# Momentum

## Momentum Revision

**Momentum**

**Momentum** is a property of all moving objects. It depends on **mass** and **velocity**. In particular we will look at** Linear Momentum** in this.

**Linear Momentum**

**Momentum** \left(\text{kg ms}^{-1}\right) is the quantity given by the product of** mass** \left(\text{kg}\right) and **velocity** \left(\text{ms}^{-1}\right) of a moving object.

As **momentum** is a **vector **quantity, the direction of motion usually determines if the** momentum** is positive or negative. Conventionally, motion to the left or downwards is considered negative, and to the right or upwards is positive.

Recall the equation for **momentum**:

\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).

Heavy, fast moving objects, such as a lorry, have higher** momentum** whilst slow and light objects have lower **momentum**. Remember if an object is **stationary** then it has no **momentum**, because it has no velocity.

**Conservation of Momentum**

The law of **conservation of momentum** states that the total **momentum** before a collision is equal to the total **momentum** after a collision (provided no external **forces** act during the collision).

When two objects collide, both of their **momentums** need to be considered before and after the collision. In general we can write this as an equation:

\begin{aligned} \text{Momentum before} &= \text{Momentum after} \\ m_1 u_1 + m_2 u_2 &= m_1 v_1 + m_2 v_2 \end{aligned}

**Example:**

Two carts are moving towards each other as shown below. They collide causing the carts to move off separately in the same direction to the right. Using the information below, find the missing speed.

**[3 marks]**

Therefore, the speed is 2 \: \text{ms}^{-1} to the** right**.

**Impulse**

**Impulse** is a term that gives a quantity to the effect a **force **has on an object over a given time. It is defined in the equation:

I = F \times t

- I is the
**impulse**in Newton-seconds \left(\text{Ns}\right). - F is the
**force**in Newtons \left(\text{N}\right). - t is the
**time**in seconds \left(\text{s}\right).

**Impulse** is also equal to the change in momentum during a collision. Therefore, impulse can also be written as:

I = mv - mu

mv represents the **final momentum**, and mu represents the **initial momentum**.

Because I = F \times t, we can find the **impulse** from a force-time graph. Here, **impulse** is equal to the area underneath the graph:

**Example: **Calculate the impulse using over the whole collision, using the force-time graph below.

**[3 marks]**

\text{Impulse} = \text{Area under graph}

\text{Smaller section:} \: \boldsymbol{\left(0.5 \times 0.5 \times 16\right)}

\text{Bigger section:} \: \boldsymbol{\left(0.5 \times 2.5 \times 16\right)}

Therefore:

\text{Impulse} = 4 + 20 = \boldsymbol{24} \: \textbf{Ns}

We know that impulse is the effect a **force **has on an object over a given time, and it can also be defined as the change in momentum. To help understand this, let’s consider these two equations for impulse:

Equating these:

F \times t = mv - muFor a given object, **mass** usually remains constant so we can rewrite this as:

Therefore, we have used **impulse** to define **force** as the **rate of change of momentum**.

From this equation we can conclude that force is **inversely proportiona****l** to time. So if a collision happens over a longer period of time, the force is **reduced**. This has many practical implications in safety and forms the basis for concepts like airbags in cars, cushioned and padded clothing etc. They all act to increase the **contact time** a collision occurs over. So the longer you are in contact with an air bag, the more the force on you is reduced, hence decreasing the chance of any injuries.

**Collisions**

**Collisions** can be classified as **elastic** or **inelastic**.

We have previously looked at an example of an **elastic collision**. This is when **kinetic energy** is conserved and usually occurs when the two objects remain separated after a collision.

However, an **inelastic collision** is where** kinetic energy is not conserved** and usually happens when the two objects combine as a result of the collision, forming one mass moving at one velocity.

Recall the equation for **kinetic energy**:

E_k = \dfrac{1}{2} \times m \times v^2

- E_k is the
**kinetic energy**in joules \left(\text{J}\right) - m is the
**mass**in kilograms \left(\text{kg}\right) - v is the
**velocity**in metres per second \left(\text{ms}^{-1}\right)

So, for an elastic collision where kinetic energy is conserved:

\begin{aligned} \text{Kinetic Energy Before} &= \text{Kinetic Energy After} \\ \dfrac{1}{2} \times m \times v^2 &= \dfrac{1}{2} \times m \times v^2 \end{aligned}**Example:** Two objects collide as shown below. After the collision they move off to the right as one mass. Estimate their velocity after the collision assuming that the collision is inelastic.

**[4 marks]**

- The collision is inelastic, so the kinetic energy is not conserved. However, momentum is still conserved:

\begin{aligned} m_1 u_1 + m_2 u_2 &= \left(m_1 + m_2 \right) \times v \\ \left(50 \times 15\right) + \left(7 \times -2\right) &= \left(50 + 7 \right) \times v \\ 750 - 14 &= 57 \times v \\ 736 &= 57 \times v \\ v &= \dfrac{736}{57} = 12.9 \: \text{ms}^{-1} \end{aligned}

- We can confirm ourselves that kinetic energy is not conserved in this collision and that it is inelastic:

Therefore some kinetic energy is lost in the collision and the collision inelastic.

## Momentum Example Questions

**Question 1:** Describe the concept of the conservation of momentum and give an example of it in action.

**[3 marks]**

The law of conservation of momentum states that the **momentum before a collision is equal to the momentum after a collision**.

An example of this is when a gun is fired. When a gun is fired, **a low mass bullet is fired at high speed in one direction**. To conserve momentum, the **gun recoils causing a small velocity of a high mass in the opposite direction**.

**Question 2:** Explain how momentum is conserved in an explosion.

**[3 marks]**

When an object explodes it fires fragments off in **all directions**. Because momentum is a vector, some fragments would be considered to have **positive momentum whilst those fired in the opposite direction, would have negative momentum**. These positive and negative momentums when added together will **cancel each other out and sum to zero, which is the momentum the object before the explosion**. Hence, momentum is conserved.

**Question 3:** Describe the difference between an elastic and inelastic collision.

**[2 marks]**

An elastic collision is where **kinetic energy is conserved**, whilst an inelastic collision is where some **kinetic energy is lost in the collision**.