1. 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.

2. 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.

3. Types of collision

4. 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

5. 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

6. 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)
= 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

7. 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

8. 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.

9. 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.

10. 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

11. 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 = v2
12 = 3v2
v2 = 12 / 3 = 4ms-1
Before
After
v = 0
v = 6ms-1 v2
5kg
2kg m2

12. 2007

13. 2007

14. 2009

15. 2008

16. Elastic and inelastic collisions

17. 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.

18. 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 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) = ( ) v
= v
v = / 16000
v = ms-1 in original direction
Before
After
4 ms-1
2 ms-1 v
4000 kg
12000 kg
4000 kg kg

19. Kinetic energy before collision:
½m1v12 + ½m2v22
= (½ x 4000 x 42) + ( ½ x x 22) =
= J
Kinetic energy after collision:
½(m1 + m2) v32
= ½ ( ) x 2.52
= J

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

21. 2012

22. 2011

23. 2005 Qu: 4

24. Past Paper Questions
2010 Qu: 22

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

26. 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