# Momentum

## Momentum Revision

**Momentum**

**Momentum** is a property of all moving objects. It depends on **mass **and **velocity**.

**Momentum**

**Momentum** is a **vector**, meaning it has magnitude and direction. It is defined by the equation:

\textcolor{aa57ff}{p = mv}

- \textcolor{aa57ff}{p} is the
**momentum**in kilogram metres per second \left(\text{kg m/s}\right). - \textcolor{aa57ff}{m} is the
**mass**in kilograms \left(\text{kg}\right). - \textcolor{aa57ff}{v} is the
**velocity**in metres per second \left(\text{m/s}\right).

This equation can be used to work out the momentum of a moving object. A stationary object has no momentum \left(\text{because} \: \textcolor{aa57ff}{v} = 0 \: \text{m/s}\right).

**Conservation of Momentum**

In a **closed system**, the **total momentum before **an event is **equal **to the** total momentum after** an event. This is the **conservation of momentum**.

**Example: **

A football is kicked at a stationary football, at a velocity of v. They both have the same mass of m. The **momentum** before they collide is just the sum of the momentum of each ball:

So the total momentum before the collision is just the momentum of the moving ball, because the other ball is not moving and so has no momentum.

The total **momentum** after they collide will be equal to mv. Ball B now has velocity and hence it has momentum. Ball A now moves with less velocity. So the sum of these two momentums will be equal to the total momentum before the collision mv.

**Changes in Momentum**

We know that when a** resultant force **acts on an object, it causes it to accelerate. **Acceleration **is a change in velocity. So if a **force** causes an object’s velocity to change, then it also changes the objects **momentum** \left(p = mv\right).

We can combine the equations for **Newton’s 2nd Law** and acceleration to express force as a **change in** **momentum**:

Newton’s 2nd Law: F = ma. Acceleration: a = \dfrac{v-u}{t}.

Combine them to get: F = \dfrac{m\left(v-u\right)}{\Delta t}

This can be written as:

\textcolor{00bfa8}{F = \dfrac{m \Delta v}{\Delta t}}

- \textcolor{00bfa8}{F} is the
**Force**in Newtons \left(N\right) - \textcolor{00bfa8}{m \Delta v} is the
**change in momentum**in kilograms \left(\text{kg m/s}\right). \textcolor{00bfa8}{m} is the**mass**and \textcolor{00bfa8}{\Delta v} is the**change in velocity**. - \textcolor{00bfa8}{\Delta t} is the
**change in time**.

A faster rate of change of **momentum** means a bigger **force**. So if an object changes velocity very quickly, it’s momentum will change very quickly. Therefore there is a larger force on the object.

Because of this, there are often** safety measures **on vehicles in order to reduce the rate of change of momentum, and hence the reduce force on passengers. This reduces injuries on passengers. Some vehicle safety measures are:

**Seat belts**– these stretch so that the time taken for the passenger to stop is longer.**Air bags**– these inflate when there is a crash, in order to slow the passenger down.

Other examples of safety measures include:

**Gymnasium crash mats**and**cushioned surfaces for playgrounds**– They are cushioned so increase the time taken to fall. This decreases the rate of change of momentum and hence decreases the force on the gymnast.**Cycle helmets**– foam inside the helmet is crushable so increases the time taken for your head to stop.

**Example: Calculations involving momentum with a collision**

Two cars crash head on into each other and both come to rest. The momentum after the crash is \textcolor{00bfa8}{0 \: \text{kg m/s}}. The diagram shows the velocities and masses of the two cars before the collision. Calculate the velocity of car A before the collision, \textcolor{aa57ff}{v_A}. Momentum is conserved.

**[3 marks]**

If the momentum after the collision is \textcolor{00bfa8}{0 \: \text{kg m/s}}, then the momentum before the collision must also be 0 \: \text{kg m/s}.

Therefore the momentum of the two vehicles must add up to zero. This means the momentum of each vehicle must be equal:

\text{Momentum of Car A} = \text{Momentum of Car B}

Using the momentum equation, p = mv:

m_A \times \textcolor{aa57ff}{v_A} = m_B \times v_B

Substitute in the values:

\textcolor{00d865}{1000 \: \text{kg}} \times \textcolor{aa57ff}{v_A} =\textcolor{f95d27}{1500 \: \text{kg}} \times \textcolor{ed1111}{9 \: \text{m/s}}

**Rearrange **to find the velocity of Car A:

\textcolor{aa57ff}{v_A} = \dfrac{\textcolor{f95d27}{1500} \times \textcolor{ed1111}{9}}{\textcolor{00d865}{1000}}

\textcolor{aa57ff}{v_A} = 13.5 \: \text{m/s}

## Momentum Example Questions

**Question 1: **A toy car of mass 0.6 \: \text{kg} travels with a velocity of 3 \: \text{m/s}. Calculate the momentum of the toy car.

**[2 marks]**

p = mv

p = 0.6 \: \text{kg} \times 3 \:\text{m/s}

p = \bold{1.8 \: }\textbf{kg m/s}

**Question 2:** Explain what is meant by the **conservation of momentum**.

**[2 marks]**

In a **closed system**, the **total momentum before** an event is **equal** to the** total momentum after** an event.

**Question 3:** A bowling ball travelling at a velocity of 3.5 \: \text{m/s} hits a stationary pin.

After the collision, the bowling ball moves at a velocity of 1 \: \text{m/s}.

Calculate the velocity of the pin after the collision. Momentum is conserved.

The mass of the bowling ball is 6 \: \text{kg} and the mass of the pin is 2 \: \text{kg} .

**[4 marks]**

\text{total momentum before} = \left( 6 \: \text{kg} \times 3.5 \: \text{m/s} \right) + \left( 2 \: \text{kg} \times 0 \: \text{m/s} \right) = 21 \:\text{kg m/s}

\text{total momentum after} = \left( 6 \: \text{kg} \times 1 \: \text{m/s} \right) + \left( 2 \: \text{kg} \times ? \: \text{m/s}\right)

\text{total momentum before} = \text{total momentum after}

21 \: \text{kg m/s} = 6 + \left(2 \times ? \right)

\text{Velocity of pin} = \dfrac{21-6}{2}= \bold{7.5 \:} \textbf{kg m/s}

**Question 4**: A car decelerates from a speed of 20 \: \text{m/s} to 10 \: \text{m/s}, in 25 \:\text{seconds}. The car has a mass of 1200 \: \text{kg}. Calculate the force felt by the car.

**[3 marks]**

F = \dfrac{m \Delta v}{\Delta t}

Change in velocity = 20 - 10 = 10 \: \text{m/s}

F = \dfrac{1200 \times 10}{25} = \bold{480 \:} \textbf{N}