1. Linear Dynamics Momentum and motion of systems. Collisions. o Linear Dynamics  Nature of Momentum (& Energy), origins  Linear Momentum Conservation  Momentum for a system of particles o Collision and Impulse  Elastic Collisions  Inelastic Collisions

2. o Aim of the lecture  Concepts in Collisions Dependence on relative motion Dependence on masses Dependence on speed  Newton’s Second Law Conservation of momentum o Main learning outcomes  familiarity with  Two body Collisions (particles!)  Use of energy and momentum conservation  Centre of Mass for a system Lecture 7

3. Linear Momentum o Conserved Quantities  If there is a symmetry in nature, then  There will be a conserved quantity associated with it. (the maths to prove this is beyond the scope of the course) o Examples:  Physics today is the same as physics tomorrow, TIME symmetry The conserved quantity is Energy  Physics on this side of the room is the same as on the other side Linear Translational symmetry The conserved quantity is called Momentum o What is momentum?  For a mass m moving at speed v, the momentum, p is p = mv o In fact momentum is a vector quantity, so we should write p = mv, where v and p are vectors. v m

4. Centre of Mass Frame of Reference o The momentum will depend on  the mass  and the speed  But speed depends on where you measure from.  The momentum is defined relative to a reference frame, here it is the woman watching the ball who is ‘stationary’ p=mv v

5. Centre of Mass Frame of Reference o The man sees the ball as stationary because he is moving with it  the mass is the same  but the speed as measured by the man is zero  The momentum is zero in this ‘reference frame’  It is called the ‘zero momentum frame’ or ‘the centre of mass’ p=mv p = m0 = 0

6. o For several objects considered together ‘a system’ o The total momentum is the sum of the individual momenta P =  p i Where:  p i is the momentum of the i th particle   means ‘sum of’  P is the total momentum P = p 1 + p 2 = 2mv The total momentum of the two balls is the sum of their individual momenta v v

7. o For several objects considered together ‘a system’ o In this case the balls move in opposite directions, so the total momentum is zero P =  p i v P = p 1 + p 2 = mv – mv = 0 The total momentum of the two balls is the sum of their individual momenta v

8. o Momentum is conserved  Must stay in one frame of reference  In collisions  Always in fact o Energy is also conserved o Together energy and momentum conservation  Solve collision problems What happens after collision? Before collision total momentum P = mv-mv=0 v v

9. Total P = 0 so the balls could just stop After Collision Total P = 0 so balls could bounce Total P = mv + 0 = mv this is not zero and so CANNOT happen

10. What happens depends on energy conservation as well as momentum conservation If all the energy remains as kinetic energy, the The buses will simply bounce back the way they came ? At the same speed they came in v v

11. If all the energy is absorbed in a crash, in sound/deformation ? v v The buses will simply crash and stop

12. ? v v P = 0, so after collision p 1 = -p 2 If energy remains as kinetic energy E 1 + E 2 = const E 1 =E 2 = mv 2 /2 = p 2 /2m so for both energy and momentum to be conserved the buses must simply bounce back at same speed

13. ? v v P = 0, so after collision p 1 = -p 2 If energy is converted into sound/deformation in collision Kinetic energy: E 1 = E 2 = 0 after collision = mv 2 /2 = p 2 /2m so for both energy and momentum to be conserved the buses must stop

14. o More generally  Consider two unequal masses and speeds  In elastic collision energy and momentum are conserve v 1i What happens after collision? Before collision total momentum P = 2mv + 0 = 2mv total energy E = 2mv 2 /2 + 0 = 2mv 2 /2 2m m v 2i =0

15. v 1f Before collision total momentum P = mv 1i + mv 2i total energy E = m(v 1i ) 2 /2 + m(v 2i ) 2 /2 2m m v 2f After the collision both energy and momentum remain the same With some algebra: v 1f = (m 1 -m 2 )v 1i /(m 1 + m 2 ) = v 1i /3 v 2f = 2m 1 v 1i /(m 1 + m 2 ) = 4v 1i /3

16. Newton’s Cradle - explain