Momentum & Safety

Momentum Basics

Momentum is a measure of how difficult it is to stop a moving object. It depends on both the mass of the object and its velocity.

Momentum (p) is defined as:

$$\text{momentum} = \text{mass} \times \text{velocity}$$

or

$$p = m v$$

where:

• p is momentum in kilogram metres per second (kg·m/s)
• m is mass in kilograms (kg)
• v is velocity in metres per second (m/s)

Momentum is a vector quantity, which means it has both size and direction. The direction of momentum is the same as the direction of the velocity. It is important to always consider the direction when calculating momentum.

For instance, if a car of mass 1000 kg is moving east at 20 m/s, its momentum is:

\[ p = 1000 \times 20 = 20,000 \text{ kg·m/s east} \]

Conservation of Momentum

In a closed system (where no external forces act), the total momentum before an event is equal to the total momentum after the event. This is called the conservation of momentum.

This principle applies in collisions and explosions:

• Collisions: Two or more objects collide and momentum is transferred between them.
• Explosions: An object breaks apart into pieces that move in different directions.

Mathematically:

$$\text{Total momentum before} = \text{Total momentum after}$$

or

$$\sum p_{\text{before}} = \sum p_{\text{after}}$$

For example, if two ice skaters push off each other on ice, their combined momentum before pushing off is zero (both at rest). After pushing, their momenta are equal in size but opposite in direction, so total momentum remains zero.

Momentum in Collisions

There are two main types of collisions:

• Elastic collisions: Objects bounce off each other without permanent deformation or loss of kinetic energy.
• Inelastic collisions: Objects stick together or deform, and kinetic energy is not conserved (some is converted to other forms).

In all collisions, momentum is transferred between objects, but the total momentum remains constant (conservation of momentum).

To calculate momentum in collisions, use:

$$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$$

where:

• $m_1, m_2$ are the masses of the two objects
• $v_{1i}, v_{2i}$ are the velocities before the collision
• $v_{1f}, v_{2f}$ are the velocities after the collision

For example, if a 2 kg ball moving at 3 m/s collides with a stationary 3 kg ball, you can use conservation of momentum to find their velocities after the collision.

Example: A 2 kg ball moving at 3 m/s hits a stationary 3 kg ball. After the collision, the 2 kg ball moves at 1 m/s. What is the velocity of the 3 kg ball?

Using conservation of momentum:

$$\text{Total momentum before} = \text{Total momentum after}$$

$$(2 \times 3) + (3 \times 0) = (2 \times 1) + (3 \times u)$$

$$6 = 2 + 3u$$

$$3u = 4$$

$$u = \frac{4}{3} = 1.33 \text{ m/s}$$

So, the 3 kg ball moves at 1.33 m/s after the collision.

Safety and Momentum

When a vehicle crashes, the momentum of the vehicle changes rapidly to zero. This sudden change in momentum causes a large force on the occupants, which can cause injury.

To reduce injury, safety features aim to reduce the force by increasing the time over which the change in momentum happens. This is because force is related to the rate of change of momentum:

$$\text{Force} = \frac{\text{Change in momentum}}{\text{Time taken}}$$

This is also called impulse, which is the product of force and time. The impulse-momentum theorem states that impulse equals the change in momentum:

$$\text{Impulse} = \text{Force} \times \text{Time} = \text{Change in momentum}$$

Increasing the collision time reduces the force experienced, making the crash less severe.

Examples of safety features that increase collision time:

• Seat belts: Stretch slightly to increase the time over which the wearers momentum changes.
• Airbags: Inflate to provide a soft cushion that slows the occupant down more gradually.
• Crumple zones: Parts of the car designed to deform and crumple during a crash, increasing the time taken to stop the car.

For example, if a car occupants momentum changes from 500 kg·m/s to 0 in 0.1 seconds, the force is:

$$F = \frac{500}{0.1} = 5000 \text{ N}$$

If the collision time is increased to 0.5 seconds by an airbag, the force reduces to:

$$F = \frac{500}{0.5} = 1000 \text{ N}$$

This much lower force reduces the risk of injury.