Additional Physics Topic 7 - Force and Momentum

The Concept of Momentum

Momentum is quite a difficult concept.  Work through these notes slowly and carefully.  If you are not sure how to tackle a question, look at the answer, and see how it works.  Then try again for yourself.  If in doubt, ASK YOUR TEACHER!

Momentum is an important Physics concept, but unlike other quantities in Physics that you have met, it is not something you can see or feel.  For example you can easily see that a moving object has kinetic energy

These three pictures show an old aeroplane being slammed into a block of high density concrete at high speed.

Pictures by NASA

The kinetic energy of the plane is so high that it is vaporised on impact, except for the wing tips.  What you can't see is the momentum of the plane.

The brakes on your bike get hot when you have converted kinetic energy to heat.  You can feel the force of a moving football as it hits you.  But you cannot see or feel momentum.  Nor can your physics teacher order a class set of momentum for your practical lesson.

However momentum as a concept helps us to understand situations in Physics like:

• Collisions

• Explosions

Both of these are difficult to explain just by studying the energy changes.  Momentum makes them easier to understand.

Working out Momentum

Momentum is the product of mass and velocity, which means mass and velocity multiplied together.

 Question 1 What is velocity?  How does it differ from speed?

Momentum is worked out by this equation which you must learn for the exam:

Learn this:

momentum (kg m/s) = mass (kg)  × velocity (m/s)

In Physics Code:

p = mv

In triangle form:

Notice the following:

• The units are kilogram metres per second (kg m/s) [or Newton seconds (Ns); they are the same];

• Momentum has a direction.  The direction is very important.

Marks are lost if you say momentum = mass × speed.

 Worked Example What is the momentum of a ball of mass 0.1 kg travelling with a velocity 5 m/s form left to right? Answer Formula first: p = mv p = 0.1 kg × 5 m/s = 0.5 kg m/s (from left to right)

 Question 2 What is the momentum of a 70 kg runner whose velocity is 8 m/s from left to right?

Collisions and Explosions

Momentum comes into its own when we use it to explain collisions and explosions.  We need to be sure what is meant by a collision or an explosion:

• When two objects come together, that is a collision;

• When two objects fly apart, we call that an explosion.  Sorry to disappoint you if you were looking for flashes and bangs.  Just for the record...

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The key concept to understand is that:

momentum is always conserved, provided no external forces act

This means that

momentum before = momentum after

We will always assume that no external forces are acting.

Explosion

Think about a cannon firing a cannon ball:

Before the cannon is fired, there is no movement, so the momentum is zero.

When the cannon is fired, the ball moves from left to right at a certain velocity.  Therefore there is momentum from left to right.  Let's say that from left to right is positive.

Since momentum has to be conserved, the cannon will have equal negative momentum from right to left.

 Question 3 What is the total value of both the momentums added together?

The cannon will recoil.  Any object that ejects another object will recoil.  If an ice-skater throws a heavy object away from them, they will recoil backwards.  A nucleus ejecting an alpha particle recoils just like a cannon.

 Worked Example The cannon ball in the example above travels at 150 m/s from left to right and its mass is 5 kg.  If the mass of the cannon is 500 kg, what is its recoil velocity? Answer The momentum before = 0 kg m/s   The momentum after = 0 kg m/s   Make left to right positive.  Velocity of cannon ball is positive; velocity of cannon is negative.   Momentum of cannon ball + - momentum of cannon = 0 kg m/s   5 kg × 150 m/s + 500 kg × - v m/s = 0 500 v = 750 kg m/s   v = 750 ÷ 500 = 1.5 m/s from right to left

Collisions

There are three possible ways of having a collision:

• One vehicle stationary and the other hitting it;

• Both vehicles travelling in the same direction;

• Two vehicles travelling in opposite directions.

We will only think about collisions in one dimension only.  As before momentum is conserved.  However this time the momentum before is not zero.

Momentum before = momentum after

Theory

There are two kinds of collision:

• Elastic, where two vehicles bounce apart after the collision. Kinetic energy is conserved only in a perfectly elastic collision.

• Inelastic, where the two vehicles stick together. Kinetic energy is lost.

Here is an inelastic collision. Wagon 1 has a mass m1 and wagon 2 has a mass m2. Wagon 1 has a velocity u1 where u is the velocity before the collision. Wagon 2 has 0 velocity.

Before:

Wagon 1 has momentum = m1u1

During:

The two vehicles collide and stick together.

After:

The wagons have stuck together, and both travel off at a common velocity v. The total mass is m1 + m2, as they have stuck together.

Momentum is conserved, which means that momentum before = momentum after.

Therefore:

m1u1 = (m1 + m2) v

Therefore the common velocity can be found:

v = (m1u1) ÷ (m1 + m2)

In an elastic collision, the vehicles bounce apart, so we get:

Again, momentum before = momentum after:

m1u1 = m1v1 + m2v2

If we add up all the kinetic energy before and after, and find it's the same, we can say that the collision was perfectly elastic. Inelastic collisions result in the loss of kinetic energy.

In these three worked examples, we are going to consider two railway wagons:

• Wagon A has a mass of 5000 kg;

• Wagon B has a mass of 8000 kg.

In each case both the wagons have automatic couplings that couple the two wagons together after the collision.  So they do not bounce off each other.

 Worked Example Wagon A is travelling at 1.5 m/s from left to right and couples up to B which is stationary.  What is the velocity of the two wagons after the collision? Answer Momentum of A before = 5000 kg × 1.5 m/s = 7500 kg m/s   Momentum of B before = 0   Total momentum = 7500 kg m/s   Momentum after = 7500 kg m/s   Mass after = 5000 kg + 8000 kg = 13 000 kg   7500 kg m/s = 13000 kg × v m/s   v = 7500 kg m/s ÷ 13000 kg = 0.58 m/s (from left to right)

 Worked Example Wagon A is travelling at 1.5 m/s from left to right while Wagon B is travelling at 0.5 m/s from left to right?  They couple together.  What is their combined velocity? Answer Momentum of Wagon A = 5000 kg × 1.5 m/s = 7500 kg m/s   Momentum of Wagon B = 8000 kg × 0.5 m/s = 4000 kg m/s   Total momentum before = 7500 kg m/s + 4000 kg m/s = 11 000 kg m/s = momentum after.   11 000 kg m/s = 13 000 kg  × v   v = 11 000 kg m/s ÷ 13 000 kg = 0.85 m/s from left to right

 Worked Example Wagon A is travelling from left to right with a velocity of 1.5 m/s while Wagon B is travelling from right to left with a velocity of 0.5 m/s.  What is the combined velocity and what is the direction?  The two wagons stay on the rails! Direction is important.  Make left to right positive.   Momentum of Wagon A = 5000 kg × +1.5 m/s = +7500 kg m/s   Momentum of Wagon B = 8000 kg × -0.5 m/s = -4000 kg m/s   Total momentum before = +7500 kg m/s + -4000 kg m/s = +3500 kg m/s = momentum after.   11 000 kg m/s = 3500  × v   v = 3500 kg m/s ÷ 13 000 kg = 0.27 m/s from left to right

The diagram shows two gliders on a linear air track.  You may have one at school.  It works with a blower (a hoover acting in reverse) that blows air through tiny holes.  Gliders are put on the track and can move up an down on a cushion of air with very little friction.  Ask your teacher about it.

This diagram is useful for Questions 4, 5, and 6.

The gliders have a magnet at one end and a bluetac buffer at the other.  The two magnets put together will repel and force the gliders apart.  The two bluetac buffers come together in collisions and make the two gliders stick together.

 Question 4 On the air track, a glider, A, of mass 0.25 kg is in contact with a glider, B, of mass 0.1 kg.  Both are held together against repelling magnets.  When they are released, the velocity of A is measured by light gates and comes up on the computer as 0.3 m/s (from right to left).  What is the velocity of B?

 Question 5 Glider A is sent down the air track at a velocity of 0.30 m/s from left to right with its bluetac buffer facing the direction of travel.  It strikes Glider B, which is stationary, which has its bluetac buffer facing that of Glider A.  This means that the two stick together.  What is their combined velocity and in which direction?

 Question 6 Glider A is sent down the air track at a velocity of 0.30 m/s from left to right with its bluetac buffer facing the direction of travel.  It strikes Glider B, which has a velocity from left to right of 0.15 m/s.  The two stick together.  What is their combined velocity and in which direction?

 Wagon A is travelling from left to right with a velocity of 1.5 m/s.  What should wagon B's velocity be from right to left if both wagons are stopped dead by the collision?

Collisions and Kinetic Energy

If two object bounce off each other, and there is no loss of kinetic energy, we say that the collision is perfectly elastic.  In reality this only happens in collisions between single atoms, although the gliders with repelling magnets collide with almost a perfectly elastic collision.

In most collisions, some (or all) of the kinetic energy is transferred.  These are called inelastic collisions.  Where the two objects stick together, these are perfectly inelastic collisions.

 Question 8 What happens to the kinetic energy?

Momentum and Road Safety

If you are in a car that suddenly stops, and you are not wearing a seatbelt, your momentum will make you carry on at the same speed as you were travelling before, until you hit a solid object like the steering wheel or the windscreen.  Either way you will suffer terrible or fatal injuries.

The seat belt will hold you against the car, and prevent these injuries.  Since seat belts were made compulsory, the number of fatalities in motor accidents has been considerably reduced.

Force and Momentum (Extension only)

Momentum and force are related by this simple equation:

force (newton, N) = change in momentum (kilogram metre/second, kg(m/s))

time taken for the change (second, s)

In Physics code:

F = Dp

t

In triangle form:

The change in momentum is called the impulse.

 Worked Example A ball of mass 0.2 kg is travelling from left to right at a velocity of 6 m/s.  It hits a wall and bounces off travelling at a velocity of 5 m/s from right to left.  The time of the collision is 0.1 s.  What force acts on the ball? Answer Make from left to right positive:   Momentum from left to right = 0.2 kg × +6 m/s = + 1.2 kg m/s   Momentum from right to left = 0.2 kg × -5 m/s = -1.0 kg m/s   Change in momentum = momentum after - momentum before                                 = -1.0 kg m/s - + 1.2 kg m/s = - 2.2 kg m/s   F =  (-) 2.2 kg m/s ÷ 0.1 s = (-) 22 N   The minus sign indicates that the force is from right to left.

 Question 9 Wagon A is travelling at 1.5 m/s from left to right and couples up to B which is stationary. If the time for the collision was 0.05 s, what force acts on wagon B?

Source not known

In a car crash, the change in momentum will always be the same.  If we can increase the time for the change in momentum, the force exerted on the people inside will be reduced, which results in fewer injuries.  Old cars were made to be very solid.  In an accident, the car may well have been hardly damaged at all, but the occupants killed.

The picture above shows how cars are now designed to crumple up, increasing the time interval for the change in momentum.  Also other safety features like air-bags prevent injuries from the steering wheel.

A great deal of expensive research is carried out to make sure that if you are unlucky enough to be involved in a car accident, the chances are that you will get up and walk away.  A car can be replaced; you cannot.

 Question 10 Do the interactive crossword

 Summary Momentum = mass × velocity Units for momentum are kilogram metre per second (or newton second) Momentum is conserved in explosions or collisions provide no external force acts on the system. In elastic collisions, kinetic energy is same at the end as it was at the start. In inelastic collisions, kinetic energy is transferred. Force = change in momentum ÷ time interval