Key Areas covered

1. Key Areas covered • Force-time graphs during contact of colliding objects. • Impulse found from the area under a force-time graph. • Equivalence of change in momentum and impulse. • Newton’s third law of motion.

2. What we will do today • State what is meant by impulse. • Investigate force-time graphs and use these to calculate impulse.

3. Impulse

4. An object is accelerated by a force, F, for a time, t. The unbalanced force is given by: Fun = ma = m(v- u) t = mv – mu t Unbalanced force = change in momentum time = rate of change of momentum

5. Impulse = change in momentum Impulse = force x time Impulse = Ft = mv – mu Units of Impulse are kgms-1 or Ns. Impulse is a vector quantity, so DIRECTION is important.

6. The concept of impulse is useful in situations where the force is not constant and acts for a very short period of time. An example of this is when a golf ball is hit by a club. During contact, the unbalanced force between the club and the ball varies with time (often ms) as shown in the graph opposite. F t 0

7. Making use of impulse • We can make use of impulse to help with safety features. • Re-arranging the equation to F = mv-mu / t shows that to decrease the force, you have to increase the time of contact. • This is why cars have a crumple zone and why helmets have padding inside. • These safety features increase the time of contact which decreases the average force applied (decreasing the damage on the human body).

8. This can be expressed in graph form. • Note the change in momentum does not change (mass, initial velocity and final velocity are all the same).

9. 2003 Qu: 5

10. Chevrolet Bel Air vs. 2009 Chevrolet Malibu crash test. - YouTube

11. Newton’s third law • For every action there is an equal and opposite reaction. • This holds true for impulse, the force created during contact between the two object (i.e. golf club and ball) is equal.

12. Bouncing Balls This effect can also be seen when using balls of different materials. A hard ball, such as a basket ball, deforms only a little.

13. Click on video to start. Click on green area to move to next slide.

14. A softer ball, such as a squash ball, deforms more and so the time of contact is longer.

15. Click on video to start. Click on green area to move to next slide.

16. Tiger Woods Slo Mo + Close up of Ball.mpg

17. In practical situations the force is not constant, but comes to a peak and then decreases. Impulse = Area under a Force-time graph In any collision involving impulse, the unbalanced force calculated is always the average force and the maximum force experienced would be greater than the calculated average value. Note: time is often given in ms, this must be converted to seconds (x10-3)

18. A snooker ball of mass 0.2 kg is accelerated from rest to a velocity of 2 ms-1 by a force from the cue which lasts for 50 ms. What size of force is exerted by the cue? Solution u = 0, v = 2 ms-1, m = 0.2 kg, t = 50 ms = 0.05 s, F = ? Ft = mv – mu F x 0.05 = (0.2 x 2) – 0 F = 0.4 / 0.05 F = 8N Example 1

19. A tennis ball of mass 100 g, initially at rest, is hit by a racket. The racket is in contact with the ball for 20 ms and the force of contact varies as shown in the graph. What was the speed of the ball as it left the racket? Solution Impulse = area under graph = ½ x (20 x 10-3) x 400 = 4 Ns Ft = mv – mu 4 = 0.1v – 0 v = 4 / 0.1 = 40 ms-1 F / N Example 2 400 t / ms 0 20

20. 2005 Qu: 5

21. 2007

22. 2008

23. 2003 Qu: 5

24. 2004 Qu: 5

25. Past Papers • 2003 Qu: 22 • 2006 Qu: 22 • 2009 Qu: 22(b) • 2010 Qu: 23 (a)(ii) & (iii)

26. Open-ended questionspecimen paper

27. Possible solution • Hard outer shell absorbs some impact from the road • Soft polystyrene foam liner increases contact time of head • Using equation F = mv – mu / t it is clear that to decrease the average force on the head, contact time should be increased • Decreasing the average force results in less damage to the head

28. Open ended question2012 Revised Higher

29. Possible solution • Compare hitting the ball with and without swinging ‘through the ball’ and using the equation Ft = mv – mu • In both cases the mass of the ball and the initial velocity are the same • If we take u = 0 then Ft = mv • Assume the racquet used is same in both cases and applies same force, F • Not going through the ball means a shorter contact time, t • If contact time is small then so is Ft and mv • As m is a constant then final velocity, v, is small • Going through increases contact time, t, increases Ft and as a result, increases v

30. Questions • Activity sheets: • Collisions and explosions • You should now be able to answer questions: 14 onwards in class jotter

31. Collisions and Explosions 14. 1·58 ms −1 15. 100 N 16. 3·0 × 10−2 s 17. 2·67 m s −1 18 (a) + 0·39 kg m s −1 if you have chosen upwards directions to be positive (b) + 0·39 N s if you have chosen upwards directions to be positive (c) 15·6 N downwards (d) 15·6 N upwards (e) 16·6 N upwards 19. (a) v before = 3·96 ms −1 downwards; v after = 2·97 ms −1 upwards (b) 9·9 × 10 −2 s 20.(a) Teacher Check (b) 0·2 s (c) 20 N upwards (or –20 N for the sign convention used in the graph) (d) 4·0 J 21. 1·25 × 103 N towards the wall 22. 9·0 × 104 N 23. Teacher Check 24. (a) (i) 4·0 m s −1 in the direction the 2·0 kg trolley was travelling (ii) 4·0 kg m s −1 in the direction the 2·0 kg trolley was travelling (iii) 4·0 kg m s −1in the opposite direction the 2·0 kg trolley was travelling (b) 8·0 N 25. Teacher Check 26. Teacher Check Answers