# Lesson Explainer: Impulse and Momentum Mathematics

In this explainer, we will learn how to relate the impulse on a body to the change of momentum of the body.

With Newton’s second law, we know that when a constant force acts on a body, it causes it to accelerate, which means that the body’s velocity changes. We will now introduce a new quantity, the impulse, which takes into account the time over which the force acts.

### Definition: The Impulse

The impulse, , is the integral of a force, , over a time interval, , for which the force acts:

Let us now define the momentum of a body.

### Definition: The Momentum of a Body

The momentum of a body is given by where is the mass of the body and is the velocity of the body.

We see that the momentum of a body is proportional to both its mass and its velocity.

Furthermore, the momentum of a body is a vector quantity that has the same direction as the body’s velocity. When a net force, , acts on a body of constant mass, , according to Newton’s second law of motion, the body’s acceleration is proportional to the force and inversely proportional to the mass, as expressed by the formula or where is the acceleration of the body.

Since , we have

And as and is constant, the above relation can be written as

Integrating both sides of the equation over the time interval gives where is the velocity of the body before the force acts on it and is the velocity of the body when the force has stopped acting on it.

### Property: Impulse and Change of Momentum

For a constant-mass body, the impulse produced by the action of a force over an interval of time equals the change in the momentum of the body:

Let us suppose that a body of constant mass moving at a constant speed collides with and rebounds from a barrier.

Perhaps the most intuitive case of an impulse on a body is that for which, after such a collision, the body moves at a constant speed in the opposite direction to that in which it moved before the collision. The following figure qualitatively illustrates such a case.

For such cases, the time in which force acts on the body does not need to be known, only the initial and final velocities of the body.

Let us look at an example of such a case.

### Example 1: Finding the Impulse Exerted on a Sphere Moving on a Horizontal Smooth Plane after Collision with a Barrier

A smooth sphere of mass 1‎ ‎412 g was moving horizontally in a straight line at 13.5 m/s when it hit a smooth vertical wall and rebounded at 9 m/s. Determine the magnitude of the impulse exerted on the sphere.

The impulse of the force that acts on the sphere as it rebounds changes the momentum of the sphere. The change in the momentum can be determined from the momentum of the sphere before and after it rebounds.

Before rebounding, the momentum of the sphere is given by

Since the sphere is moving in a horizontal line, the component of the momentum along the horizontal axis, , is where is the component of the velocity of the body along the horizontal axis before it hit the wall.

The unit kilogram-metres per second (kg⋅m/s) is typically used for momentum, so the mass of the sphere is converted from 1‎ ‎412 grams to 1.412 kilograms to be consistent with this.

The direction of the velocity of the sphere before rebounding is chosen to be positive, giving an initial momentum for the sphere of

After rebounding, the component of the momentum of the sphere along the horizontal axis is given by where is the component of the velocity of the sphere along the horizontal axis after rebounding. But as the direction of the sphere has reversed, the velocity now has a negative value, given by

The impulse of the force is equal to the change in the momentum of the sphere, given by

The direction of the impulse is opposite to the direction of the initial velocity of the sphere. The magnitude of the change in the momentum of the sphere equals the magnitude of the impulse on the sphere, 31.77 kg⋅m/s.

It is worth noting that the time interval for which the sphere was in contact with the barrier is not given and does not need to be known to determine the impulse, as the impulse is equal to the change in momentum, which can be determined from the change in the velocity of the sphere.

Another case of an impulse on a body is that of a body traveling in a resistive medium. The impulse from the resistive medium accelerates the body in the opposite direction to the direction of its velocity but can only result in the velocity of the body tending to some nonnegative minimum value in its initial direction rather than making the body reverse. The impulse from the medium is applied to the body throughout the motion of the body in the medium. The following figure qualitatively illustrates such a case.

Let us now look at an example where a resistive medium produces an impulse on a body.

### Example 2: Finding the Change in the Momentum of a Body as a Result of the Resistance of Water

A body of mass 5 kg fell vertically from rest. It reached the water’s surface after 2.2 seconds. Then, as it moved through the water, it descended vertically with an average speed such that it covered 3.9 m in 1.5 seconds. Find the magnitude of the change in its momentum as a result of the resistance of the water.

In this question, the body moves in a straight line, which means that vector quantities such as forces, displacement, velocity, and acceleration are one-dimensional (along the motion axis). We can therefore use the one-dimensional equations of motion involving the single components of these vectors. We take here the downward direction as positive. The velocity of the body when it reaches the water is given by where is the initial vertical velocity of the body, is its vertical acceleration, and is the time for which it accelerates.

The body is initially at rest, so is zero. The value of is the acceleration due to gravity on Earth, which is approximately equal to 9.8 m/s2. Considering this, we have that

When the body moves in the water, the velocity of the body changes. The question states a vertically downward displacement of the body of 3.9 m occurs in a time interval of 1.5 seconds once the body reaches the water.

As the body descends 3.9 m in 1.5 seconds, it has an average velocity in that time interval given by

The impulse on the body is, therefore, given by

Substituting the values of the mass, the velocity of the body when it reaches the water, and the average velocity of the body in the water, we obtain

The impulse on the body is negative even though the body continues to move in the same direction before and after the force acts on it. The force producing the impulse reduces the velocity of the body in the direction of its initial velocity, and so the impulse has the opposite sign to that of the initial velocity.

The magnitude of the change in the momentum of the body equals the magnitude of the impulse on the body, 94.8 kg⋅m/s.

In the first example, an impulse was delivered by a force that acted during the contact between a body and a barrier. The force acted for an undetermined time interval. In the second example, an impulse was delivered by a force that acted due to the resistance of a medium. The force acted throughout the motion of the body in the medium.

It is a valid method to posit a definite time interval for the contact between a body and a barrier, even if the time interval is small. A value for a time interval for contact between a body and a barrier is actually necessary to determine the force that acts on the body due to contact with the barrier.

Suppose that a body travels from a point to a point and then to a point , as shown in the following figure.

It is undeniable that the body must accelerate at some point in its motion, as it reverses direction. Knowing only the displacements of and of from , the magnitude of the acceleration of the body cannot be determined. The rate of change of displacement must be known to determine the acceleration of the body.

Similarly, the magnitude of the force acting on a body to change the momentum of a body cannot be determined only by reference to the change of the momentum of the body alone; the rate of change of the momentum of the body must be known to determine the magnitude of the force.

If the time interval over which the force acts is known, the average force acting over that time interval can be determined. Let us look at an example where an average force over a time interval is determined.

### Example 3: Finding the Force of Impact of a Body Falling Vertically on the Ground and Rebounding

A sphere of mass 83 g fell vertically from a height of 8.1 m onto a section of horizontal ground. It rebounded and reached a height of 3.6 m. Given that the duration of the impact was 0.42 seconds, and the acceleration due to gravity is 9.8 m/s2, find the average impact force to the nearest two decimal places.

In this question, the sphere moves in a straight line (vertically), which means that vector quantities such as forces, displacement, velocity, and acceleration are one-dimensional (along the motion axis). We can therefore use the one-dimensional equations of motion involving the single components of these vectors.

For this question to be solvable, we must assume that the sphere falls from rest. By making this assumption, we are able to determine the velocity of the sphere as it reaches the ground, using the formula where is zero and is 9.8 m/s2. We note that vertically downward acceleration is taken as positive.

The value of obtained is given by

The same formula can be used to determine the upward velocity of the sphere at the ground, as the velocity of the sphere is instantaneously zero when it reaches its maximum rebound height. When the sphere rebounds, we can define vertically upward as the positive direction.

The value of is given by

We have defined the vertically upward direction as positive; hence, is negative. We have then that

The change in the velocity of the sphere due to the force of its impact is given by

The initial velocity is in the opposite direction to the final velocity, and the final velocity has been defined as positive; hence, the initial velocity is defined as negative.

The momentum of the sphere is proportional to its mass. The unit for momentum required where time is in seconds and displacement is in metres that can be directly used to determine a force in newtons is kilogram-metres per second (kg⋅m/s), and so the 83-gram mass is converted to a 0.083 kg mass. The change in the momentum of the ball is given by

Using the formula and recalling that impulse equals the change in momentum, we see that where the question states that the duration of the impact, , is 0.42 s.

The value of the average force is given by

Let us now consider an example where the impulse due to the action of forces expressed in vector component form is determined.

### Example 4: Finding the Impulse on a Body Where Three Forces in Vector Form Are Acting on It for a Given Time

Three forces, , , and , where , , and are three mutually perpendicular unit vectors, acted on a body for 3 seconds. Find the magnitude of their combined impulse on the body.

Three forces act on the body, and these can be added to determine the resultant force on the body.

In the direction of , the net force is given by

In the direction of , the net force is given by

In the direction of , the net force is given by

The resultant of these forces is, therefore, the force , given by

The magnitude of is given by

The impulse due to is given by

The time interval for which acts is 3 seconds, so the magnitude of the impulse is given by

The same result could have been obtained by multiplying each component of by , as follows:

The impulse of force is the product of the force and the time for which the force acts, and so the impulse of a force corresponds to the area under the line of a force–time graph. Let us look at an example involving a force–time graph.

### Example 5: Calculating the Magnitude of the Impulse on a Body given Its Force–Time Graph

The given figure shows a force–time graph for a force acting in a constant direction on a body moving along a smooth horizontal plane. Using the information provided, calculate the magnitude of the force’s impulse.

The magnitude of the impulse of the force is the product of the force and the time for which it acts.

The area of a trapezoid is given by where and are the lengths of the parallel sides and is the perpendicular distance between them.

By inspection of the graph, we see that

The area of the trapezoid is given by

The impulse of the force is equal to this area, and so

Let us now look at an example where the impulse of a force is determined, where the force varies as a function of time.

### Example 6: Finding the Magnitude of the Impulse of a Force Acting on a Body for a Given Time Using Force–Time Expression

The given figure shows a force–time graph. At time seconds, where , the force is given by . Find the impulse over the first four seconds.

The impulse of the force in the first four seconds corresponds to the area between the line representing the change in the value of the force with time and the force and time axes between and . This area can be found by integration, using the formula where in this case

Using these values, we find that

This integral can be solved by substitution.

To do this, we let

We see then that

The limits of the integral can be expressed in terms of , such that when and when

We can then see that

### Key Points

• The impulse, , is the integral of a force, , over a time interval, , in which the force acts:
• For a constant force, , acting in a time interval , the impulse is given by
• For a force with an average value, , acting in a time interval , the impulse is given by
• For a constant-mass body, we have where is the velocity of the body before the force acts on it and is the velocity of the body when the force has stopped acting on it.