Key Areas covered Explosions and Newton’s third law. Conservation of momentum in explosions in one dimension only. Kinetic energy in elastic and inelastic collisions.

What we will do today Take notes and carry out calculations concerned with a variety of collisions. State what is meant by an elastic or inelastic collision. Use values for kinetic energy to determine if a collision is elastic or inelastic.

Types of collision

Solving Problems 1. Always make a sketch of the system before and after the collision or explosion. 2. Mark all masses and velocities (with direction!!) on the sketch. 3. You will need to allocate a positive direction for vector quantities – mark this also on the sketch. 4. Use the rule: total momentum before = total momentum after

In general, there are three types of problem: 1. Two masses collide and move apart with different velocities after the collision: m1v1 + m2v2 = m1v1 + m2v2 Before After v1 v2 v1 v2 m1 m2 m1 m2

Total p before = total p after Example: Find the velocity of m1 after the collision Total p before = total p after m1v1 + m2v2 = m1v1 + m2v2 (3x4) + (2x2) = (3v1) + (2x4) 12 + 4 = 3v1 + 8 16 – 8 = 3v1 v1 = 8 / 3 v1 = 2.67 ms-1 Before After 4ms-1 2ms-1 v1 4ms-1 3kg 2kg 3kg 2kg

2. Two masses collide and stick together: m1v1 + m2v2 = (m1 + m2)v3 where v3 is the velocity after collision. Before After v1 v2 v3 m1 m2 m1 + m2

Example Solution + ve Tot mom Before = Tot mom After m1v1 + m2v2 = m3v3 (3 x 5) + (2 x -5) = (3 + 2) v3 15 + (-10) = 5v3 v = 5 / 5 v = 1 ms-1 to right 5 ms-1 5 ms-1 3 kg 2 kg Find the velocity of the trolleys when they stick together after colliding.

Explosions (Newton’s third law) In an explosion, one body, originally at rest, explodes into two parts, moving in opposite directions. This follows Newton’s third law – for every action there is an equal but opposite reaction. Therefore the momentum at the beginning is zero so the momentum at the end must also be zero. Care must be taken to use –ve values for any objects moving to the left to allow this law to hold true. Examples include a gun shooting a bullet and a cannon firing a cannon ball.

3. An explosion. In this case: (m1+ m2) v = m1(v1) + m2v2 If initially at rest (e.g. gun before firing a bullet), then: 0 = m1(-v1) + m2v2 As objects will move in opposite directions so –ve value required Note: this is not always the case i.e. if the joined masses are already moving at a velocity before explosion. Before After v v1 v2 m m1 m2

Example: Find the velocity of m2 after explosion Need to work out m2 first, m2 = (m1+ m2) – m1 = 5-2 = 3kg total p before = total p after (m1+ m2) v = m1(v1) + m2v2 Note v1 is –ve 5 x 0 = 2x(-6) + 3v2 0 = -12 + 3v2 12 = 3v2 v2 = 12 / 3 = 4ms-1 Before After v = 0 v = 6ms-1 v2 5kg 2kg m2

2007

2007

2009

2008

Elastic and inelastic collisions

Elastic and Inelastic Collisions An Elastic Collision is one in which both kinetic energy and momentum are conserved. An Inelastic Collision is one in which only momentum is conserved. NB: In any collision all energy is conserved (cons. of energy), elastic and inelastic only deals with kinetic.

Example Before After Solution m1v1 + m2v2 = (m1 + m2) v3 During a space mission, it is necessary to ‘dock’ a space probe of mass 4000 kg onto a space ship of mass 12000 kg. The probe travels at 4 ms-1, and the ship travels at 2 ms-1 ahead of the probe, but in the same direction. (a) What is the velocity of the ship after the probe has ‘docked’? (b) Is this collision elastic or inelastic? Solution m1v1 + m2v2 = (m1 + m2) v3 (4000 x 4) + (12000 x 2) = (4000 + 12000) v 16000 + 24000 = 16000v v = 40000 / 16000 v = 2.5 ms-1 in original direction Before After 4 ms-1 2 ms-1 v 4000 kg 12000 kg 4000 kg + 12000 kg

Kinetic energy before collision: ½m1v12 + ½m2v22 = (½ x 4000 x 42) + ( ½ x 12000 x 22) = 32000 + 24000 = 56 000 J Kinetic energy after collision: ½(m1 + m2) v32 = ½ (4000+12000) x 2.52 = 50 000 J

Ek before does not equal Ek after. Therefore collision is inelastic.

2012

2011

2005 Qu: 4

Past Paper Questions 2010 Qu: 22

Questions Activity sheets: Collisions and explosions You should now be able to answer questions: 1-13 in class jotter

Answers Collisions and Explosions 1. (a) 20 kg ms−1 to the right (b) 500 kg ms−1 down (c) 9 kg ms−1 to the right 2. (a) 0·75 ms−1 in the direction in which the first trolley was moving (b) Teacher Check 3. 2·4 ms−1 4. 3·0 kg 5. (a) 2·7 ms−1 (b) 0·19 J 6. 8·6 ms−1 in the original direction of travel 7. (a) 23 ms−1 (b) Teacher Check 8. 8·7 ms−1 9. 0·6 ms−1 in the original direction of travel of the 1·2 kg trolley 10. 16·7 ms−1 in the opposite direction to the first piece 11. 4 kg 12. 0·8 ms−1 in the opposite direction to the velocity of the man 13. 1·3 ms−1 in the opposite direction to the velocity of the first trolley