6.2 Apply the linear impulse–momentum equation | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Impulse, Momentum, and Impact

How a force acting over time changes a particle’s momentum

The linear impulse–momentum equation connects what a force does over a time interval to the change in an object's momentum. It is especially useful when a problem gives forces and a time interval but does not require the object's acceleration at every instant. In this lesson, the system is a particle, and motion is measured by an observer in an inertial reference frame. We choose axes and positive directions before assigning signs to forces and velocities. The equation is a vector relationship, so direction matters as much as magnitude.

What you will learn

Define a particle system, observer, reference frame, axes, and positive direction for an impulse–momentum problem.
Calculate impulse from a constant or time-varying force.
Apply the linear impulse–momentum equation to find an unknown velocity or impulse.
Check the units, direction, and physical reasonableness of a result.

1. Define the system and the time interval

A system is the object, or group of objects, we choose to study. For the particle problems here, the system is one object whose size and rotation are not needed to answer the question. The observer measures its motion relative to a reference frame. We use a fixed, inertial frame, such as one attached to the ground, and assume Newton’s laws apply in that frame.

Identify the initial time and final time of interest. The initial and final states specify the particle's velocities at the ends of this interval. Position tells where the particle is; displacement is its change in position; path length is the distance travelled along its route. Velocity is the rate of change of position and has a direction. The impulse–momentum equation uses velocity, not path length or displacement.

Choose coordinates to match the motion. For a straight-line problem, define one axis along the line and state which direction is positive. A velocity or force in the positive direction has a positive component; one in the opposite direction has a negative component. For a planar problem, use perpendicular

x

and

y

axes and resolve each vector into components.

\mathbf{p}=m\mathbf{v}

State the system, observer, frame, time interval, axes, and positive directions.
Treat mass as a scalar and momentum as a vector.
Keep the chosen signs consistent for forces, impulses, and velocities.

2. Relate force, impulse, and momentum

Momentum is mass times velocity. For a particle of constant mass, Newton’s second law states that the resultant external force equals the time rate of change of momentum. Integrating this relationship from the initial time to the final time gives the linear impulse–momentum equation.

The impulse of a force is its time integral. It measures the accumulated effect of that force over the interval. On a force-versus-time graph, the signed area under the curve is the impulse component. A force that changes with time must be integrated or represented by its average over the interval; using its peak value for the whole interval generally gives the wrong impulse.

The equation uses the resultant external force on the selected system. If several forces act, add their impulses as vectors. Internal forces within a chosen system are not included in the net external impulse. For a single particle, list all forces acting on it and include each relevant contribution.

In one dimension, a positive impulse increases positive momentum; a negative impulse decreases it. A negative final velocity means motion opposite the chosen positive direction. In two dimensions, apply the equation separately to the

x

and

y

components, while preserving the same time interval.

\int_{t_1}^{t_2}\sum\mathbf{F}\,dt=m\mathbf{v}_2-m\mathbf{v}_1

Impulse has units of newton-seconds and the same dimensions as momentum.
The area under a signed force–time graph gives impulse.
Use the resultant external force, not an arbitrarily selected force.

3. A reliable solution plan

Start by listing the known initial velocity, mass, force information, and time interval. Mark the unknown, such as final velocity or impulse. Draw a simple particle diagram when force directions need to be made clear. It is a force diagram: arrows represent forces on the particle, not its velocity or its momentum.

Choose the impulse–momentum method when the information naturally describes forces acting over a time interval and the desired result is a momentum change or velocity. If a force is constant, its impulse is the force multiplied by the duration. If it varies, integrate its components or calculate the signed area under its force–time graph.

Write the vector equation first, then resolve it into components if needed. Keep units attached when substituting: mass in kilograms, velocity in metres per second, force in newtons, and time in seconds. Since one newton is one kilogram metre per second squared, impulse and momentum both have units of kilogram metres per second.

Check the result before accepting it. Confirm the sign against the chosen positive direction, verify units, and ask whether the momentum change points with the net impulse. If the net impulse is zero, the final velocity must equal the initial velocity. These checks can expose a sign or arithmetic error.

\mathbf{J}=\int_{t_1}^{t_2}\sum\mathbf{F}\,dt

Use a force diagram to clarify directions, then use signed components in the equation.
Solve symbolically before substituting values.
Check dimensions, direction, and the zero-impulse limiting case.

Worked example

Constant force over a time interval

A

2.0\,\mathrm{kg}

cart moves along a straight, level track. In a ground-fixed inertial frame, its initial velocity is

1.0\,\mathrm{m/s}

to the right. A constant net horizontal force of

10\,\mathrm{N}

acts to the right for

3.0\,\mathrm{s}

. Find its final velocity.

Cart and net force

The force and initial velocity are both in the positive direction.

Define states and signs
The system is the cart, the observer is fixed to the ground, and the reference frame is inertial. Let right be positive. The initial state has velocity $v_1=+1.0\,\mathrm{m/s}$ ; the final state is after $3.0\,\mathrm{s}$ .
Calculate the impulse
The net force is constant and positive, so its impulse is the force multiplied by the duration.
$J=(10\,\mathrm{N})(3.0\,\mathrm{s})=30\,\mathrm{N\,s}$
Apply the equation
Use the one-dimensional equation and solve for the final velocity. The initial momentum is positive, and the positive impulse increases it.
$v_2=v_1+\frac{J}{m}=1.0+\frac{30}{2.0}=16\,\mathrm{m/s}$

Answer: The final velocity is

16\,\mathrm{m/s}

to the right.

Check: Impulse has units

\mathrm{N\,s}=\mathrm{kg\,m/s}

, so dividing by mass gives velocity. The positive force produces a positive momentum change, consistent with the answer.

Worked example

Impulse from a triangular force–time history

A

4.0\,\mathrm{kg}

particle moves on a straight line with initial velocity

3.0\,\mathrm{m/s}

to the right. During a

2.0\,\mathrm{s}

interval, the net force is to the left: its magnitude rises linearly from zero to

12\,\mathrm{N}

at the midpoint, then falls linearly to zero. Find the final velocity.

Particle and force direction

The force is opposite the initial velocity; the force magnitude varies with time.

Define states and signs
The system is the particle, viewed in a ground-fixed inertial frame. Take right as positive. Initially $v_1=+3.0\,\mathrm{m/s}$ , and the final state is at the end of the stated $2.0\,\mathrm{s}$ force history.
Find signed impulse
The force–time graph is a triangle with base $2.0\,\mathrm{s}$ and height $12\,\mathrm{N}$ . Its area is negative because the force points left.
$J=-\frac{1}{2}(2.0\,\mathrm{s})(12\,\mathrm{N})=-12\,\mathrm{N\,s}$
Solve for final velocity
The negative impulse reduces the particle's positive momentum. Apply the linear impulse–momentum equation and divide by the mass.
$v_2=3.0+\frac{-12}{4.0}=0\,\mathrm{m/s}$

Answer: The particle is momentarily at rest at the end of the interval:

v_2=0\,\mathrm{m/s}

.

Check: The initial momentum is

12\,\mathrm{kg\,m/s}

to the right, exactly cancelled by the

12\,\mathrm{N\,s}

leftward impulse. The result is consistent with both the signs and units.

Worked example

A ball reverses direction

A

0.15\,\mathrm{kg}

ball travels to the right at

20\,\mathrm{m/s}

. During contact with a surface, the net force on the ball is to the left and gives it an impulse of magnitude

4.0\,\mathrm{N\,s}

. In a ground-fixed inertial frame, find the ball's velocity immediately after contact. Treat the ball as the system and neglect any other impulse during contact.

Ball and contact impulse

The impulse acts opposite the ball's initial direction.

Set the frame and states
The system is the ball; an observer fixed to the ground measures its motion. Use a ground-fixed inertial frame and take right as positive. The initial velocity is $+20\,\mathrm{m/s}$ ; the final state is immediately after contact.
Assign the impulse sign
The stated impulse points left, so its component is negative in the chosen coordinate system.
$J_x=-4.0\,\mathrm{N\,s}$
Find the final velocity
Add the signed impulse to the initial momentum and divide by the ball's mass. A negative result indicates motion left, not a negative speed.
$v_{2x}=20+\frac{-4.0}{0.15}=-6.7\,\mathrm{m/s}$

Answer: Immediately after contact, the ball moves at approximately

6.7\,\mathrm{m/s}

to the left.

Check: The impulse needed to stop the ball is

m v_1=(0.15)(20)=3.0\,\mathrm{N\,s}

. The actual leftward impulse is larger, so reversal is expected. The remaining momentum is approximately

1.0\,\mathrm{kg\,m/s}

to the left.

Common mistakes and how to avoid them

Using the force magnitude without its direction.

Correction: Choose a positive direction first, then assign each force and impulse a signed component.

Using the peak of a varying force as though it acted throughout the interval.

Correction: Calculate the signed area under the force–time graph, or integrate the force over time.

Confusing impulse with force.

Correction: Force is measured in newtons; impulse includes time and is measured in newton-seconds.

Treating a negative velocity as a negative speed.

Correction: Velocity's sign gives direction relative to the chosen axis; speed is a nonnegative magnitude.

Including only one force when the equation requires the resultant external impulse.

Correction: Account for all relevant external forces acting on the selected system over the time interval.

Lesson summary

Impulse is the time integral of the resultant external force.
The linear impulse–momentum equation equates net impulse to the change in momentum.
Use signed components so that force, impulse, and velocity directions remain consistent.
Verify units and check whether the impulse changes momentum in the expected direction.

Check your understanding

Question 1

A particle has zero resultant impulse over an interval. What follows from the linear impulse–momentum equation?

Its final velocity equals its initial velocity.
Its final velocity must be zero.
Its path length over the interval must be zero.
correctIndex 0

Show answer and explanation

Its final velocity equals its initial velocity.

Zero net impulse means zero change in momentum. For constant mass, the initial and final velocities are equal.

Question 2

A net impulse of

-6\,\mathrm{N\,s}

acts on a particle whose positive direction is right. What is the direction of its momentum change?

Right
Left
It cannot be determined without knowing the initial velocity.
correctIndex 1

Show answer and explanation

Left

The momentum change equals the net impulse, so it points left. Initial velocity is not needed to determine the direction of the change.

Question 3

A constant force of

5\,\mathrm{N}

acts in the positive direction for

0.40\,\mathrm{s}

. What impulse does it give?

$2.0\,\mathrm{N\,s}$ in the positive direction
$12.5\,\mathrm{N\,s}$ in the positive direction
$2.0\,\mathrm{N\,s}$ in the negative direction
correctIndex 0

Show answer and explanation

2.0\,\mathrm{N\,s}

in the positive direction

For a constant force, impulse is force multiplied by duration:

5(0.40)=2.0\,\mathrm{N\,s}

, in the force's positive direction.

Key terms

System: The object or group of objects selected for analysis.
Momentum: The vector quantity equal to mass times velocity.
Impulse: The time integral of force; it equals the change in momentum for the selected system.
Resultant external force: The vector sum of the forces from outside the selected system.
Inertial reference frame: A reference frame in which Newton's laws apply without adding fictitious forces.

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6.2 · Apply the linear impulse–momentum equation

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

How a force acting over time changes a particle’s momentum

What you will learn

1. Define the system and the time interval

2. Relate force, impulse, and momentum

3. A reliable solution plan

Constant force over a time interval

Impulse from a triangular force–time history

A ball reverses direction

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review