Mechanics Tutorial 12 - Conservation
This is quite a long tutorial. You may find it easier to work through it in several sessions.
There is an extension after Question 10 which is challenging, suited to the A* or Pre-U student.
We have seen how momentum is the product between mass and velocity. We cannot see a momentum. There is no school or college (or university) that has a class set of momenta (plural of momentum). But it is a concept that is essential in explaining what happens in collisions and explosions.
This has important implications in the study of collisions. In simple terms, we can say that the total momentum before = total momentum after. The key thing is that share of the momentum may change.
The system may consist of several elements, each of which has its own momentum. The total momentum is the sum of all these different momenta. As long as the total momentum remains the same, momentum can be shared out differently between each element. In other words, each body can exchange momentum with the other bodies, and get a different share, as long as the total momentum stays the same.
If one element hogs most of the momentum, the others won’t have much. We can model a closed system using the dodgems in a fairground.
Photo by Andrew Dunn. Wikimedia Commons.
Each car has the same mass, but we can imagine them having different velocities. Remember that velocity has a value (the speed) and a direction. In the diagram below, we can see that the velocities are all random, but the sum of all the momenta is zero.
If it were not, the change in momentum would result in an overall force, resulting in movement (Newton II).
A few moments later, we might see:
You can see that all the different cars have different velocities, hence different momenta. But the overall momentum remains the same. The difference in momentum is zero. This arises because forces act in pairs (Newton III).
We will only consider the momentum of two objects, acting in a straight line. This has important implications in the study of collisions. Remember:
The total momentum of a system remains constant provided that no external forces act on the system.
What do you understand by the statement above?
Collisions in 1 dimension
Momentum is always conserved in collisions.
objects bounce off each other, the collision is elastic.
If the total kinetic energy is the same (conserved)
at the end as it is at the start, then the collision is perfectly elastic.
The rebound of particles against each other tends to be perfectly
elastic. A tennis ball bouncing
off the floor is not perfectly elastic as it can lose up to 25 % of its kinetic
energy in doing so.
If some kinetic energy is lost, converted into heat or light, then the collision is inelastic.
Think about two objects travelling in the same direction. The table below shows the properties of the objects:
We can show this as a diagram:
are two important principles here:
1. Conservation of momentum:
momentum before = total momentum after
+ mu2 = Mv1 + mv2
Energy is conserved
energy before = total energy after
E is the energy that is lost
in the collision. In a perfectly
E = 0.
When doing momentum calculations, always be careful about the directions you are using.
The diagram shows two cars at a fairground, before and after bumping into each other. One car and driver has a total mass of 500 kg, while the other car and driver has a total mass of 400 kg.
(a) What is (i) the total kinetic energy before the collision;
(ii) the total kinetic energy after the collision.
(iii) the total loss in kinetic energy.
(b) Is this an elastic collision? Explain your answer.
A second collision is shown below:
What is the speed of the 500 kg car after the collision?
A bullet of mass 45 g is travelling horizontally at 400 m/s when it strikes a wooden block of mass 16 kg suspended on a string so that it can swing freely. The bullet is embedded in the block.
a) The velocity at which the block begins to swing;
b) The height to which the block rises above its initial position;
c) How much of the bullet’s kinetic energy is converted to internal energy.
This question is a typical question you will find in the exam. It is often called synoptic, because it tests several different concepts at the same time.
An explosion not just where something goes BANG. For all those who are disappointed with this, here is an explosion:
Explosions in physics are much tamer affairs. The following are explosion systems:
Two physics trolleys pushed together on their spring plungers and then released;
A gun firing a bullet;
A student on a skateboard throwing a medicine ball (a heavy ball).
A heavy nucleus ejecting an alpha particle.
In physics it is any situation where there is zero momentum at the start. Since linear momentum is conserved, it means that the total momentum at the end must be zero.
The general principles to work with are:
Momentum is always conserved in explosions.
Momentum before = momentum after = 0.
There is zero velocity before.
Then the two components move in opposite directions with equal momentum.
The momenta add up to 0 because the directions
Consider two railway wagons that are buffered up very tightly and the springs in the buffers are ready to push them apart.
When the wagons are released, they fly apart in opposite directions as in the picture below:
Since momentum is conserved and the momentum at the start was zero, we can write:
0 = m1v1 + m2v2
which we can rearrange to:
m1v1 = -m2v2
Wagons 1 and wagon 2 are buffered up tightly together, but NOT coupled together. The brakes on the wagons are released at the same time. The release of the springs makes wagon 2 move to the right at a velocity of 0.10 m s-1. What is the velocity of wagon 1?
Momentum at start = 0:
Momentum at end = 0
Momentum of wagon 2 = 20000 kg × +0.1 m s-1 = +2000 kg m s-1
Momentum of wagon 1 = 0 kg m s-1 - +2000 kg m s-1 = -2000 kg m s-1
Velocity of wagon 1 = -2000 kg m s-1 ÷ 15 000 kg = -0.13 m s-1 from right to left.
In the physics lab the situation can be recreated using physics trolleys.
The trolleys are pushed together and then spring apart:
Let the big trolley have mass M and the small trolley have mass m. At the start, the momentum of both trolleys is 0.
We can write down the momenta at the end:
Momentum of the big trolley = MV
Momentum of the small trolley = mv
We know that the momentum at the start will be 0. Therefore the momentum at the end will be zero due to the conservation of momentum. So we can now write:
MV + mv = 0
We rearrange this to:
-MV = mv
The minus sign is there to tell us that the big trolley is going from right to left.
The big trolley has a mass of 1.2 kg and the small trolley has a mass of 0.80 kg. The velocity of the small trolley is measured at 2.0 m s-1 from left to right. Calculate the velocity of the big trolley.
0.8 kg × +2.0 m s-1 = 1.2 kg ×
In these cases, the energy for the explosion was provided by springs which were part of the vehicle. The mass of each vehicle was the same after as before.
We can derive a single expression for the recoil velocity. Consider an unstable nucleus of mass M. It emits an alpha particle of mass m. Immediately after the explosion event, the daughter nucleus has a mass N where:
N = M - m
This is shown below:
The nucleus has a velocity of V from right to left, while the alpha particle has a velocity v from left to right.
We can derive an expression for the velocity V for the nucleus. This is an explosion, so the initial momentum = 0. By the conservation of momentum:
0 = NV + mV
mv = -NV
We can write this as:
Note that in this case, N is the mass of the nucleus after it has ejected the alpha particle. If it had a mass M before the explosion, we would have to take the loss of mass into account:
N = M - m
Therefore we have to be careful to make sure that we read the question carefully - RTQ, Read The Question. (I can't talk - I have often not read the question. When Mr Roscoe got mad at my homework once, he wrote RTFQ. I was assured by my housemaster that it meant Read the Full Question...)
In the case above, the unstable nucleus lost mass. In the next question, the energy for the explosion comes from a charge in the behind the shell. When a reaction happens, there is negligible loss of mass. The mass of the reactants is the same as the mass of the products.
A cannon has a mass of 1500 kg. It fires a shell of mass 25 kg. The gunpowder charge is 10 kg. The cannon recoils at a velocity of 10 m/s. What is the velocity of the shell? (The answer is NOT 600 m s-1)
Collisions and Energy
Energy is always conserved as it must be in any physics system. All moving objects have kinetic energy, which we know from GCSE has the relationship:
Total energy before = total energy after.
While all the energy of the moving objects before the collision is assumed to be kinetic energy, the energy after the collision may NOT be all kinetic energy. If we do the momentum calculation and work out the velocity of each object, we will find that the momentum is conserved, but the kinetic energy before is greater than the kinetic energy after. Some kinetic energy is lost. It ends up as noise and internal energy. This is shown in the equation:
The term E is the difference between the kinetic energy before and the kinetic energy after. When E = 0, we have a perfectly elastic collision.
Since the velocity is squared when calculating the kinetic energy, the kinetic energy is always positive, regardless of the direction of the movement.
Most collisions are inelastic, which means that kinetic energy is transferred to other sorts of energy.
Kinetic energy after cannot be greater than before, unless external work is done on the system.
A large railway wagon of mass 35 000 kg is travelling from left to right at a velocity of 5.0 m s-1. A small wagon of mass 25 000 kg is travelling at a velocity of 2.0 m s-1 from right to left. They collide. The larger wagon then moves off at a velocity of 1.5 m s-1 from left to right. Calculate:
(a) The velocity of the smaller wagon;
(b) The energy transferred in the collision.
The question above showed that a lot of energy can be lost in collisions. In the case above, there would have been quite a loud bang as the two wagons collided. There also was a lot of energy dissipated as heat.
Kinetic energy in elastic collisions
Perfectly elastic collisions only happen in particle physics. They can be modelled with a linear air track in which the vehicles carry mutually repelling magnets:
Newton's cradle is a toy that also models conservation of momentum and conservation of kinetic energy:
Drop one ball and one ball will swing up at the other end:
We would never get two balls swinging up. Consider the rules:
Momentum has to be conserved;
Kinetic energy is conserved.
Let the velocity of the ball be v, and the mass be m. If one ball comes up when two balls are dropped, to conserve momentum we need:
The velocity at the end is half what it was at the start. There is no reason why this couldn't happen, but look at the kinetic energy. Kinetic energy at the start is:
Kinetic energy at the end is:
This means that the kinetic energy is doubled. To achieve this, the speed is has to be 1.4 (Ö2) times what it was at the start. This is not consistent with the momentum. Therefore it cannot happen.
In reality, the collisions are not perfectly elastic. Some of the kinetic energy is dissipated as sound (you hear the balls clicking as they collide).
Kinetic energy in Explosions (A level only)
Earlier on in the tutorial, we considered the case of an unstable nucleus emitting an alpha particle. The unstable nucleus has a mass M. It emits an alpha particle of mass m. Immediately after the explosion event, the daughter nucleus has a mass N where:
N = M - m
This is shown below:
The nucleus has a velocity of V from right to left, while the alpha particle has a velocity v from left to right. Now let's think about the kinetic energy. The kinetic energy is a scalar so the directions of the velocities are not important. Let's assume that the unstable nucleus is stationary, so that the original kinetic energy is zero.
Where does the kinetic energy come from?
We are going to derive an expression that shows how the kinetic energy of the emitted particle is linked with the total kinetic energy. If we know the energy of the decay, we can use it to predict the kinetic energy of the alpha particle (hence its speed). Measuring the speed of an alpha particle by direct measurement is not easy.
So we can write:
Total energy = kinetic energy of the recoiling nucleus + kinetic energy of the alpha particle
We have an expression for V derived from the momenta of the two particles:
So we can substitute:
We can look at the kinetic energy of the alpha particle, which we will call Ea which is given by:
So we can now write:
Therefore rearranging to make Ea the subject:
This rearranges to:
I have included this seemingly rather contrived derivation as it formed part of a question in an A2 paper. (Open confession: It stumped me, which irritated me intensely. It also stumped my colleague.)
At university you will study the Q-value, which is the same as kinetic energy of the decay, Ek tot here. It is the difference of the rest energy of the parent nucleus, EP, and the combined rest energies of the daughter nucleus, ED, and the emitted particle, Ea. A version of this equation appears HERE.
Q = (ED + Ea) - EP
The energies are the binding energies of the nuclei.
We can use the integer mass numbers for the masses m and N. For the alpha particle, the mass number is 4.
(Challenge for A level only)
Consider the decay of polonium-214 to lead-210:
The binding energy of the polonium is 1666.01 MeV
The binding energy of the lead is 1645.55 MeV
The binding energy of the alpha particle is 28.296 MeV
(a) Calculate the Q-value.
(b) Calculate the kinetic energy of the alpha particle in MeV and J.
(c) Calculate the speed of the alpha particle.
Momentum in Two Dimensions (OCR Syllabus)
This is a requirement of the OCR, Cambridge International, and Cambridge Pre-U syllabuses. It is not on other syllabuses.
Consider a ball bearing of mass m1 kg travelling at a velocity u1 m s-1 at an angle of q1 to the horizontal. Its total momentum is p1 N s.
Since momentum is a vector, we can resolve it into vertical and horizontal components, as shown on the picture. Now consider a second ball bearing of a different mass m2 kg travelling at a different velocity u2 m s-1 at a different angle of q2 to the horizontal. Its total momentum is p2 N s.
Let's make the two ball bearings collide:
After the collision we see:
We will model the two ball bearings as point masses.
It all looks rather complicated, but if we follow through logically, we will find a way through. We need to look at some important rules in handling problems like this:
Momentum is conserved in both horizontal and vertical directions.
We must keep the horizontal components separate from the vertical components.
We need to keep to the convention that left to right is positive, and that going upwards is positive.
Before the collision:
Total horizontal momentum = p1 cos q1 + p2 cos q2 = m1u1 cos q1 + m2u2 cos q2
Total vertical momentum = p1 sin q1 + p2 sin q2 = m1u1 sin q1 + m2u2 sin q2
After the collision:
Total horizontal momentum = p3 cos q3 + p4 cos q4 = m1v1 cos q3 + m2v2 cos q4
Total vertical momentum = p3 sin q3 + p4 sin q2 = m1v1 sin q3 + m2v2 cos q4
The worked example below shows how these equations are used with numbers in.
Let's make m1 = 1.5 kg, u1 = 2.0 m s-1, and q1 = 25o. Let's make m2 = 2.0 kg, u2 = 1.8 m s-1, and q2 = 34o. The two objects collide.
After the collision, mass m1 has a velocity of 1.5 m s-1 and an angle of 40o and mass m2 has a velocity of v m s-1 at an angle of 20o. Work out the velocity of mass m2.
Work out the momentum vectors before:
Momentum of mass m1 = 1.5 kg × 2.0 m s-1 = 3.0 N s
Momentum of mass m2 = 2.0 kg × 1.8 m s-1 = 3.6 N s
Now work out the horizontal and vertical momenta by summing the horizontal and vertical components of the momentum vectors:
Total horizontal momentum = (3.0 N s × cos 25) + (3.6 N s × cos 34) = 2.719 N s + 2.985 N s = 5.704 N s (Both are from left to right so are positive.)
Total vertical momentum = (3.0 N s × sin 25) + -(3.6 N s × sin 34) = 1.268 N s + -2.013 N s = -0.745 N s (Remember the minus sign as m2 is moving downwards.)
We will round to an appropriate number of significant figures later.
Work out the momentum vectors after:
Momentum of mass m1 = 1.5 kg × 1.5 m s-1 = 2.25 N s
Momentum of mass m2 = 2.0 kg × v m s-1 = 2.0 v N s
Now work out the horizontal and vertical momenta by summing the horizontal and vertical components of the momentum vectors:
Total horizontal momentum = (2.25 N s × cos 40) + (2.0 v N s × cos 20) = 1.930 N s + 1.879 v N s = 5.704 N s (Both are from left to right so are positive.)
Therefore we can work out the horizontal component:
1.879 v = 5.704 N s - 1.930 N s = 3.774 N s
v = 2.009 m s-1.
This gives us the horizontal component.
Now we can work out the vertical component:
Total vertical momentum = -(2.25 N s × sin 40) + (2.0 v N s × sin 20) = -1.446 N s + 0.6840 v N s = -0.745 N s (Remember the minus sign as m1 is moving downwards.)
0.6840 v = -0.745 N s - -1.4460 N s = 0.7010 N s
v = 1.025 m s-1
This gives us the vertical component.
So the resultant velocity is summarised here:
So the resultant velocity is the vector sum:
vres2 = (2.343 m s-1)2 + (1.025 m s-1)2 = 6.5402 m2 s-2
v = 2.557 m s-1 = 2.6 m s-1 (to 2 s.f.)
The main difficulty here is making sure that the signs are right.
Black ice is very dangerous for motorists. The mixture of water and water ice makes for a very slippery surface. Cars A and B are travelling too fast for the conditions and, because of they are travelling on black ice, the drivers cannot control their vehicles. The two cars are approaching a fork where two roads merge into one, as shown on the diagram:
The two vehicles collide and travel in a horizontal line (due East) as shown in the diagram.
Car A has a mass of 2500 kg. Before the collision, it is travelling at 12 m s-1 at an angle to the horizontal line of 20o.
Car B has a mass of m kg. Before the collision, it is travelling at 15 m s-1 at an angle to the horizontal line of 25o.
After the collision both cars stick together, as shown.
(a) Show that the mass of Car B is about 1600 kg.
(b) Calculate the magnitude of the velocity of the two cars after the collision.
A rocket moves because fuel is burned to gases which move with a certain velocity. Mass is conserved in the reaction, so that if 10 kg of fuel is burned, 10 kg of gases are produced. As the gases move with a velocity, there has to be momentum. As momentum is conserved there has to be an equal and opposite momentum applied to the rocket.
If we bring time in, we can convert the momentum to a force. So if 10 kg s-1 of fuel is burned, then 10 kg s-1 of gas is produced. If we know the velocity, we can work out the change in momentum per second, which is force (Newton II).
Forces act in pairs (Newton III) so there is an equal and opposite force on the rocket.
Remember that the upwards thrust FE from the engines is opposed by the weight W of the rocket, which acts vertically downwards. Therefore the resultant force FR is given by:
FR = FE - W
Always take into account the weight of the rocket. Failure to do so will give you the wrong answer. You will get the acceleration that would be achieved in outer space.
Change in momentum of
kg m s-1 every second. Since
, this results in a force.
Forces act in pairs (Newton III)
Force results in acceleration (Newton II).
A rocket of mass 1200 kg
accelerates vertically upwards off the launch pad with a constant thrust
from its engines of 16 kN.
(Use g = 9.8 m s-2)
Modelling the Motion of the Rocket (Extension)
Using a spreadsheet, we can model the acceleration of the rocket. We use the following equations:
Mass of rocket (kg) = Original mass (kg) - [rate of fuel use (kg s-1) × time (s)]
Weight (N) = Mass of Rocket (kg) × 9.81 N kg-1
Resultant force (N) = Thrust (N) - Weight (N)
Acceleration (m s-2) = Resultant force (N) ÷ Mass (kg)
If you are a dab hand at spreadsheets, you can try it for yourself. In generating the spreadsheet, I used the data from Question 9. I have used a sampling time of 0.1 s. Here are the data for the first second:
And here are the formulae:
Here is the acceleration-time graph produced by the spreadsheet:
You can see that the acceleration is not constant. It increases considerably as the fuel is used up:
The mass is decreasing as fuel is used up;
The weight is decreasing as the fuel is used up;
The resultant force (= thrust - weight) is increasing.
The model is limited in that it does not take into account the air resistance, which will come into effect very quickly to act in the same direction as the weight. Also it works by using a 0.1 s sampling method. This gives and approximation that is not so far out. The true answer would be obtained if we knew the function that links the acceleration of the rocket with time.
You can have a play with the data on your own copy of this model which you can download HERE. It is an Excel file.
Try this out for yourself - Use the spreadsheet to generate data for a speed-time and a distance time graph. We know that the area under the acceleration-time graph gives the speed. In calculus notation:
If a is constant, the solution is:
However in this case, the acceleration is not constant, as we have seen above. This is because the mass is decreasing as the fuel in the rocket is used up. As a result, the weight is decreasing. Therefore the resultant force is increasing.
The speed can be worked out using:
The t term is the time interval, not the time value. If you use the time value, your speed values will be wrong. This relationship works because it is a formula that sums the previous speed value:
To work out the speed time graph below, I have used a smaller time interval sample (0.01 s rather than 0.1 s as above). There is not much difference between the highest speeds obtained. With the 0.1 sample time, the speed at 61.5 s = 746 m s-1 compared with 747 m s-1 at 61.50 s with the 0.01 s sample time. Not much difference.
The speed time graph looks like this:
The distance would be worked out by counting the squares. This is tedious, especially if you lose your place. Here is the same graph showing the area under the graph being used to work out the distance. The area has been broken up into rectangles and triangles.
If we add all the small areas together, we get:
s = 4000 m + 2900 m + 2000 m + 1400 m + 800 m + 400 m + 1050 m + 37.5 m + 200 m + 200 m + 300 m + 300 m + 450 m + 550 m + 1000 m = 15640 m = 15600 m (3 s.f.)
A more precise answer can be gained by breaking up the graph into little strips of time interval 0.1 s or 0.01 s and working out the area of each strip using:
Either of these equations gives you the area of each little strip, i.e. the distance travelled in 0.1 s or 0.01 s, depending on your chosen time interval. It does not give you the sum total of the strips so far. The maximum value (according to the data that I used) was about 7.5 m. You need to make a little summing formula in the next spreadsheet column.
Then you have add up all the areas of the little strips together:
For the 0.1 s interval, the distance = 15504 m
For the 0.01 s interval, the distance = 15528 m.
The distance time graph shown here is generated using the 0.01 s time interval.
The most precise value would obtained by using calculus integration. This would depend on knowing the function by which the acceleration was related to the time.