Chapter 8

1. Chapter 8 Momentum, impulse and yet another conservation principle

2. Learning Goals for Chapter 8 Looking forward at … the momentum of a particle, and how the net force acting on a particle causes its momentum to change. the circumstances under which the total momentum of a system of particles is constant (conserved). elastic, inelastic, and completelyinelastic collisions. what’s meant by the center of mass of a system, and what determines how the center of mass moves. how to analyze situations such as rocket propulsion in which the mass of a body changes as it moves.

3. Momentum and Newton’s second law The momentum of a particle is the product of its mass and its velocity: Newton’s second law can be written in terms of momentum:

4. Impulse The impulse of a force is the product of the force and the time interval during which it acts: On a graph of versus time, the impulse is equal to the area under the curve:

5. Impulse

6. Impulse

7. Impulse and momentum Recall that, in Latin, Newton’s second law actually reads: “The change of momentum of a body is proportional to the impulse impressed on the body.” Impulse–momentumtheorem: The change in momentum of a particle during a time interval is equal to the impulse of the net force acting on the particle during that interval:

8. Impulse and momentum • When you land after jumping upward, your momentum changes from a downward value to zero. • It’s best to land with your knees bent so that your legscan flex. • You then take a relatively long time to stop, and the force that the ground exerts on your legs is small. • If you land with your legs extended, you stop in a short time, the force on your legs is larger, and the possibility of injury is greater.

9. Rocketry motor classes illustrate the difference between impulse and thrust (force)

10. Rocketry motor classes illustrate the difference between impulse and thrust (force) For a given impulse...

11. Rocketry motor classes illustrate the difference between impulse and thrust (force) For a given impulse... ...there can be varying thrust

13. An example of a high-power rocket motor

14. An example of a high-power rocket motor In June 2019, the North Seattle College Rocketry Club will send a 27 kg rocket 10,000 feet in the air using an M2500 motor as part of the International Rocket Engineering Competition (IREC).

15. The “impulse curve” is a plot of force vs. time. Note that the area under the curve is the impulse the motor provides; in other words, the integral

16. This is the impulse curve for the M2500 motor, which shows that it averages about 2500 N of thrust over roughly 3.1 s, which is an impulse (the shaded area) of 7750 Ns, well within the 5120 to 10240 Ns range of M-class motors. This is what an M1850 motor looks and sounds like https://youtu.be/ZoiCritiFRI (start at 44:22)

17. From the impulse curve, you can estimate the initial acceleration of the rocket. According to the curve, the motor hits 2000 N of thrust within the first 0.1 s. Using F = ma, with the mass of rocket at 27 kg, the acceleration is about 74 m/s2 or nearly 7 ½ g. Humans would pass out under these accelerations! Of course, you don’t need to do these calculations, practically. With an Arduino microprocessor board (left), accelerometer board (right) and a lithium-ion battery (black case), you can measure the performance of your rocket with a package that fits in the palm of your hand.

18. Compare momentum and kinetic energy • The kinetic energy of a pitched baseball is equal to the work the pitcher does on it (force multiplied by the distance the ball moves during the throw). • The momentum of the ball is equal to the impulse the pitcher imparts to it (force multiplied by the time it took to bring the ball up to speed).

19. An isolated system Two astronauts push each other as they float freely in the zero-gravity environment of space. There are no external forces; when this is the case, we have an isolatedsystem.

20. Conservation of momentum External forces (the normal force and gravity) act on the skaters shown, but their vector sum is zero. Therefore the total momentum of the skaters is conserved. Conservationofmomentum: If the vector sum of the external forces on a system is zero, the total momentum of the system is constant.

21. Remember that momentum is a vector! When applying conservation of momentum, remember that momentum is a vector quantity! Use vector addition to add momenta, as shown at the right.

22. Elastic collisions: Before In an elasticcollision, the total kinetic energy of the system is the same after the collision as before.

23. Elastic collisions: During In an elasticcollision, the total kinetic energy of the system is the same after the collision as before.

24. Elastic collisions: After In an elasticcollision, the total kinetic energy of the system is the same after the collision as before.

25. Completely inelastic collisions: Before In an inelasticcollision, the total kinetic energy after the collision is less than before the collision.

26. Completely inelastic collisions: During In an inelasticcollision, the total kinetic energy after the collision is less than before the collision.

27. Completely inelastic collisions: After In an inelasticcollision, the total kinetic energy after the collision is less than before the collision.

28. Collisions In an inelasticcollision, the total kinetic energy after the collision is less than before the collision. A collision in which the bodies stick together is called a completelyinelasticcollision. In any collision in which the external forces can be neglected, the total momentum is conserved. In elastic collisions only, the total kinetic energy before equals the total kinetic energy after.

29. Inelastic collision example Cars are designed so that collisions are inelastic—the structure of the car absorbs as much of the energy of the collision as possible. This absorbed energy cannot be recovered, since it goes into a permanent deformation of the car.

30. Elastic collision example Billiard balls deform very little when they collide, and they quickly spring back from any deformation they do undergo. Hence the force of interaction between the balls is almost perfectly conservative, and the collision is almost perfectly elastic.

31. Elastic collisions in one dimension Let’s look at a one-dimensional elastic collision between two bodies A and B, in which all the velocities lie along the same line. We will concentrate on the particular case in which body B is at rest before the collision. Conservation of kinetic energy and momentum give the result for the final velocities of A and B:

32. Elastic collisions in one dimension When B is much more massive than A, then A reverses its velocity direction, and B hardly moves.

33. Elastic collisions in one dimension When B is much less massive than A, then A slows a little bit, while B picks up a velocity of about twice the original velocity of A.

34. Elastic collisions in one dimension When A and B have similar masses, then A stops after the collision and B moves with the original speed of A.

35. Center of mass We can restate the principle of conservation of momentum in a useful way by using the concept of center of mass. Suppose we have several particles with masses m1, m2, and so on. We define the center of mass of the system as the point at the position given by:

36. Center of mass of symmetrical objects

37. Center of mass of symmetrical objects

38. Motion of the center of mass The total momentum of a system is equal to the total mass times the velocity of the center of mass. The center of mass of the wrench at the right moves as though all the mass were concentrated there.

39. External forces and center-of-mass motion When a body or a collection of particles is acted on by external forces, the center of mass moves as though all the mass were concentrated at that point and it were acted on by a net force equal to the sum of the external forces on the system.

40. Example of conservation of momentum: rockets A rocket has three parts: Payload (science, human, life support, communications…) Structure (rocket, pumps, tanks, structural support…) Fuel (expelled for propulsion) We want a way to calculate the change in speed of the spacecraft (v = “delta-vee")due to firing a rocket. Momentum is conserved: total initial momentum equals total final momentum

41. Example of conservation of momentum: rockets initial momentum (Σpi) = final momentum (Σpf) So, for the situation where the rocket starts at rest but ends up moving So this shows the basic tradeoff for rockets. To go faster you must: Maximize fuel speed Maximize fuel mass Minimize craft mass

42. Conservation of momentum for rockets The equation on the previous slide seems to explain rockets. However, it’s more complicated than it appears. Rockets don’t work by throwing all of their fuel off at once. To get into orbit would kill an astronaut due to extreme accelerations! Each bit of fuel ejected is pushing against the payload and the remaining fuel. In other words, mfuel is continually changing.

43. Rocket equation To correctly compute the change in speed, we need to use calculus. The differential form of the rocket equation: Has the solution: or mfuel= fuel mass mcraft = spacecraft mass ve = exhaust speed v = change in spacecraft’s speed Posed and solved by the Russian rocketry pioneer KonstatinTsiolkovsky in 1903.

44. The graph of the rocket equation mfuel= fuel mass mcraft = spacecraft mass ve = exhaust speed v = change in spacecraft’s speed

45. The semi-logarithmic plot of the rocket equation In this graph, the y-axis values arelogarithmic. However, since the x-axis is still linear, the overall graph is called “semi-logarithmic”. It’s an effective way of compressing the y-axis.

46. Rocket propulsion As a rocket burns fuel, its mass decreases. To provide enough thrust to lift its payload into space, this AtlasV launch vehicle ejects more than 1000 kg of burned fuel per secondat speeds of nearly 4000 m/s.

47. The rocket equation: an example How much fuel do we require to send a 1000 kg payload to the Moon (v = 16 km/s) using a chemical rocket (ve = 4 km/s)?

48. The rocket equation: an example How much fuel do we require to send a 1000 kg payload to the Moon (v = 16 km/s) using a chemical rocket (ve = 4 km/s)?

49. The rocket equation: an example Which is nearly 54 times the (non-fuel) spacecraft mass! How much fuel do we require to send a 1000 kg payload to the Moon (v = 16 km/s) using a chemical rocket (ve = 4 km/s)?