1. We define the Momentum of an object as: Momentum = mass x velocity p = m v Momentum is measured in kg ms -1 Momentum is a vector quantity. (size and direction) The LAW OF CONSERVATION OF MOMENTUM is “In the absence of external forces, in any collision, the total momentum before the collision is equal to the total momentum after the collision.” Momentum p m v Notes p.26

2. Worked Example 1: 20 m s -1 1500 kg p =mv p = 1500 x 20 p = 30 000 kg m s -1 Worked Example 2: ? m s -1 p =mv 50kg70kg momentum = 360 kg m s -1 v = 3 ms -1

3. Worked Example 3A – Sticking together A trolley of mass 4kg travelling at 2 ms -1 collides and sticks to a 3kg stationary trolley. Determine the velocity of the two trolleys after the collision. 4kg 2 ms -1 3kg 0 ms -1 4kg v = ? ms -1 3kg p = m 1 v 1 + m 2 v 2 = (4 x 2) + (3 x 0) = 8 + 0 = 8 kg ms -1 p = m v v = p / m = 8 / 7 = 1.14 ms -1

4. Example 3B – Sticking together (no need to copy – discuss & try!) What would be the velocity of the two trolleys after the collision had the 3kg trolley been moving towards the 4kg one at 2 ms -1 ? 4kg 2 ms -1 3kg 2 ms -1 4kg v = ? ms -1 3kg p = m 1 v 1 + m 2 v 2 = (4 x 2) + (3 x -2) = 8 + (-6) = 2 kg ms -1 p = m v v = p / m = 2 / 7 = + 0.29 ms -1 SIGN CONVENTION! … + Right

5. Types of Collisions This is very rare (eg. In Molecular collisions with no contact!) This is much more realistic! (eg. Car crash!)

6. Additional Questions Revision Questions in Higher Physics, page 21, Q 1 & 4. Problems to do Q. 76- 84 76a) 20 kgm/s, right b) 500kgm/s, down c) 9kgm/s, left 77 0.75 m/s, Ek before = 2.25J, Ek after = 1.125J so INELASTIC 78 2.4 m/s 79 3 kg 80 a) 2.7m/s b) 0.2J 81 8.6 m/s 82a) 23m/s 82 b) INELASTIC … Ek before = 730000J, Ek after = 702000J. 83 8.67 m/s 84 0.6m/s in initial direction of 2 nd vehicle