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EDD 5165A Presentation Package PHYSICS SUBJECT Conservation of Momentum TOPIC
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)
Responsibility
About the Package ,[object Object],[object Object],[object Object],[object Object],[object Object]
Previous Knowledge ,[object Object],[object Object],[object Object]
Objectives ,[object Object],[object Object],[object Object]
Shooting gun ,[object Object]
Table of content ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
Collision ,[object Object],[object Object],[object Object],[object Object]
Elastic Collision ,[object Object],Animation of two balls
Partially Elastic Collision The balls bounce back to a lower level
Inelastic Collision ,[object Object],[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object],[object Object],Momentum
Animation of  a football and a tennis move at  same velocity Which one has  greater  momentum? Football / Tennis ball Tennis ball
Animation of two  identical  cars move at  different velocity Which one has greater momentum? Car A   Car B Car B Car A Car B / Car A Car B
Conservation of Momentum Conservation of momentum Conservation of momentum
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
When F ext  = 0, = m 1 u 1  +  m 2 u 2   Before collision m 1 v 1  +  m 2 v 2 After collision
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. PROBLEM
Solution Step 1 Make a  sketch  showing the  direction ,  masses  and  velocities  of each object  before  collision. Before  collision: 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 0.02kg 0.06kg
Step 3 Make a  sketch  showing the  direction ,  masses  and  velocities  of each object  after   collision. After  collision + ve 0.02kg 0.06kg v 0.1ms -1 Before  collision: 0.02kg 0.06kg 0.4ms -1 0.2ms -1 + ve
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. m 2 u 2 m 1 u 1 m 1 v 1 m 2 v 2 + + =
START!! MC EXERCISES
15 cm s -1 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 kg 2 kg QUESTION 1
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 left 3 m s -1  towards the right 4 m s -1  towards the right 2 ms -1 4 ms -1 A B QUESTION 2
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
6 reversed 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 6 same as before 12 same as before 12 reversed 21 reversed   Speed   Direction QUESTION 4
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
Application of  conservation of momentum
Explanation Rocket gains momentum in the up direction The hot gases gain momentum in the down direction
WATER ROCKET
Shooting gun ,[object Object],Remember the question asked at the beginning of this lesson?
[object Object],[object Object],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 The Gun Recoil
THE END
Sorry, you are wrong Don’t disappointed TRY AGAIN
Congratulations

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Conservation Of Momentum

  • 1. EDD 5165A Presentation Package PHYSICS SUBJECT Conservation of Momentum TOPIC
  • 2. 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.
  • 5.
  • 6.
  • 7.
  • 8.
  • 9.
  • 10.
  • 11. Partially Elastic Collision The balls bounce back to a lower level
  • 12.
  • 13.
  • 14. Animation of a football and a tennis move at same velocity Which one has greater momentum? Football / Tennis ball Tennis ball
  • 15. Animation of two identical cars move at different velocity Which one has greater momentum? Car A Car B Car B Car A Car B / Car A Car B
  • 16. Conservation of Momentum Conservation of momentum Conservation of momentum
  • 17. 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
  • 18. When F ext = 0, = m 1 u 1 + m 2 u 2 Before collision m 1 v 1 + m 2 v 2 After collision
  • 19. 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. PROBLEM
  • 20. Solution Step 1 Make a sketch showing the direction , masses and velocities of each object before collision. Before collision: 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 0.02kg 0.06kg
  • 21. Step 3 Make a sketch showing the direction , masses and velocities of each object after collision. After collision + ve 0.02kg 0.06kg v 0.1ms -1 Before collision: 0.02kg 0.06kg 0.4ms -1 0.2ms -1 + ve
  • 22. 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. m 2 u 2 m 1 u 1 m 1 v 1 m 2 v 2 + + =
  • 24. 15 cm s -1 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 kg 2 kg QUESTION 1
  • 25. 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 left 3 m s -1 towards the right 4 m s -1 towards the right 2 ms -1 4 ms -1 A B QUESTION 2
  • 26. 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
  • 27. 6 reversed 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 6 same as before 12 same as before 12 reversed 21 reversed Speed Direction QUESTION 4
  • 28. 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
  • 29. Application of conservation of momentum
  • 30. Explanation Rocket gains momentum in the up direction The hot gases gain momentum in the down direction
  • 32.
  • 33.
  • 35. Sorry, you are wrong Don’t disappointed TRY AGAIN