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**NEWTON’S LAWS**

- Unless a resultant, external force acts on a body, its velocity will not change
- If a resultant, external force acts on an object, the rate of change of the object’s momentum is directly proportional to the force and acts in the direction of the force
- Every action has an equal and opposite reaction, i.e. if a body A exerts a force on another body B, that body exerts a force equal in magnitude but opposite in direction on body A

**1ST LAW**

- A stationary object will not move unless a resultant external force acts on it
- An object moving with constant velocity in a straight line will do so forever unless a force acts on it.
- All objects have a refusal to change their state of motion, called inertia, mass (kg) is a measure of inertia.
- The brakes on your bike overcome the inertia of the bike and you by applying a resultant force on the edges of the wheels to resist their motion, which brings us to friction

Friction as a force:

**Friction is a force that opposes the relative motion of two objects in contact.**

- According to Law 1, all moving objects should stay moving forever, which obviously isn’t true, this is because the force of friction opposes the motion
- Friction causes many nuisances but is also beneficial
- It causes wear between parts that rub together
- This can be reduced by reducing friction using lubricants/polish etc
- It also allows brakes to stop bikes/cars/trains

Momentum

**A body’s momentum is the product of its mass (m) and its velocity (v)**

This is a vector quantity with units of \(kg ms^{-1}\)

**Example: **A body with mass 500kg is travelling at \( 10ms^{-1} \). What is its momentum?

\( text{Momentum} = m times v \)

\( = 500 times 10 = 500kgms^{-1} \)

Momentum is usually conserved in collisions:

**The principle of conservation of momentum (PCM) states that in a closed system where no external forces act, total momentum is conserved, i.e. total momentum before = total momentum after. For two bodies:**

\( m_{1}u_{1}+m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} \)

\(m_{1}, m_{2}\) are the masses of each

\(u_{1}, u_{2}\) are their respective velocities before collision and:

\(v_{1}, v_{2}\) are their respective velocities after the collision.

Example: A body of mass 4kg moving at \( 4ms^{-1} \) to the right collides with a body of mass 5kg moving at \( 1ms^{-1} \) to the left. After the collision the 10kg particle moves at \( 2ms^{-1} \) to the right. Calculate the velocity of the 5kg particle after the collision.

\( m_{1}u_{1}+m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} \)

\( 10 times 4 + 5times -1 = 4 times 2 + 5 times v_{2} \)

\(35 = 20 + 5v_{2} \)

\( 5v_{2} = 15 \)

\( v_{2} = 3ms^{-1} \) to the right

**LAW 2**

**Force is proportional to rate of change of momentum**

F propto mtimes frac{v-u}{t}

F = kma

F=ma \)

- We can verify experimentally that k = 1
- Force is a vector quantity
- Its unit is \( kg ms^{-2}\) or the Newton (N)
- 1N is the force that gives a mass of 1kg an acceleration of \(1ms^{-2}\)

Mass and Weight

Outside of physics we use these words similarly but in physics they are very different

**Mass **is a scalar quantity of constant value, measured in kg

**Weight **is a vector, it is variable depending on height above ground/size of planet etc, it’s unit is the newton

\( Weight = mass times g \)

**A body’s mass is a measure of its inertia, i.e. its resistance to change in its velocity**

**Weight is the force of gravity acting on a body**

**Example**: Calculate the weight of a 200g body on earth

\( F=ma \) or \( W = mg \)

\( 0.2 x 9.8 = 1.96N \)

**LAW 3**

Examples:

- A rocket engine expels exhaust gas by exerting a force on it, the gas exerts a reaction force on the rocket, propelling the rocket upwards
- The force of your weight gives rise to a force acting on the ground. The ground exerts an equal and opposite reaction force on you.

Example Question: A person of mass 70kg is standing in a lift, calculate the force the floor exerts on them when:

- The lift is at rest
- The lift is moving with uniform speed \( 3ms^{-1} \)
- The lift is accelerating upwards at \( 3ms^{-2} \)
- The lift is accelerating downwards at \( 3ms^{-2} \)
- The lift is in freefall

**1, 2: **The normal reaction force occurs, the person is at equilibrium so there is no net force on them, therefore the force the lift exerts is just equal to the persons weight. It is the same for constant velocity due to Newton’s first law (scroll up)

\( F = ma = W = mg \)

\( W = 70 times 9.8 = 686N \)

**3: **Now there is the reaction force and the force caused by the acceleration to deal with, let F be the reaction

\( F=ma\)

\( F – W = ma \)

\( F -686 = 2times 70 \)

\( F = 826N \)

**4**: The opposite of above

\( F=ma\)

\( F – W= -ma \)

\( F -686 = -2times 70 \)

\( F = 546N \)

5: The same as part 4 but acceleration of lift equals acceleration of person, therefore

\( F=ma\)

\( F – W= -ma \)

\( F -686 = -9.8 times 70 \)

\( F = 0N \)

The lift and person are in freefall, the person would experience apparent weightlessness

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