Conservation of Momentum. Last modified: 08/05/2018

Transcription

1 Conservation of Momentum Last modified: 08/05/2018

2 Links Momentum & Impulse Momentum Impulse Conservation of Momentum Example 1: 2 Blocks Initial Momentum is Not Enough Example 2: Blocks Sticking Together What Do Before and After Mean Exactly? 2 and 3 Dimensions Example 3: Collision in 2D Elastic Collisions Example 4: Are Previous Examples Elastic? Example 5: Elastic Collision Example 6: Collision with Known Loss of KE Not Just Collisions Momentum Recipe Rocket Equation (*) Conservation Laws & Symmetry (*) Starred (*) sections will not be included in exam.

3 Momentum Contents We have already seen the definition of the momentum vector p for an object of mass m moving with velocity v: p = mv v m p and the definition of force in terms of the change in momentum (Newton s Second Law): F = p t By looking at the forces acting on an object, we have been able to determine the motion of that object. This was relatively easy in the cases we have looked at, as the forces, and hence the accelerations, were constant in time. This method becomes much more difficult if this is not the case.

4 Impulse Consider a rubber ball bouncing off a wall - shown in slow motion in the diagram at right. We expect the force on the ball to vary quite dramatically over the duration of the collision as the ball is deformed. If we take the moment of first contact of the ball with the wall to be t = 0 and the duration of the contact with the wall to be t, then we can calculate the total effect of the force to be: t t=0 F(t)dt = t t=0 dp dt = p dt This integral (equal to the change in momentum) is called the impulse of the wall on the ball. Actually calculating this integral requires a detailed knowledge of how the force changes with time. Knowledge that we will rarely have. p A F F F p A + p A F F Contents t = 0 time t = t

5 Let s now consider the collision between two balls A and B, shown at right. Balls A and B experience forces F A and F B respectively. The magnitudes of these forces will vary with time. p A F A F B p B t = 0 Calculating the impulse of each force gives: and t t=0 t t=0 F A (t)dt = p A F B (t)dt = p B But from Newton s Third Law, we know that F A = F B and thus that: p A = p B F A F A F A F A p A + p A F B F B F B F B p B + p B t = t

6 Conservation of Momentum Contents If we now consider the two balls to form a system (as with energy, the system is an imaginary box used for calculations), then we can determine the total momentum in this system before and after the collision: Before Collision After Collision p A p B p A + p p B p p before = p A + p B p after = (p A + p) + (p B p) = p A + p B Momentum of the system is conserved (i.e. p before = p after ).

7 This simple example involved only two objects, but the conservation law will hold for a system containing any number from two up to many billions of objects. Momentum conservation is used in a similar way to energy conservation to determine (usually) unknown velocities. Remember the conservation of energy equation: E start + W external forces = E end If there are forces external to the system, then momentum conservation equation is similar: p start + I external forces = p end where I is the impulse of the external forces. This impulse is generally difficult to calculate, so choosing the right system, in order to eliminate the effect of these forces is very important (in the previous example, the system is taken to be the two objects, rather than one).

8 A block of mass m = 1 kg travels at u = 10 m/s, before colliding with a stationary block of mass M = 5 kg. After this collision, the block m has a speed of 1 m/s in the original direction. Calculate the speed V of the 5 kg block after the collision. The appropriate system in this case contains both of the blocks. Draw diagrams showing the before and after positions: Before Collision After Collision u = 10 m/s m = 1 kg M = 5 kg v = 1 m/s m V M p = mu + M0 before = (1)(10) = 10 p = mv + MV after = (1)(1) + 5V = 1 + 5V

9 Next, use Conservation of Momentum to determine the speed V : p = p before after 10 = 1 + 5V V = 1.8 m/s Notes: We have been a bit lazy with vectors in this calculation. Remember momentum is a vector, so p before = 10i, etc. Beause all of the motion is in the same direction in this case, this was OK, but this won t always be true. To successfully solve this problem, we needed to be told something about the motion after the collision (i.e. v = 1 m/s). This will always be the case.

10 Initial Momentum is Not Enough Contents Note that in this example, we needed to be given one of the final velocities. Collisions with the same initial velocities can have very different outcomes. For instance, if we have two equal masses colliding as below, any of the situations shown at right will conserve momentum ( p = 0 in each case) and are therefore possible. 10 m/s 10 m/s m m 10 m/s 7 m/s 5 m/s 10 m/s 7 m/s 5 m/s There will be literally millions of possible outcomes for the same initial momenta. We don t even require the same number of objects afterward: one or both of the masses could break into pieces, or they might stick together. Without some additional information about the collision, we cannot calculate the final outcome.

11 A block of mass m = 1 kg travels at u = 10 m/s, before colliding with a stationary block of mass M = 5 kg. After this collision, the two blocks are stuck together. Calculate the speed V of the combined masses. The method is exactly the same as used in the previous example: u = 10 m/s m = 1 kg M = 5 kg Before V m + M After p = mu + M0 before = (1)(10) = 10 p = (m + M)V after = (6)V Cons. of momentum: p = p before after 10 = 6V V = 5 3 = 1.67 m/s

12 What Do Before and After Mean? Contents What if the 5 kg mass in the previous example was hanging by a rope before the collision, and we are asked to calculate how high the combined mass will swing after the collision, before stopping? We might at first think that the before and after positions should be those at right. We then have p before = 10 and p after = 0. Momentum is not conserved!! Why? 10 m/s 1 kg 5 kg Before h After 6 kg There is an external force (gravity) acting in the time interval between the two points. The impulse of this force causes a change in momentum, which is not easy to calculate. The correct approach is to break this into two sub-problems.

13 The first problem is a momentum problem. One that we have already solved! 10 m/s 1 kg 5 kg Before 1.67 m/s 6 kg After The second is an energy problem. E start = E end 1 2 mv 2 = 1 2 6(1.67)2 = mgh = 60h h = 0.14 m 1.67 m/s 6 kg Start h End 6 kg In momentum calculations, the Before and After positions should always be chosen to be immediately before and immediately after the collision.

14 Momentum in 2D and 3D Contents The previous examples have shown collisions where all motion before and after the collision is in the same direction. Of course the real world is three-dimensional and so in general, a collision will involve two or even three-dimensional momentum vectors. This does not change the logic of the problem at all, but as always, vectors make the maths a little more complicated. As we ve seen previously (with projectiles and forces), it can make life easier if we consider each direction separately. So, for example in two dimensions, we could use one equation with momentum vectors: p before p x = after p x before p = after or two equations using vector components: and before p y = after p y

15 A block of mass m = 1 kg travels at u = 10 m/s, before colliding with a stationary block of mass M = 5 kg. After the collision the velocity of the 1 kg block is 5 m/s at 37 to its original direction. Calculate the speed V of the 5 kg block immediately after the collision. First we must draw diagrams showing velocities before and after: v = 5 m/s u = 10 m/s m = 1 kg M = 5 kg Before m 37 θ M After V

16 Method 1: Using full vectors Before u = 10i m/s m = 1 kg M = 5 kg p = mu + M0 before = (1)(10i) = 10i Cons. of momentum: p = p before after 10i = 4i + 3j + 5V After v = 4i + 3j m/s 3 m 4 p = mv + MV 37 θ M V after = (1)(4i + 3j) + (5)V = 4i + 3j + 5V V = 1 (6i 3j) 5 = 1.2i 0.6j m/s so the speed is: V = (1.2) 2 + (0.6) 2 = 1.34 m/s We can also calculate the angle θ = tan 1 ( ) = 26.6.

17 Method 2: Using components Before u = 10i m/s m = 1 kg M = 5 kg After v = 4i + 3j m/s m 37 V = V xi + V y j 4 θ M 3 p x = mu x + M0 before = (1)(10) = 10 p y = mu y + M0 before = 0 p x = mv x + MV x after = (1)(4) + (5)V x p y = mv y + MV y after = (1)(3) + (5)V y p x p x = before after 10 = 4 + 5V x V x = 6 5 = 1.2 p y p y = before after 0 = 3 + 5V y V y = 3 5 = 0.6 V = 1.2i 0.6j m/s Which is the same answer found using Method 1. You can use which ever method you prefer when solving this type of problem.

18 Elastic Collisions Contents A collision in which kinetic energy is also conserved is called an elastic collision. Consider the previous example of two equal mass objects colliding. 10 m/s 10 m/s m m 10 m/s 10 m/s KE = 100m 7 m/s 7 m/s KE = 49m KE = 1 2 m m102 = 100m 5 m/s KE = 25m 5 m/s We saw previously that, among other possibilities, all of the final states at right conserve momentum. KE = 0 But calculations show that only the top case is elastic.

19 Most collisions are NOT elastic (i.e. are inelastic). In a collision, total energy must be conserved, but kinetic energy being conserved means that there is no conversion between different forms of energy. i.e. no work is done when the objects collide. Considering a real-world collision to be elastic is usually an approximation, similar to assuming no friction in a forces problem. You should be very careful about assuming that a collision is elastic. Sometimes this can be a very useful approximation. For example, an ideal gas is one in which the collisions between the gas molecules are elastic. As you may know from Chemistry, for gases at low temperatures and pressures this is very accurate.

20 Are the collisions seen in the previous examples elastic or inelastic? All three examples have the same before diagram, and so the same initial kinetic energy: KE = 1 2 (1) (5)02 = 50 J before Before 10 m/s 1 kg 5 kg Example 1 After 1 m/s 1.8 m/s 1 kg 5 kg In Example 1, we found the final speeds shown at left, and can calculate the kinetic energy at this moment to be: KE = 1 2 (1) (5)(1.8)2 = 8.6 J after KE after < KE before, which indicates that some of the initial kinetic energy has been converted to other forms of energy - heat, sound etc. The collision is inelastic.

21 Example 2 After 5 3 m/s 6 kg In Example 2, the KE is again less after the collision: KE = 1 ( ) (6) = 8.33 J after and so this collision is also inelastic. In this example, we could have predicted this result and so not needed to calculate the kinetic energy. How? If the blocks are stuck together, then clearly there must be forces acting between them. These forces will do work, changing the kinetic energy. Similarly if one of the blocks had broken into pieces, forces will have acted, and work has been done. If the objects in a collision change shape in any way, the collision cannot be elastic. No calculation is necessary to demonstrate this.

22 In Example 3 (like Example 1), the blocks have not changed shape in the collision, so it could possibly be an elastic collision. The only way to determine if such a collision is elastic, is to calculate the kinetic energies before and after and then compare them. Example 3 After For Example 3: 1 kg 5 m/s KE = 1 2 (1) (5)(1.34)2 = 17 J after 5 kg 1.34 m/s Again, KE after < KE before, so this collision is also inelastic.

23 A block of mass m = 1 kg travels at u = 10 m/s, before colliding with a stationary block of mass M = 5 kg. Given that the collision is elastic, calculate the speeds v and V of the two blocks immediately after the collision. 10 m/s 1 kg 5 kg Before v V m M Conservation of momentum: p = p before after mu + M 0 = mv + MV 10 = v + 5V v = 10 5V After

24 Collision is elastic: KE KE = before after 1 2 mu2 = 1 2 mv MV 2 [m=1 kg, u=10 m/s, M=5 kg] 100 = v 2 + 5V 2 [v=10 5V ] 100 = (10 5V ) 2 + 5V 2 0 = 10 0V + 3 0V 2 = V (3V 10) V = 0 or 10 3 and v = 10 or 20 3 Two mathematical solutions, but only one is physically correct: 10 m/s m M Solution m/s m/s 3 3 m M Solution 2

25 As we see in this example, an incoming object colliding elastically with a second, stationary object will lose kinetic energy, while the stationary object gains energy. u m M 1 KE after KE before = v 2 u 2 Before v V 0 10m 20m 30m 40m 50m m = M M m M After 1 v u The general collision, above left, can be solved for the final velocities, and the graph, above right, shows a plot of the kinetic energy and velocity of block m as ratios to their before values, for varying mass M.

26 From this plot, we can see that: Block m loses more energy when m M, and all of its energy when m = M When M m, then energy loss is small, and block m will bounce backward with almost the same speed as it started with. An elastic collision has no loss of kinetic energy, but more generally, if know the exact amount of kinetic energy lost in a collision, then we can again form two equations: mv mv = before after 1 2 mv 2 = 1 2 mv 2 + KE lost after before which we can solve for the unknown speeds.

27 A block of mass m = 1 kg travels at u = 10 m/s, before colliding with a stationary block of mass M = 5 kg. If 30% of kinetic energy is lost in the collision, calculate the speeds v and V of the two blocks immediately after the collision. 10 m/s 1 kg 5 kg Before v V m M After Conservation of momentum gives the same equation as the previous example: before p = after p mu + M 0 = mv + MV 10 = v + 5V v = 10 5V

28 30% of kinetic energy is lost: [m=1 kg, u=10 m/s, M=5 kg] KE = KE + ( 3 ) 10 before after ( )( 2 mu2) = 1 2 mv MV 2 ( = v 10) 2 + 5V 2 [v=10 5V ] 70 = (10 5V ) 2 + 5V 2 V = 1 3 before 0 = V + 3 0V 2 0 = (3V 1)(V 3) or 3 and v = 25 3 or 5 Again, two mathematical solutions exist, only one is physically correct: KE Contents m/s 3 m/s m M Solution 1 5 m/s 3 m/s m M Solution 2

29 Not Just Collisions Contents So far, we have only considered the conservation of momentum in collisions, but it can be applied more broadly to other interactions of objects. For example, consider a student standing on a stationary skateboard, holding a basketball of mass m. If the student throws the ball horizontally, at speed v, then the skateboard will move away in the opposite direction to the ball, at speed V. Conservation of momentum allows us to connect the two speeds v and V. V v p before = p after 0 = mv MV Before After V = mv M where M is the combined mass of the student and skateboard.

30 Momentum Problems Recipe Contents Read the question carefully, and draw two diagrams - one for the situation immediately before the collision and the other immediately after Write down expressions for the total momentum at each of these points and use conservation of momentum (possibly in 2D or even 3D): p before p = after IF you are given information about the kinetic energy lost in the collision, then write down the extra equation: KE + KE lost ( elastic KE lost = 0) before KE = after Solve these equations.

31 Rocket Equation (non-examinable) Contents A rocket engine works by ejecting hot gas backward at a speed v e relative to the engine. Consider a spaceship immediately before and after the ejection of one molecule of this gas: v m Before dm v e v }{{} relative to ground v + dv m + dm After Applying conservation of momentum: p before = p after mv = (m + dm)(v + dv) + ( dm)(v v e ) mv = mv + mdv + vdm + dmdv vdm + v e dm dv v e = dm m

32 If the spaceship has an initial mass m i and speed v i and finishes with mass m f and speed v f, then: vf v i dv v e = mf m i dm m 1 (v f v i ) = ln m f + ln m i = ln v e ( ) mi v = v e ln m f ( mi This equation is known as the rocket equation and allows the calculation of the mass of fuel required by a rocket to reach a particular speed. The thrust force can be calculated using Newton s Second Law: F = ma = m dv dt = m dt v dm e m = v dm e dt m f )

33 A spaceship has a rocket engine which ejects particles at a speed of 5 km/s, relative to the engine. Starting from, what speed will the spaceship reach after half its initial mass has been burned as fuel, at the rate of 2 kg/s? What is the thrust force? Using the rocket equation: ( ) mi v f = v = v e ln = ln m f ( m i 1 2 m i ) = m/s And to find the thrust force: dm F = v e dt = = N = 10 kn

34 Noether s Theorem Contents As we ve seen previously with Energy and now with Momentum, knowing that a quantity is conserved can make calculations much easier, especially with systems of multiple objects. Only some physical quantites are conserved - force for example is not. This raises a couple of related questions: What is special about Energy and Momentum? Why are they conserved? Are there any other quantites that are conserved, and so can be similarly useful? These questions are answered (very mathematically) by what is called Noether s Theorem: For every continuous symmetry of a set of equations, there will be a conserved quantity.

35 Continuous Symmetry Contents A symmetry which is only true for some specific values is called a discrete symmetry. Consider rotating a square about its centre: Rotation through 90 Rotation through 45 Square looks the same Square looks different The square will look the same after the rotation only for some rotation angles: 90, Rotation is a discrete symmetry of the square. A circle on the other hand, can be rotated about its centre by any angle at all and will look identical. Rotation is a continuous symmetry of the circle. Rotation through any angle Circle looks the same

36 Symmetries of Nature Contents The universe (and so the Laws of Physics) have some very basic continuous symmetries: 1. Time Translation Invariance (t t + t for any t) This is a technical way of saying that we can choose our zero of time to be anywhere we like - the equations don t change. The conserved quantity associated with this symmetry is: ENERGY 2. Space Translation Invariance (x x + x for any x) Similarly, this means that we can take the origin of the x-axis (also y and z) to be anywhere, without changing our equations. The conserved quantity is: MOMENTUM (Note a 3D vector, because there are three spatial directions) 3. Rotation Invariance (axes can be rotated by any angle) We can choose the directions of our x,y,z axes to be whatever we like (as long as they remain perpendicular to each other!). In this case, the conserved quantity is: ANGULAR MOMENTUM