A.2.2 Linear momentum and impulse

Understanding Momentum and Impulse

You're playing a game of billiards.
You strike the cue ball, and it collides with another ball, sending both rolling across the table.
What determines how these balls move after the collision?

The answer lies in two fundamental concepts: momentum and impulse.

What is Momentum?

Definition

Momentum

Momentum is a measure of how difficult it is to stop a moving object. It depends on two factors: the object's mass and its velocity.

Definition of Momentum

Momentum is defined as the product of an object's mass and its velocity:

$p = m v$

where:

$p$ is the momentum (in kg m/s)
$m$ is the mass (in kg)
$v$ is the velocity (in m/s)

Hint

Momentum is a vector quantity.
It means that it has both magnitude and direction. Its direction is the same as the object's velocity.

Why is Momentum Important?

Momentum helps us understand how objects behave during interactions like collisions or explosions.
It is a conserved quantity, meaning the total momentum of a system remains constant if no external forces act on it.

Example

Consider a 2 kg ball moving at 3 m/s. Its momentum is:

$p = m v = 2 kg \times 3 m/s = 6 kg m/s$

If the ball's velocity doubles, its momentum also doubles, illustrating how momentum depends on both mass and velocity.

Impulse: Changing Momentum

Definition

Impulse

Impulse describes how a force applied over a period of time changes an object's momentum.

Impulse-Momentum Theorem

Definition

Impulse-momentum theorem

The impulse-momentum theorem states that the impulse on an object is equal to the change in its momentum

It is expressed by:

$J = Δ p = F Δ t$

where:

$J$ is the impulse (in N s)
$F$ is the force applied (in N)
$Δ t$ is the time interval over which the force acts (in s)

Hint

Impulse is also a vector quantity, sharing the same direction as the force applied.

Calculating Impulse

Impulse can be calculated in two ways:

Using the formula $J = F Δ t$ for constant forces.
By finding the area under a force-time graph for variable forces.

Example question

A 0.5 kg ball moving at 4 m/s hits a wall and rebounds at -2 m/s. Calculate the average force exerted by the wall on the ball.

Solution

The change in momentum is:

$Δ p = m Δ v$ $= 0.5 kg \times (- 2 m/s - 4 m/s) = - 3 kg m/s$

If the ball is in contact with the wall for 0.15 s, the average force exerted is:

$F = \frac{Δ p}{Δ t}$ $= \frac{- 3 kg m/s}{0.15 s} = - 20 N$

The negative sign indicates the force is in the opposite direction of the initial motion.

An example of the force versus time graph to calculate impulse.

Conservation of Momentum

Definition

The law of conservation of momentum

The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it.

Applying Conservation of Momentum

In a collision or explosion, the total momentum before the event equals the total momentum after the event.

Example question

Elastic Collision

Consider two blocks colliding elastically:

Block 1: mass = 2 kg, velocity = 5 m/s
Block 2: mass = 3 kg, velocity = 0 m/s

Calculate their velocities after collision.

Solution

Before Collision:

Total momentum:

$p_{initial} = (2) (5) + (3) (0) = 10 kg·m/s$

After Collision:

Using momentum conservation:

$2 u_{1} + 3 u_{2} = 10 (1)$

Using energy conservation:

$\frac{1}{2} (2) (5^{2}) = \frac{1}{2} (2) u_{1}^{2} + \frac{1}{2} (3)$

$25 = u_{1}^{2} + \frac{3}{2} u_{2}^{2} (2)$

Solving Equations:

From (1):

$u_{1} = 5 - \frac{3}{2} u_{2} (3)$

Substitute (3) into (2):

$25 = (5 - \frac{3}{2} u_{2})^{2} + \frac{3}{2} u_{2}^{2}$

Solve to get:

$u_{1} = - 1 m/s, u_{2} = 4 m/s$

Example question

Inelastic Collision

Consider two blocks colliding and sticking together:

Block 1: mass = 4 kg, velocity = 6 m/s
Block 2: mass = 8 kg, velocity = 0 m/s

Calculate their common velocity after collision.

Solution

Before Collision:

$Total momentum = (4 k g \times 6 m / s) + (8 k g \times 0 m / s) = 24 k g m / s$

After Collision:

The blocks move together with a common velocity $v$ :

$Total momentum = (4 k g + 8 k g) \times v = 12 v$

Setting the total momentum before and after the collision equal:

$12 v = 24 ⟹ v = 2 m/s$

Note

Momentum is always conserved in collisions, but kinetic energy may not be.
In this example, kinetic energy is lost as the blocks stick together.

Common Mistake

A common mistake is to assume that a larger force always produces a greater impulse.

Remember, impulse depends on both the force and the time duration.
A smaller force applied over a longer time can produce the same impulse as a larger force applied briefly.

Reflection and Connections

Momentum and impulse are foundational concepts in physics, helping us understand interactions between objects.

Self review

How is impulse related to the change in momentum?
Why is momentum conserved in an isolated system?
How do airbags and crumple zones use the impulse-momentum theorem to enhance safety?

Theory of Knowledge

How do the principles of momentum and impulse apply to other fields, such as economics or biology?
Can you think of examples where conservation laws play a role outside physics?

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A.2.2 Linear momentum and impulse

Understanding Momentum and Impulse

What is Momentum?

Definition of Momentum

Why is Momentum Important?

Impulse: Changing Momentum

Impulse-Momentum Theorem

Calculating Impulse

Conservation of Momentum

Applying Conservation of Momentum

Elastic Collision

Inelastic Collision

Reflection and Connections

You've read 2/2 free chapters this week.

Linear Momentum and Impulse