Conservation of Momentum & Collisions

1. Conservation of Momentum & Collisions (Chapter 7) Recap: • For situations involving an impactor a collision, where large forces exists for a very small time we define: • Impulse = F x Δt(units:N.s) - avector where F is the force and Δt is the time of action. • By Newton’s 2nd law (F = m.a) we determined: • Impulse = F. Δt = m. Δv or Impulse = Change in momentum (ΔP) Result:Large impulses cause large changes in motion! impulse (F. Δt) momentum change m (ΔP = m. Δv)

2. Momentum • Momentum (P)is the product of themassof an object anditsvelocity: P = m.v(units:kg.m/s) • Momentumis avectoracting in samedirectionasvelocityvector. Example 1:A 100kg boulder rolling towards a castle gate at 3m/s. Momentum boulder:P = m.v=100 x 3=300 kg.m/s Example 2: A 1 kg missile flying towards the castle gate at 300 m/s (speed of sound). Momentum missile:P = m.v= 1 x 300 = 300 kg.m/s • Result: Different objects can have the same momentum (quantity of motion) • But it’s the change in momentum of an object that is important as this equals the impulse: ΔP = m. Δv= impulse

3. Impulse - Momentum Principle Δv impulse F. Δt m • An impulse acting on an object causes a change in its momentum. • The change in momentum is equal in magnitudeand direction to the applied impulse. • Impulse= m. Δv • A new way of looking at Newton’s 2nd law! Example: A 50kg rock is hurled by a giant catapult with a force of 400 N applied for 0.5 sec. Impulse = F. Δt = 4000 x 0.5 = 2000 N.s Thus: Change in momentum: = 2000 Ns = m. Δv or Δv = = 40 m/s ( ≈ 140 km/hr) Note: When the initial velocity is zero: • change in momentum = objects momentum • change in velocity = objects velocity 2000 50

4. Assume no energy is lost, therefore KE of the ball is the same before and after impact. (KE = ½m.v2) • On impactthe momentum of ball is decreased to zero. ΔP P2=m.v P1=m.v Ex: Ball’s Change in Momentum impulse before during after 3. The total change in momentum ΔP is: ΔP = P2 – P1 = m.v – (- m.v) ΔP = 2 m.v • The impulse required to change the direction of ball is therefore equal to twicemomentumof the impacting ball. (ie. Twice as large as what is needed to simply stop the ball.) (As P1 opposite to P2)

5. Summary: • Rewriting Newton’s 2nd law shows that: Impulse = Change in momentum (ΔP) Impulse = F. Δt = m. Δv • Large impulses produce large changes in momentum -resulting in large velocity changes. Where: Impulse = F x Δt(units:N.s) - avector Momentum: P = m.v (units: kg.m/s) - a vector.

6. Conservation of Momentum • A new principle for studying collisions which results from Newton’s 3rd law applied to the impulse/ momentum equation. • Conservation of momentum enables us to understand collisions and to predict many results without a detailed knowledge of time varying forces. Consider:Two sticky objects moving towards each other…they meet in mid-air and after colliding stick together and move as one body. • During the moment of impact there is a strongforce acting for a time Δt. • By Newton’s 3rd law, an equal and opposite force ‘F’ acts back (remember forces occur in pairs).

7. Conservation of Momentum v2 v1 • As Δt is the same for both forces, the impulses they produce (= F.Δt) are the same magnitude but in opposite directions. • By Newton’s 2nd law: - impulse impulse Impulse = Change in momentum (ΔP = m. Δv) - Thus the change in momentum experienced by both objects must be the same…but in opposite directions. - The total change in momentum of the system (i.e. both objects combined) is therefore ZERO! - In other words , the total momentum of thesystem is conserved (i.e. changes of momentum within system cancel each other out).

8. Conservation of Momentum • The total momentum of the system is conserved (if no other external forces acting). • However, different parts of the system canexchange momentum (but the total remains the same). Note:If a net external force acts on system, then it will accelerate and its momentum will change. • Conservation of momentum allows us to examine interesting impact situations…

9. mc = 1 kg mt = 100 kg positive direction Example: Custard pie fights! vc = -10 m/s vt = 0.5 m/s Pc = mc.vcPt = mt.vt Pc = -1 x 10 = -10 kg.m/sPt= 100 x 0.5 = 50 kg.m/s (negative sign as opposite direction) Total momentum = Ptarget + Pcustard pie= 50 - 10 = 40 kg.m/s Question: What is the velocity of target and pie after impact? Total mass = mt + mc = 101 kg Total momentum = 40 kg.m/s (P = m.v) v = P / m = 40 / 101 ≈ 0.4 m/s Result: The unsuspecting target has a larger initial momentum, so his direction of motion prevails but the pie reduces his forward velocity (briefly)!

10. Recoil • A special case of conservation of momentum when the initial velocity of the interacting bodies is often zero. E.g.: - two ice skaters “pushing off”- firing a gun - rocket propulsion… • We have already looked at what happens to the ice skaters motion (using Newton’s 3rd law), but now we can use conservation of momentum to determine their velocities… Example: Initial momentum = 0 Thus: total momentum after “push off” = 0 The momentum of each person must therefore be equal but opposite in direction and P2 = -P1. But, as P1 = m1.v1 and P2 = m2.v2, the velocities will be in opposite directions and will depend on their masses. Eg. If m1 = 3m2 then v2 will be 3v1 inopposite direction!

11. Firing a Gun (initial momentum = zero) Momentum of bullet = Momentum of gun m1v1 = - m2v2 • Mass of bullet is small but its velocity is high…creating a large recoil. • To reduce velocity of recoil (v2), hold gun with locked arms so the mass m2 becomes mass of gun + your body. • Similarly a very massive cannon will “jump back” much less than a light one…for the same shot. Rocket propulsion: • Exhaust gasses have large momentum (light molecules but very high velocity). • Momentum gained by rocket in forward direction equals momentum of exhaust gasses in opposite direction. • This is why rockets (i.e. recoil) work in outer space… as gasses and rocket push against each other as gasses expelled.

12. Collisions • Two main types: Elastic and Inelastic… • Different kinds of collisions produce differentresults…e.g. sometimes objects stick together and other times they bounce apart! • Key to studying collisions is conservationof momentum and energy considerations… Questions: - What happens to energy during a collision? - Is energy conserved as well as momentum?

13. Perfectly Inelastic Collisions (Sticky ones!) 15x103 kg 2x104 kg 5x103 kg 104 kg • E.g. Two objects collide head on and stick together, moving as one after collision (only one final momentum / velocity to compute). • Ignoring external forces (which are often low compared with large impact forces), we use conservation of momentum. E.g. Coupling train trucks(low rolling friction) Before:System momentum: mv = 2x104x10 = 2x105 kg.m/s After:Final system mass = (20+10+5+15)x103 kg = 50x103 kg As final momentum = initial momentum vfinal = • Thus total momentum of system has remained constant but the colliding truck’s velocity has reduced (i.e. momentum shared). v = 10 m/s 3 stationary trucks Pfinal total mass 2x105 5x104 = 4 m/s =

14. Question: What happens to the energy of this system? Total energy = Kinetic Energy = ½.m.v2 (i.e. no PE change) Before impact: KEtot = KEtruck + KE3trucks = ½(2x104)(10)2 + 0 = 106 Joules (1MJ) After impact: KEtot = ½.mtot.v2tot = ½(5x104)(4)2 = 2x105 J Energy differences = (10 - 2)x105 J = 8x105 J (i.e. 80%loss) Results:Energy is lost in an inelasticcollision (heat, sound…) and the greatestportion of energy is lost in a perfectly inelastic collision when objects stick together! Extreme example: Pie (or bullet) hitting a wall… All KE is lost on impact!

15. Bouncing Collisions • If objects bounce off one another rather than sticking together, less energy is lost in the collision. • Bouncing objects are called either “elastic” or “partially inelastic”. The distinction is based on energy. • Elastic Collisions: • No energy is lost in an elastic collision. E.g. A ball bouncing off a wall / floor with no change in its speed (only direction). • Partially Inelastic Collisions: • In general most collisions are “partiallyinelastic” and involve some loss of energy… as they bounce apart. • Playing pool: • Very little energy is lost when balls hit each other and the collision is essentially elastic. In such cases: Momentum and energy are conserved.

16. In an elastic collision we need to find the final velocity of both colliding objects. • Use conservation of momentum and conservation of energy considerations… Example: • Answer: The cue ball stops dead on impact and red ball moves forward with the same velocity (magnitude and direction) as that of the cue ball prior to impact! • Why?...Because both KE(= ½.m.v2) and momentum (m.v) are conserved on impact. • As the masses of both balls are the same the only solution to conserve both KE and momentum is for all the energy and momentum to be transferred to the other (red) ball. • It’s a fact…try it for yourself!!! Question: What happens to red ball and cue ball? P1 cue ball (no spin) P2

18. Impulse • Impulse is the average force acting on an object multiplied by its time interval of action. Impulse = F . Δt (units = N.s) Note:Since instantaneous force may vary during impact we must use average force. • Impulse is a vector acting in thedirection of average force • Thelargertheforce(F) and thelonger itacts(Δt) thelargertheimpulse. ( Impulse is therefore a measure of the overalleffect of the force.) However:Impulse =F. Δt = m. Δv • So an impulse causes a change in velocity (Δv) in magnitude and direction. • In Newton’s words the product “m.Δv” is the change in the “quantity of motion”. • We now term this the change in momentum.

19. Other forms of transport: - Aircraft move by pushing against air. - Boats push against sea. - Walking / driving pushing against land. - Rockets: self contained and pushes against its own exhaust gasses. Note: - Rockets gain momentum gradually (rather than through a single brief impulse)… but can be thought of as a continuous series of small impulses. - As their mass decreases during flight, the resultant velocity is more difficult to calculate.