Further Mechanics Tutorial 2 - Conservation of Momentum

 

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.

 

An important principle:

  The total momentum of a system remains constant provided that no external forces act on the system. 

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. 

Question 1

What do you understand by the statement above?

Answer

 

 

 

Collisions

Momentum is always conserved in collisions.

If 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:

Property

Large Object

Small Object

Mass

M

m

Initial velocity

u1

u2

Final velocity

v1

v2

  We can show this as a diagram:

There are two important principles here:

1.      Conservation of momentum:

Total momentum before = total momentum after

Mu1 + mu2 = Mv1 + mv2

  1. Energy is conserved

Total energy before = total energy after

½Mu12 + ½mu22 = ½Mv12 + ½mv22 + E

The term E is the energy that is lost in the collision.  In a perfectly elastic collision E = 0.

When doing momentum calculations, always be careful about the directions you are using.

 

Question 2

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. 

 

Answer
Question 3

A second collision is shown below:

What is the speed of the 500 kg car after the collision?

 

Answer
Question 4

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.  

Calculate:

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. 

 

Answer

 

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. 

 

Momentum in Explosions

An explosion not just where something goes BANG.  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.  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

Worked example

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.  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 = +2000 kg m/s

Momentum of wagon 1 = 0 kg m/s - +2000 kg m/s = -2000 kg m/s

Velocity of wagon 1 = -2000 kg m/s ÷ 15 000 kg =  -0.13 m/s from right to left.

Explosions explain the way the guns and rockets work.

Question 5

A cannon has a mass of 1500 kg.  It fires a shell of mass 25 kg.  It recoils at a velocity of 10 m/s.  What is the velocity of the shell?

Answer