1. Momentum Imagine a cannon: When the cannon fires it applies a force to the cannon ball of mass m B given by the equation: F = m B A Boom But we have another formula for Acceleration : So by substituting for A we can say: mBmB or In Words....

2. It is now convenient to introduce a new quantity called ‘Momentum’. It is a measure of the mass and the velocity of an object so.. Momentum = Mass × Velocity Now we can simplify the word equation above: Now remember Newton’s 3 rd Law. ‘Whenever one object exerts a force on another, the other object exerts the same force back on the first object in the opposite direction’. So if the canon exerts a force on the cannon ball, the cannon ball must exert an equal and opposite force on the cannon. (This is why the cannon ‘recoils’)

3. The time for which this force is applied is the same so the cannon must undergo an equal and opposite change in momentum. What was the combined momentum of the cannon and the ball before the explosion ? Answer....0 (because neither were moving) What was the combined momentum of the cannon and the ball after the explosion ? Answer....0 (because they both started off stationary and experienced the same force for the same time but in the opposite direction) Conclusion. Momentum is a quantity which is ‘CONSERVED’ within a ‘system’. The momentum of all the objects within a system is the same before and after an event like an explosion or a collision. This is best understood by looking at the diagram again.

4. Question 1. The cannon ball had a mass of 80 kg and velocity of 40 m/s after being fired. The Cannon had a mass of 1600 kg. What was its recoil velocity ? Boom Ball is pushed this wayNewton’s 3 rd law tells us the cannon is pushed this way. Momentum is conserved, so... Mass of canon  velocity of cannon backwardsMass of ball  velocity of ball forwards = Mass of canon  velocity of cannon backwardsMass of ball  velocity of ball forwards = 1600 × v80 × 40 = Before the canon fires. Both canon and ball are stationary After the canon fires.

5. Question 2. The cannonball took 0.2 seconds to travel along the barrel what was the force on it ? Question 3. What was the force on the cannon ? Explain your reasoning Equal and opposite (Newton’s 3 rd Law) Question 4. What are the units of momentum ? Momentum = Mass × Velocity kgm/s Units of momentum are... kg m/s

6. Impulse We now go back to the basic formula linking Force, Mass, and Acceleration And to the definition of acceleration which is: This can be written: Substituting: But m  v is the change in momentum of an object, so we can say: or We call the quantity Force × Time (F  t) ‘Impulse’, so in words, the equation above says: Impulse = Change in Momentum

7. Using the formula Rewrite it as Then it can be used in two ways For situations where velocity changes but mass stays the same (collisions, explosions etc) it becomes For situations where mass changes but velocity stays the same rockets, jet engines) it becomes A final comparison with GCSE work. We know that Distance = velocity  Time, and that distance can be found from a velocity-time graph by calculating the area under the graph. By comparison we can see that change in momentum = Force × Time, so change in momentum can be found by calculating the area under the force-time graph Velocity Time Force Time Area = Distance Area = Change in momentum

8. Questions Page 7. 1. a. b. 2. a. 3. a.

9. 4 a.Force (N) Time (s) 400 20 40 b.

10. Impulse and Momentum Imagine a cannon: When the cannon fires it applies a force to the cannon ball of mass m B given by the equation: Boom But we have another formula for Acceleration : So by substituting for A we can say: mBmB or In Words....

11. It is now convenient to introduce a new quantity called ‘Momentum’. It is a measure of the mass and the velocity of an object so.. Now we can simplify the word equation above: Now remember Newton’s 3 rd Law. So if the canon exerts a force on the cannon ball,

12. The time for which this force is applied is the same so the cannon must undergo an equal and opposite What was the combined momentum of the cannon and the ball before the explosion ? Answer.... What was the combined momentum of the cannon and the ball after the explosion ? Answer.... Conclusion. This is best understood by looking at the diagram again.

13. Question 1. The cannon ball had a mass of 80 kg and velocity of 40 m/s after being fired. The Cannon had a mass of 1600 kg. What was its recoil velocity ? Boom Momentum is conserved, so... = Mass of canon  velocity of cannon backwardsMass of ball  velocity of ball forwards = Before the canon fires. After the canon fires.

14. Question 2. The cannonball took 0.2 seconds to travel along the barrel what was the force on it ? Question 3. What was the force on the cannon ? Explain your reasoning Question 4. What are the units of momentum ?

15. Impulse We now go back to the basic formula linking Force, Mass, and Acceleration And to the definition of acceleration which is: This can be written: Substituting: But m  v is the change in momentum of an object, so we can say: or We call the quantity Force × Time (F  t) ‘Impulse’, so in words, the equation above says:

16. Using the formula Rewrite it as Then it can be used in two ways For situations where velocity changes but mass stays the same (collisions, explosions etc) it becomes For situations where mass changes but velocity stays the same (rockets, jet engines) it becomes A final comparison with GCSE work. We know that Distance = velocity  Time, and that distance can be found from a velocity-time By comparison we can see that change in momentum = Force × Time, so change in momentum can be found by Velocity Time Force Time

17. Questions Page 7