into a large block, mass M, which is suspended like a pendulum. After the collision, pendulum & projectile swing up to a maximum height h. Find the relation between the initial horizontal speed of the projectile, v & the maximum height h.

Example: Inelastic Collisions BeforeAfter Momentum Conservation mv = (m + M)v´ Mechanical Energy (½)(m +M)(v´) 2 = (m + M)gh Conservation  v = [1 +(M/m)](2gh) ½ a v = 0 a a a a a a  ℓ - h  ℓ ℓ

Example A bullet, m = kg hits & is embedded in a block, M = 1.35 kg. Friction coefficient between block & surface: μ k = Moves d = 9.5 m before stopping. Find v of the bullet it before hits the block. Multi-step problem! 1. Find V using Work-Energy Principle with friction. 2. Find v using momentum conservation. But, to find V, first we need to 3. Find the frictional force! F fr = μ k F N = μ k (M+m)g

1. Friction force: F fr = μ k F N = μ k (M+m)g 2. The Work- Energy Principle: W fr = -F fr d =  KE = 0 – (½)(M+m)V 2 OR: -F fr d = - (½)(M+m)V 2 μ k (M+m)gd = (½)(M+m)V 2 (masses cancel!) Stops in distance d = 9.5 m  V = 6.82 m/s 3. Momentum conservation: mv + 0 = (M+m) V  v = (M+m)V/m = 375 m/s (bullet speed)

Summary: Collisions Basic Physical Principles: Conservation of Momentum: Rule # 1: Momentum is ALWAYS conserved in a collision! Conservation of Kinetic Energy: Rule # 2: KE is conserved for elastic collisions ONLY !! –Combine Rules #1 & #2 & get relative velocity before = - relative velocity after. As intermediate step, might use Conservation of Mechanical Energy (KE + PE)!! v A – v B = v B – v A

Collisions in Two or Three Dimensions Conservation of energy & 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.

Elastic Collisions in 2D qualitative here, quantitative in the text Physical Principles: The same as in 1D 1. Conservation of VECTOR momentum: P Ax + P Bx = P Ax + P Bx P Ay + P By = P Ay + P By 2. Conservation of KE (½)m A (v A ) 2 + (½)m B (v B ) 2 = (½)m A (v A ) 2 + (½)m B (v B ) 2