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Non-conservation

By Maria Perry,2014-12-19 20:02
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Non-conservation

Episode 221: Elastic collisions

    This episode extends the idea of conservation of momentum to elastic collisions, in which, because KE is conserved, useful information can also be found by calculating the changes in KE of the colliding objects.

Summary

    Demonstration + Discussion: To introduce totally elastic collisions. (20 minutes) Student experiment: To test conservation of momentum and kinetic energy in an elastic collision. (20 minutes)

    Worked examples + student questions: Calculations of final velocities in elastic collisions and loss of KE in inelastic collisions. (20 minutes):

    Discussion: More abstract problems and situations which commonly cause difficulties. (15 minutes)

    Demonstrations: Showing some applications such as catching a ball or finding the speed of an air rifle pellet. (15 minutes)

Demonstration + Discussion:

    To introduce totally elastic collisions

    This follows a similar approach to the demonstration in Episode 220.

TAP 220-1: Observing collisions

    Demonstrate „springy‟ (elastic) collisions using trolleys, one of which has its spring-load released so its spring can “soften the collisions” (Alternatively, use air-track gliders with repelling magnets

    attached.)

    Direct a single trolley at a second, stationary trolley. The first trolley stops, the second moves off at the speed of the first. Momentum is conserved.

    Now try a light trolley colliding with a heavy one, and vice versa. What pattern is seen? A light trolley bounces back from a heavier one (its momentum is negative); a heavier one moves on, but at a slower speed.

    Students may accept that momentum is conserved; alternatively, with suitable light gates you should be able to measure the initial speed of the one trolley and the final speeds of both. Momentum conservation can be shown. (Or try using a computer and motion sensor.) Now ask: How do the trolleys know at what speed they must move? There are many combinations of velocity which conserve momentum; there must be something else going on here. Introduce the idea that kinetic energy (KE) is also involved, and that, in a springy collision, there is as much KE after as before; in other words, KE is conserved.

    Ask whether KE is conserved in an explosion (obviously not; it is “created” in the explosion in the change from chemical PE to KE) and in an inelastic collision (the total amount decreases, but you may need to work through a numerical example to show this).

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    Discuss where KE comes from in an explosion (from energy stored in a squashed spring, chemical explosive or whatever), and where it goes to in an inelastic collision (work is done in deforming material leads to heating; some sound).

    Terminology: Usually, „elastic‟ is taken to imply that KE is conserved. In some texts, this is written as „perfectly elastic‟. „Inelastic‟ describes a collision in which some KE is lost. Students should learn to use these terms, rather than „springy‟ and „sticky‟.

Student experiment:

    To test conservation of momentum and kinetic energy in an elastic collision Ask students to carry out an experiment to determine whether, in a springy collision between trolleys or gliders, momentum and KE are both truly conserved.

    They can use the same approach as the experiment in Episode 220, but they will have to calculate values of KE as well as momentum.

Worked examples + student questions:

    Calculations of final velocities in elastic collisions and loss of KE in inelastic collisions

    Work through examples involving elastic collisions to show:

     Conservation of KE and momentum in an elastic collision, when all values of mass and

    velocity are known.

     Calculation of final velocities in an elastic collision.

    Note that the second type requires solution of simultaneous equations, one of which is a quadratic; students may find this difficult, so check that it is required by the specification that you are following.

     Non-conservation of KE in an inelastic collision.

    Students can now work through more examples.

TAP 221-1: Worked examples in momentum: elastic and inelastic

    collisions

TAP 221-2: Student Questions

Discussion:

    More abstract problems and situations which

    commonly cause difficulties

    Reinforce students‟ ideas by discussing some of the following:

    How a rocket ship works, as a controlled explosion in which reaction

    mass travels backwards. The rocket needs nothing to lift off except the

    expended fuel.

    There is a video of this with a man sitting on a cart firing a fire

    extinguisher at

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http://www.wfu.edu/Academic-departments/Physics/demolabs/demos/ (1N22.10 [M-21 W] Fire

    Extinguisher Wagon )

    The same site shows another video of two people on roller-blades throwing a cushion back and forth.

    Discuss situations in which the Earth is involved, as it may appear that momentum is not conserved. Where does momentum come from or go to in these situations? It helps to think about the forces involved.

     You push a car to start it moving. (Your feet push back on the Earth, so that its momentum

    also changes, in the opposite direction. This is equivalent to an explosion.)

     When a ball falls, it accelerates, i.e. it gains momentum. (The Earth is also accelerated

    minutely in the opposite direction, so momentum is conserved. The force is gravity.)

     When a ball bounces off a wall, its momentum is reversed. (Momentum is transferred to the

    wall + Earth by the contact force.)

     When a ball rolls to a halt, it loses momentum. (Its momentum is transferred to the Earth via

    friction).

    These all emphasise the need to think of the closed system with which we are concerned.

    Emphasise that momentum is always conserved, but KE is not. One way to think of this is that KE is just one form of energy, so it can be transformed; there is only one form of momentum, so it cannot be transformed into anything else.

Demonstrations:

    Showing some applications such as catching a ball or finding the speed of an air rifle pellet

TAP 221-3: Momentum demonstrations

    The following is a brief summary of the Resourceful Physics activities given in the link in TAP 221-3:

    RP13: Pendulum on a trolley to show conservation of momentum. This could be filmed with a

    webcam and the film studied in slow motion to great effect. There is a link here between the motion of the trolley and the motion of a rowing boat, which shoots forwards when the rowers move back on their slides, rather than when they pull on their oars.

    RP18: Momentum in catching compare to deforming barriers and elastic ones. This contrasts the momentum change on catching something and the greater change on reflecting it back. RP 10: If you have access to an air rifle demonstration, this is a lovely way to round off the topic. RP 24: Speed of a cricket ball can be used in place of the air rifle experiment

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    TAP 221-1: Worked examples in momentum: elastic and inelastic collisions

    1. When an object of mass M and velocity u collides head-on elastically with a stationary

    object of mass m then mass m moves off with velocity v given by: - b

    2M vubMm

    and the mass M changes speed from u to vwhere v is given by:- a a

    Mm vuaMm

    a) A golf club head has a mass very much larger than that of the ball. Use the above formula

    to show the golf ball will be driven from the tee at about twice the speed of the golf club

    head.

    b) An alpha particle makes a head on collision with a stationary helium atom. Show the alpha

    particle will stop and the helium atom will move off with the original speed of the alpha

    particle. (Assume an alpha particle and helium nucleus have the same mass)

    c) A fast moving electron collides with a stationary heavy atom. Show the electron bounces

    back with nearly the same speed

    -12. A cricket ball, mass 0.5 kg, was bowled at 50 m s at a batsman who misreads the ball and -1the 5 kg bat is knocked out of his hands, the ball rebounds at 25 m s.

    a) What is the initial momentum of the ball?

    b) What is the momentum of the ball just after it hit the bat?

    c) How much momentum is transferred to the bat?

    d) Calculate the velocity of the bat as it leaves the batsman.

    e) Calculate the kinetic energies before and after the collision of bat and ball. Is the collision

    elastic or inelastic?

3. A ball of mass 0.20 kg is dropped from a height of 3.2 m onto a flat surface which it hits at -1-28.0 m s. It rebounds to 1.8 m. (g = 9.8 m s)

    a) What is the rebound speed just after impact?

    b) What is the change in energy of the ball?

    c) What momentum change has the ball between just touching the surface and leaving it?

4. In October 1999 the United Nations announced that the world‟s population had reached

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    96 billion (6 x 10) and that the population of the People‟s Republic of China was rapidly

    approaching 1.3 billion. Imagine a situation in which the entire population of China were gathered together and, all at the same time, they each jumped as high as they were able. Estimate the speed that would be (temporarily) imparted to the Earth as a result of this jump. Show all your calculations and reasoning and state clearly any assumptions you 24have made. Take the mass of the Earth to be 6 x 10 kg.

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    Answers and Worked Solutions

    2M1. a) The ball has a much smaller mass than the club head so it can be ignored in vubMm

    M2so and. So the speed is twice that of the club head. As the ball does v2uvubbM

    not have zero mass the ball speed will be slightly under 2u

    Mm b) The alpha particle speed after the collision is given by as M = m the speed vuaMm

    2Mwill be zero. The helium nucleus will move off with speed given by and as M vubMm

    2M= m then so . Therefore the alpha particle stops and the helium vuvubb2M

    nucleus moves off with the same speed.

    MmM c) m can be ignored so and vuvuvuaaaMmM

    The larger mass will hardly move

    -12. a) Initial momentum of the ball = 0.5 x 50 = 25 kg m s

    -1b) Momentum of the ball just after it hit the bat = 0.5 x (-25) = -12.5 kg m s (Taking positive

    velocity to the right.)

    -1 c) Momentum transferred to the bat =25 - (-12.5) = 37.5 kg m s

    -1-1 d) Momentum of bat = 37.5 kg m s so velocity of bat = (37.5)/5 = 7.5 m s

     e) Calculate the kinetic energies before and after the collision of bat and ball. Is the collision

    elastic or inelastic?

    1122 so before collision and KE = 625 J KEmvKE0.5(50)22

    12After the collision KE of ball is so KE = 156.25 J = 156 J to 3sf KE0.5(25)2

    12After collision Bat KE = so KE = 140.63 J = 141 J to 3sf KE5(7.5)2

Total KE after collision is 156.25 + 140.63 =296.9 J = 297 J to 3sf so the collision is inelastic

    as KE before > KE after the collision

    -13. a) This is an energy reminder so and speed = 5.9 m s so v2ghv2x(9.8)x(1.8)

    -1velocity = - 5.9 m s

    12 b) KEmv so before impact KE = 0.5 x 0.20 x 8.0 x 8.0 = 6.4 J 2

    after impact KE = 0.5 x 0.20 x (-5.9) x (-5.9) = 3.5 J so KE change is 6.4 - 3.5 = .2.9J

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    -1 c) Momentum before collision = 0.2 x 8.0 = 1.6 kg m s

     Momentum of ball after = 0.2 x (-5.9) = -1.2

    -1Change in momentum = 1.6 - (-1.2) = 2.8 kg m s

    4. This question shows the effect of many small masses on a very much larger mass

    Estimate average mass of person m = 50 kg (NB Population will contain many children)

    -1Jump height _h = 0.2 m Take-off speed so = 2 m s (accept any v2ghv2x10x0.2-1reasonable method or blind guess of about 1 to 3 m s)

    Total upward momentum of all jumpers:

    9-111-1p =1.3 x 10 x 50 kg x 2 m s = 1.3 x 10 kg m s

    To conserve momentum, Earth must acquire equal momentum in the opposite direction.

    1124-14-1Earth, mass M, acquires speed v = p/M = 1 x 3 x 10 / 6 x 10 = !;2;x 10 m s i.e. velocity

    of the Earth is negligible. An elastic or inelastic jump?

External References

    Question 1 was an adaptation of question 55(L) from Revised Nuffield Advanced Physics Unit A Question 3 was an adaptation of question 50(I) from Revised Nuffield Advanced Physics Unit A Question 4 is taken from Salters Horners Advanced Physics, Section TRA, Additional sheet 8 & 9

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TAP 221- 2: Student Questions

Elastic Collision Questions

    -11. A trolley of mass 1 kg rolls along a level, frictionless ramp at a speed of 6 m s. It collides

    with a second trolley of mass 2 kg which is initially at rest. The first trolley rebounds at a -1speed of 2 m s.

     a) Find, by conservation of momentum, the velocity of the second trolley after the collision.

     b) Compare the kinetic energy before and after the collision. Is the collision elastic?

    2. A bowling ball of mass 5 kg strikes a skittle of mass 1 kg. The bowling ball is travelling at -13 m s before the collision, which is elastic. Find the velocity of the ball and the skittle after

    the collision. (NB in this question ignore the rolling motion of the ball assume it is sliding).

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    Practical Advice

    These are quite advanced and may not be appropriate for all specifications.

    Answers and Worked Solutions

    1. a) Let us use the initial direction of the first trolley as the positive direction. Then momentum -1before = mu = 1 × 6 = 6 kg m s. Afterwards let the velocities be v and v and the masses 1112

    are still m and m. Then: 12

    6 = mv+mv1122

    6 = 1 x (-2) + 2 x v 2

    -1v= 4 m s (Note the care with the signs) 2

    22 b) KE before = ?m = 1/2 ×1 × 6 = 18 J u11

    22 2 2KE after = ?mv+ ?m v = ? × 1 × (-2) + ? ×2 × (4)1122

    = 2 + 16 = 18 J

    So KE is conserved and the collision is elastic.

    2. Conservation of momentum: mu = mv + mv where the subscripts s and b refer to the bbbbss

    skittle and the ball. This gives:

    -115 kg m s = 5v + v bs

    222 Conservation of KE: ?mu = ? mv + ? mvbbbbss

    222? × 5 × (3) = ? × 5 × v + ? × 1 × v bs

    2222.5 J = 2.5v + 0.5v bs

    To solve this we use the first equation to eliminate v, v= (15 - 5v), and substitute in the ss b

    second equation:

    2 222.5 = 2.5 v + 0.5 (15 - 5v) bb

    We‟ll multiply through by 0.4 to keep the numbers easy:

     229 = v + 5(3 - v) bibe

    229 = v + 45 + 5v - 30vbbb

    2 6v- 30v + 36 = 0 bb

    2 v- 5v + 6 = 0 bb

    (v - 3)(v -2) = 0 bb

    -1-1v = 2 m s (or 3 m s but that was the original value) b

    -1Substituting back in the first equation, v = 1 m s s

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TAP 221- 3: Momentum demonstrations

13. Pendulum on a trolley - momentum conservation

    Hang a pendulum on a dynamics trolley or on the rider of a linear air track and observe its motion as the trolley moves along.

Apparatus required:

    ; Linear air track

    ; Pendulum supported on an air-track rider

18. Momentum in catching

    This simple demonstration emphasises the vector nature of momentum. Use a ball (soccer or netball) and throw it to a pupil. Get them first to catch it and throw it back and then to punch it backwards without catching it first. Ask students which require the biggest force - catching or punching it back. It should be clear that it is second and that this gives the biggest change of momentum. This demonstration hopefully emphasises the velocity change from u to -u and so a greater change of momentum than simply stopping the ball.

Safety

    Consider whether or not this activity can be done safely in your lab with your pupils.

Theory:

    Momentum change = mu - (-mu) = 2mu when it is punched back [NB this fact will be used when calculating the pressure due to a gas using the kinetic theory of gases.]

    and momentum change = mu when it is caught

Apparatus required:

    ; Netball or football

10. The air rifle and momentum

    How safe are these two experiments? The writer believes that if done with care and due regard to everyone‟s safety they are fine! So here they are.

    (a) Fix two timing gates - simply a plastic frame with a strip of aluminium foil across the centre

    - a metre apart and connected to a scalar. Fire an air rifle pellet through the two gates so

    that it starts the timer when the first strip of aluminium foil is broken and stops it when the

    second is broken. The time for the pellet to travel the one metre between the two pieces of

    foil is given directly and the speed of the bullet can easily be worked out Safety - the rifle should be clamped to a baseboard and a means of catching the pellet should be found - box of polystyrene backed with a wooden board is suitable for this. Use a safety screen.

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