2. SUBJECT TOPIC

3. Group 2 Team Members Au Yeung Sum Yee Licia (99197960) Chung On Wing (99163700) Lee Chun Kit (99197390) Leung Siu Fai (99269080) Mak Ka Man (99163620)

4. Responsibility

5. About the Package Target Audience  Band 3  Form 4 Students Function of this Package as an auxiliary teaching aids

6. Previous Knowledge Newton's Second Law Momentum Impulse

7. Objectives Recall 3 kinds of collisions and momentum Define the law of conservation of momentum Apply the law of conservation of momentum to solve problems

8. Shooting gun Why does the gun recoil(move backwards) after firing?

9. Table of content  Collision  Elastic collision  Inelastic collision  Partially elastic collision  Momentum  Conservation of momentum  Example  MC Exercise  Explosion  Demonstration of water rocket  Problem solving strategy of firing a gun

10. Collision There are three kinds of collision –Elastic collisionElastic collision –Inelastic collisionInelastic collision –Partially elastic collisionPartially elastic collision

11. Elastic Collision The two identical balls hit each other and bounce back to the same level

12. Partially Elastic Collision The balls bounce back to a lower level

13. Inelastic Collision The balls do not bounce i.e. they stick together

14. Momentum = mass (kg)  velocity (ms -1 ) p = m  v the quantity of motion how much stuff is moving (mass) how fast the stuff is moving(velocity)

15. same velocity Animation of a football and a tennis move at same velocity Which one has greater momentum? Football/Tennis ball

16. Animation of two identical cars move at different velocity Which one has greater momentum? Car ACar B Car A Car B / Car A Car B

17. Conservation of Momentum

18. What is conservation of momentum? In any collision, the total momentum before collision is equal to the total momentum after collision, provided that there is no external force acting. If F ext = 0 P before = P after

19. When F ext = 0, = m 1 u 1 + m 2 u 2 Before collision m 1 v 1 + m 2 v 2 After collision

20. A 20 g marble travels to the right at 0.4 ms -1 on a smooth, level surface. It collides head- on with a 60 g marble moving to the left at 0.2 ms -1. After collision, the 20 g marble rebounds at 0.1 ms -1. Find the velocity of the 60 g marble.

21. Solution Step 1 Make a sketch showing the direction, masses and velocities of each object before collision. Before collision: 0.02kg 0.06kg 0.4ms -1 0.2ms -1 Step 2 Assign one direction as the positive direction. Assume that the 0.06 kg marble continues to move to the left after the collision at a velocity v. Take the direction to the left as positive. + ve

22. Step 3 Make a sketch showing the direction, masses and velocities of each object after collision. 0.02kg 0.06kg v0.1ms -1 After collision Before collision: 0.02kg 0.06kg 0.4ms -1 0.2ms -1 + ve

23. Step 4 Write down the equation for conservation of momentum and substitute the known values of each object after collision. As no external force exists during the collision, by the law of conservation of momentum m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 (0.06 kg * 0.2 ms -1 ) + (0.02 kg*-0.4ms -1 ) = (0.06 kg * v) + (0.02 kg*0.1ms -1 ) v = 0.033 ms -1 Step 5 Solve the equation to find out the unknown value. The velocity of the 60 g marble is 0.033 ms -1 to the left. m2u2m2u2 m1u1m1u1 m1v1m1v1 m2v2m2v2 + +=

25. 15 cm s 10 cm s -1 20 cm s -1 25 cm s -1 30 cm s -1 In the following figure, two particles of masses 1kg and 2kg are moving in the same direction at speed of 30 cms -1 and 15 cms -1 respectively, If they stick together after collision, the final speed of the particles is 30 cms -1 15 cms -1 1 kg2 kg QUESTION 1

26. 3 m s -1 towards the left Two objects A and B of masses 2 kg and 1 kg respectively move in opposite directions. They collide head on. After the collision, the velocity of A becomes 1 m s -1 towards the left. What would be the velocity of B ? 2 m s -1 towards the right 4 m s -1 towards the left3 m s -1 towards the right 4 m s -1 towards the right 2 ms -1 4 ms -1 AB QUESTION 2

27. 1 m s -1 A trolley of mass 1 kg travelling at 3 m s -1 collides with a stationary trolley of mass 2 kg. If the two trolleys remain together after collision, their combined speed immediately after collision, is 10 m s -1 1.5 m s -1 2 m s -1 3 m s -1 QUESTION 3

28. 6reversed Trolley A towards a stationary trolley B ahead. After collision, it is found that trolley B moves at a speed of 18 m s -1. The final velocity of trolley A is 6same as before 12same as before 12reversed 21reversed Speed Direction QUESTION 4

29. moving at a constant speed of 4 m s -1 Stationary moving at a constant speed less than 4m s -1 decelerating from a speed of 4m s -1 decelerating from a speed less than 4 m s -1 QUESTION 5 After collision, both trolleys sticked together. They are

31. Rocket gains momentum in the up direction The hot gases gain momentum in the down direction

33. Shooting gun Why does the gun recoil(move backwards) after firing? Remember the question asked at the beginning of this lesson?

34. By the law of conservation of momentum, m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 Before firing, both the gun and the bullet have zero momentum i.e. m 1 u 1 + m 2 u 2 = 0 So, the gun must have a backward momentum After firing, the bullet moves forward, and have a forward momentum

36. Sorry, you are wrong Don’t disappointed TRY AGAIN