# Interactive text book: Newton's laws and momentum

### Using laws

Apart from the subject of law itself, physics is probably the subject with the most laws. Laws are used to make strong predictions and solve problems. It is always good to quote a law when answering a question as it gives the answer validity.

When asked "What is the resultant force on a helicopter as it moves with constant velocity from A to B?", you could answer "I think the resultant force is zero." This is not a strong answer as someone else could have a different opinion. A better answer would be "Newton's first law states that a body remains at rest or travelling with constant velocity unless acted upon by an unbalanced force. The helicopter has constant velocity so the forces are balanced, therefore the resultant is zero." To disagree with that answer one would have to disagree with the laws of physics.

The IB Diploma endorses critical thinking... but is physics uncritical thinking? In some ways yes; we accept the laws to be true and apply them without criticism. If we questioned the laws we wouldn't be able to solve any problems. This could also be seen as critical thinking. Comparing the two answers to the helicopter question we are able to deduce that one has more validity than the other.

The laws of physics often provide a simple solution without having to go into detail. We don't need to resolve all the forces on the helicopter to show they are balanced we just have to say that Newton's first law says they must be. What if the laws are wrong? Don't even think about it. It's like asking whether the rules of chess are wrong - they're not! Not in the game of chess at least. If the rules are wrong the game falls apart.

So how does science progress? In an IB physics class it doesn't. You are just learning the rules of the game. We assume that a ball falling in a vacuum accelerates with constant acceleration. If you take measurements in the lab you will find this to be true. But cutting edge physics is different. If you take very accurate measurements you find it's not true. Should we then abandon the notion of constant acceleration? No, it works well enough, and we can never get an exact answer anyway. Most of the physics covered by this course is not true for all examples but the predictions that result from the application of laws are good enough provided we stay within the constraints of the original assumptions. There would be little point in applying Einstein's theory of general relativity to find how long it takes for a ball to fall one metre.

#### Newton's laws of motion

Newton's laws of motion are three laws used to model the motion of bodies:

- A body will remain at rest or moving with constant velocity unless acted upon by an unbalanced force.
- The acceleration of a body is directly proportional to the unbalanced force and inversely proportional to its mass.
- If body A exerts a force on body B then body B will exert an equal and opposite force on body A.

When two bodies of different mass collide we can observe a change in the velocity of both bodies. What do the three laws tell us?

- 1st law ⇒ there must be an unbalanced force on each body
- 3rd law ⇒ the force on each body is equal and opposite
- 2nd law ⇒ the acceleration of the body with smaller mass is greater than the big one. The time of the collision is the same for both so the change of velocity is greater for the small mass.

### Applying Newton's 2nd law

**Elevator**

A \(1\text{ kg}\) ball is placed on the floor of an elevator accelerating up at \(6\text{ m s}^{-2}\) (that's high!).

Note that the cables here are invisible

We can draw a free body diagram for the forces acting on the ball:

Applying Newton's 2nd law we know that the unbalanced force is \(ma\). We also know the acceleration is upwards so \(N > mg\):

\(N-mg=ma\)

If we reveal the size of the forces we see this is true:

\(16-10=1\times 6\)

Let's now consider the elevator accelerating down with acceleration \(6\text{ m s}^{-2}\):

The weight is now greater than the normal force.

\(mg-N=ma\)

Before the elevator hits the ground the tension in the supporting cables is increased causing an upward acceleration.

The acceleration is upwards so we return to \(N > mg\) and \(N-mg=ma\).

A hotel guest stands on a weighing machine as an elevator travels from ground floor to the 10 th floor. As they travl up the reading on the weighing machine changes.

Which of the following gives the reading in kg in the correct order?

Upward acceleration - constant velocity - downward acceleration

**Sliding mass**

This is a common experiment used to verify Newton's 2nd law often seen in exam questions. In this problem there are two bodies.

Lets consider the ball first. Two forces act on the ball: weight and tension.

The ball accelerates downwards so \(mg > T\). Applying Newton's second law gives:

\(mg-T=ma\)

\(\Rightarrow T=mg-ma\)

There is no friction on the table so only one force acts on the box: tension.

\(T=Ma\)

Equating the two equations for tension gives:

\(mg-ma=Ma\)

\(mg=Ma+ma=a(M+m)\)

\(a={mg\over M+m}\)

If the hanging mass is 2 kg and the acceleration 0.1 ms^{-2}, what is the mass of the block?

0.1 = 2/(M +2)

**Pulley**

Another common experiment, and another instance where the magnitude of acceleration is common to two bodies.

The red ball has a larger mass than the blue one so accelerates downwards. Drawing free body diagrams for the balls will help us to write equations according to Newton's 2nd law.

The blue ball accelerates up so \(T > mg\):

\(T-mg=ma\Rightarrow T=mg+ma\)

The red ball accelerates down so \(Mg > T\):

\(Mg-T=Ma\Rightarrow T=Mg-Ma\)

Equating the two equations:

\(mg+ma=Mg-Ma\)

\(ma+Ma=Mg-mg\)

\(a(m+M)=g(M-m)\)

\(a={M-m\over M+m}g\)

An experiment is carried out with two hanging masses resulting in an acceleration of 1m s^{-2}. The apparatus is moved into an elevator that accelerates upwards at 5 m s^{-2}.

What is the acceleration of the balls in the elevator?

Upward acceleration of elevator increases the tension in the strings the same as if g = 15 m s^{-2}

### Applying Newton's 3rd law

Newton's 3rd law is probably the most misunderstood of Newton's 3 laws and that has a lot to do with an abbreviated version that is often quoted.

This bad version of the law is hidden so that anyone scan reading this page will not see this and think it's the proper version.

The problem is that this doesn't say anything about the fact that there must be two bodies. Applying this to a typical example of a ball on a table you would see that there are two equal and opposite forces acting on the ball.

You might then conclude that these are the action and reaction but they are not. The reason they are equal and opposite is because the ball is at rest so the forces acting on it are balanced, a consequence of Newton's 1st law not 3rd.

Newton's 3rd law is about the interaction of bodies:

**If body A exerts a force on body B then body B will exert an equal and opposite force on body A.**

Applying this to the ball on the table, if the table exerts a force on the ball then the ball must exert an equal and opposite force on the table.

Furthermore we can also deduce that if the Earth exerts a gravitational force on the ball then the ball must exert an equal and opposite gravitational force on the Earth.

Another common misunderstanding brought about by the short version of the law involves unbalanced forces. If a person pushes a car then the car must push the person with an equal and opposite force. How then can the car move? Well, the two forces act on two different bodies and only one acts on the car. If this force is bigger than friction the car will start to move.

**Ball on a rope**

The right ball exerts a force on the string (and the string exerts an equal and opposite force on the right ball). The string pulls the left ball up (and the left ball exerts an equal and opposite force on the string).

**Collisions**

The red ball pushes the blue ball to the right and the blue ball pushes the red one to the left with equal and opposite force.

**Recoil**

The toy canon pushes the ball forwards and the ball pushes the canon back.

Note that the forces shown at the start are the forces just after the spring is released.

A toy canon is fired without a canonball.

The canon will:

There is no external force so the centre of mass does not move but it shifts to the right so the canon moves left.

### Momentum

We have seen that the size of the force exerted by one ball colliding with another depends on the mass and velocity of the ball, it is convenient to combine these quantities into one new quantity, the **momentum**:

momentum = mass x velocity

\(p=mv\)

Since mass is a scalar and velocity a vector, momentum is a vector. Its direction is the same as velocity.

The units of momentum are \(\text{kg m s}^{-1}\).

A body of mass 2 kg travels left at 5 ms^{-1}.

What is its momentum?

left is normally negative

Let's have another look at the constant acceleration example.

From the definition of acceleration, \(a={v-u\over t}\).

Newton's 2nd law says \(F=ma\).

Combining:

\(F=m{v-u\over t}={mv-mu\over t}\)

Resultant force is equal to the rate of change of momentum:

So Newton's second law equivalently states that the rate of change of momentum of a body is directly proportional to the unbalanced force acting on that body and takes place in the same direction.

This is a more general statement of Newton's 2nd law that applies even if the mass is changing.

This equation shows that momentum can also be expressed in terms of the units \(\text{N s}\).

A truck is filled with sand as it drives under a sand dispenser at constant velocity.

We could look at the motion of each grain of sand but that would be complicated. Instead, let's look at the whole situation.

Considering only the horizontal motion, the velocity of the sand changes from zero to the velocity of the truck, \(v\). The sand therefore experiences an unbalanced force to the right. The force equals the rate of change of momentum:

\(F={\Delta mv\over\Delta t}\)

In this example, the change in velocity is always \(v\) and it is the mass that is always changing with time. We can write the equation in another way:

\(F={\Delta m\over \Delta t}\times v\)

This is the mass added to the truck per unit time multiplied by the change in velocity.

According to Newtons 3rd law, the truck will experience an equal and opposite force to the left. This equals the force F pulling the truck forward. That's why the truck has constant velocity even though there is a force pulling it forward. Once the truck is full of sand it starts to accelerate.

A 10 litre bucket of mass 500 g is filled with water in 20 s.

What is the rate of change of mass of the bucket?

The mass of the bucket is irrelevant.

#### Impulse

**Impulse** is the change of momentum. From the equation:

\(F={\Delta mv\over \Delta t}\)

We can see that \(\Delta mv=F\Delta t\), the area under a graph of force vs time:

This graph gives a visual clue as to the effect of the time taken to change velocity and the force required. Consider a car travelling at a given velocity slowing down gradually to stop in \(5\text{ s}\) compared to the same car hitting a wall and coming to rest in \(0.5\text{ s}\).

The impulse for each is the same because the change in momentum is the same:

If the impulse = 5 kN s what is the force when the car hits the wall?

Area = F x 0.5 = 5000

Both graphs have the same area. In order to reduce the size of force (and therefore to increase safety), objects should be designed to stop over long time periods. This can be demonstrated by accelerating a mass hanging on a string.

If the upward acceleration is \(a\), applying Newton's 2nd law gives:

\(T-mg=ma\)

\(T=ma+mg\)

A greater acceleration means a bigger tension is needed. If the tension is bigger than the force required to break the rope, the rope breaks. To prevent things breaking the time taken to change momentum should be as long as possible, this is the principle behind crash mats in the gym, stretchy climbing ropes, car crumple zones and bubble wrap.

#### Conservation of momentum

Consider a collision between two balls:

According to Newton's 1st law, since the velocity of the balls change they must experience an unbalanced force.

According to Newton's 2nd law, the force is equal to the rate of change of momentum. If the time of the collision is \(Δt\), the force experienced by each ball is:

- Red \(F_\text{r}={m_1v_1-m_1u_1\over \Delta t}\)
- Blue \(F_\text{b}={m_2v_2-m_2u_2\over\Delta t}\)

According to Newton's 3rd law these forces are equal and opposite:

\({m_2v_2-m_2u_2\over\Delta t}=-{m_1v_1-m_1u_1\over\Delta t}\)

\(\Rightarrow m_1u_1+m_2u_2=m_1v_1+m_2v_2\)

The total momentum after the collision is equal to the total momentum before. In other words, momentum is conserved.

For an isolated system of bodies the total momentum is always the same.

The pink ball in the collision below has a mass of 1 kg.

What is the mass of the yellow ball?

From the velocity vectors we can see that the velocities before are v and -v and after the velocity is -v

momentum before = 1 x v + m x -v = v - mv

momentum after = 1 x -v

momentum is conserved so v - mv = -v

mv = 2v

m = 2 kg

The principle of conservation of momentum is a consequence of Newton's three laws. It's not a new law, however it does give a simpler way of solving problems involving isolated bodies.

Example: Find the velocity of the blue ball in the collision involving the three balls of mass \(m\) shown below.

According to the principle of conservation of momentum:

momentum before = momentum after

\(20m-10m+0=-2.5m+3.12m+vm\)

\(10=0.62+v\Rightarrow v=9.38\text{ m s}^{-1}\)

The three balls in this animation have the same mass.

What is the velocity of the middle ball after the collision?

momentum before = 2m

momentum after = 0.5m + mv + 1.28m

momentum is conserved so 2m = 1.78m + mv