C2.5 Solve momentum, impulse, mass, velocity, and kinetic-energy problems

Ontario Grade 12 Physics

Energy and Momentum

SPH4U C2.5 — solving for momentum, impulse, mass, velocity, and kinetic energy

In earlier physics courses, you used velocity to describe motion and mass to describe an object's inertia. This lesson combines those ideas with momentum, impulse, and kinetic energy. Before calculating, identify the physical system, the reference frame, and a positive direction. The system is the object or group of objects being considered. The reference frame is the viewpoint used to describe their motion. A positive direction is the chosen direction that counts as positive in calculations. These choices make vector signs meaningful.

What you will learn

Distinguish vector quantities, which have direction, from scalar quantities, which have magnitude only.
Use momentum and kinetic-energy relationships to solve for mass or velocity.
Use impulse to connect a force acting over a time interval with a change in momentum.
Show units, direction, significant figures, and a reasonableness check.

1. Choose a system and describe motion

A scalar has magnitude only. Mass, time, and kinetic energy are scalars. A vector has both magnitude and direction. Velocity, momentum, impulse, and force are vectors. In one-dimensional problems, choose a positive direction and use a positive or negative sign to show direction. For example, if right is positive, a velocity to the left is negative.

Use the same reference frame for all velocities in a problem. For ordinary classroom problems, this is often the ground or the room. State the system clearly: for example, “the cart” when finding its momentum, or “the cart and ball” when considering both together. If a problem asks for a change in momentum, use the momentum before and after for that same system.

Keep units in SI form. Mass is measured in kilograms, velocity in metres per second, time in seconds, force in newtons, momentum and impulse in kilogram metres per second or newton seconds, and kinetic energy in joules.

A sign gives a vector's direction relative to the chosen positive direction.
A negative velocity or momentum does not mean a negative speed or mass.
Speed is the magnitude of velocity.

2. Momentum and kinetic energy

Momentum describes motion using both mass and velocity. A more massive object or an object moving faster has a greater momentum magnitude. Momentum points in the same direction as velocity. In one dimension, retain the velocity's sign when calculating momentum.

Kinetic energy is the energy of motion. It depends on mass and on the square of speed, so it is never negative. Because velocity is squared, the direction of travel does not affect kinetic energy. Do not use a negative velocity as a reason to report negative kinetic energy.

These relationships can be rearranged with ordinary algebra. To find mass from momentum, divide momentum by velocity. To find velocity from momentum, divide by mass. To find speed from kinetic energy, rearrange the kinetic-energy relationship and take the positive square root. Kinetic energy gives speed, not direction; direction must come from other information in the problem.

\vec p=m\vec v,\qquad E_k=\frac{1}{2}mv^2

Momentum is a vector; kinetic energy is a scalar.
Momentum units can be written as $\mathrm{kg\,m/s}$ .
The joule is equivalent to $\mathrm{kg\,m^2/s^2}$ .

3. Impulse and change in momentum

Impulse describes how a force acting over a time interval changes momentum. In this relationship, use the net force along the chosen direction. A net force is the overall force after the forces on the system are combined. When the force is constant, multiply it by the time interval. When a problem gives an average net force, use that average for the interval.

The change in momentum is final momentum minus initial momentum. Because momentum is a vector, the subtraction must keep the signs. A force in the negative direction gives a negative impulse for the chosen convention. The units of impulse, newton seconds, are equivalent to the units of momentum.

Impulse is useful when the problem gives force and time, or asks for the force or time needed to produce a stated momentum change. It does not mean force and momentum are the same quantity: force is measured in newtons, while momentum is measured in kilogram metres per second.

\vec J=\vec F_{\mathrm{net}}\Delta t=\Delta\vec p=\vec p_f-\vec p_i

Use the net force component in the direction being analysed.
Use final minus initial momentum to find the change.
Keep signs through the calculation, then state the direction in words.

4. A reliable solving method

First name the system and reference frame. Choose and state the positive direction. List known values with units and identify the unknown. Next, choose the relationship that contains the known values and the unknown. Rearrange it before substituting when that makes the steps clearer.

Substitute values with their units. For vector quantities in one dimension, include the signs. Keep extra digits during the calculation, then round the final value to a sensible number of significant figures based on the given data. State a direction for any vector answer.

Finally, check whether the units match the requested quantity and whether the result makes sense. For example, momentum should grow in magnitude if mass increases while velocity stays the same. Kinetic energy should not be negative. A very large or very small result may be reasonable, but it should follow from the given mass and speed.

Write the relationship before inserting numbers.
Do not discard direction signs before calculating a momentum change.
Check both the units and the physical meaning of the result.

Worked example

1. Momentum and kinetic energy from velocity

A 0.80 kg ball moves at 6.0 m/s to the right. In the ground frame, take right as positive. Find its momentum and kinetic energy.

Define the system and known values
The system is the ball. Its mass is $0.80\,\mathrm{kg}$ and its velocity is $+6.0\,\mathrm{m/s}$ . The positive sign means right.
Calculate momentum
Momentum is mass times velocity. The positive result means the momentum points right.
$p=(0.80\,\mathrm{kg})(+6.0\,\mathrm{m/s})=+4.8\,\mathrm{kg\,m/s}$
Calculate kinetic energy
Kinetic energy uses speed squared, so its value is positive.
$E_k=\frac{1}{2}(0.80\,\mathrm{kg})(6.0\,\mathrm{m/s})^2=14.4\,\mathrm{J}$

Answer: The ball's momentum is

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

to the right. Its kinetic energy is

14\,\mathrm{J}

to two significant figures.

Check: Momentum has units of mass times velocity. Kinetic energy has units of mass times velocity squared, which is a joule. Both values are reasonable for a moving ball with the stated mass and speed.

Worked example

2. Impulse changes momentum

A 0.50 kg cart initially moves right at 2.0 m/s. A constant net force of 3.0 N acts left for 0.40 s. Find the cart's final velocity. Use the ground frame and take right as positive.

Set the signs and identify the system
The system is the cart. Its initial velocity is positive, and the leftward net force is negative. The initial momentum is found from mass times initial velocity.
$p_i=(0.50\,\mathrm{kg})(+2.0\,\mathrm{m/s})=+1.0\,\mathrm{kg\,m/s}$
Find the impulse
The force is constant, so impulse is net force multiplied by the time interval. Its negative sign indicates a leftward impulse.
$J=(-3.0\,\mathrm{N})(0.40\,\mathrm{s})=-1.2\,\mathrm{N\,s}$
Find final momentum and velocity
Impulse equals final momentum minus initial momentum. Add the impulse to the initial momentum, then divide by mass to obtain final velocity.
$p_f=p_i+J=+1.0-1.2=-0.20\,\mathrm{kg\,m/s}$
$v_f=\frac{-0.20\,\mathrm{kg\,m/s}}{0.50\,\mathrm{kg}}=-0.40\,\mathrm{m/s}$

Answer: The final velocity is

0.40\,\mathrm{m/s}

to the left.

Check: The impulse has units of newton seconds, equivalent to momentum units. The cart first loses its rightward momentum and then moves left, consistent with the negative final velocity.

Worked example

3. Find speed from kinetic energy

A 1.5 kg object has kinetic energy of 27 J. Find its speed. The reference frame is the ground. The direction is not specified.

Identify the known values
The system is the object. Its mass and kinetic energy are known. Since kinetic energy contains speed squared, the calculation can determine speed but not direction.
Rearrange the relationship
Solve the kinetic-energy relationship for speed by multiplying by two, dividing by mass, and taking the positive square root.
$v=\sqrt{\frac{2E_k}{m}}$
Substitute and evaluate
Use joules and kilograms. The result is a speed, so report a positive magnitude.
$v=\sqrt{\frac{2(27\,\mathrm{J})}{1.5\,\mathrm{kg}}}=6.0\,\mathrm{m/s}$

Answer: The object's speed is

6.0\,\mathrm{m/s}

. Its direction cannot be determined from kinetic energy alone.

Check: A joule is equivalent to

\mathrm{kg\,m^2/s^2}

, so dividing by kilograms and taking the square root gives metres per second. Substitution confirms that this speed gives

27\,\mathrm{J}

.

Common mistakes and how to avoid them

Treating momentum as a scalar and dropping a negative sign.

Correction: Momentum has direction. Choose a positive direction and keep signs during calculations.

Using initial momentum minus final momentum for the change.

Correction: The change is final momentum minus initial momentum.

Reporting negative kinetic energy because velocity is negative.

Correction: Kinetic energy uses velocity squared, so it is non-negative.

Claiming that kinetic energy alone gives the direction of travel.

Correction: Kinetic energy gives speed. Use other information to determine direction.

Leaving out units or reporting too many digits.

Correction: Carry units through substitutions and round the final result to match the precision of the data.

Lesson summary

Momentum is a vector calculated from mass and velocity.
Kinetic energy is a scalar calculated from mass and speed squared.
Impulse equals the change in momentum and, for constant net force, equals net force multiplied by elapsed time.
State the system, reference frame, and positive direction before solving.
Check units, signs, significant figures, and whether the answer is physically reasonable.

Check your understanding

Question 1

A 2.0 kg object moves at 3.0 m/s left. If right is positive, what is its momentum?

$+6.0\,\mathrm{kg\,m/s}$
$-6.0\,\mathrm{kg\,m/s}$
$-1.5\,\mathrm{kg\,m/s}$
$+1.5\,\mathrm{kg\,m/s}$

Show answer and explanation

-6.0\,\mathrm{kg\,m/s}

Left is negative, so

p=mv=(2.0\,\mathrm{kg})(-3.0\,\mathrm{m/s})=-6.0\,\mathrm{kg\,m/s}

.

Question 2

A net force of

+5.0\,\mathrm{N}

acts for

0.20\,\mathrm{s}

. What impulse does it provide?

$+1.0\,\mathrm{N\,s}$
$+25\,\mathrm{N\,s}$
$-1.0\,\mathrm{N\,s}$
$+0.040\,\mathrm{N\,s}$

Show answer and explanation

+1.0\,\mathrm{N\,s}

For constant force,

J=F_{\mathrm{net}}\Delta t=(+5.0\,\mathrm{N})(0.20\,\mathrm{s})=+1.0\,\mathrm{N\,s}

.

Question 3

An object has mass

4.0\,\mathrm{kg}

and kinetic energy

32\,\mathrm{J}

. What is its speed?

$2.0\,\mathrm{m/s}$
$4.0\,\mathrm{m/s}$
$8.0\,\mathrm{m/s}$
$16\,\mathrm{m/s}$

Show answer and explanation

4.0\,\mathrm{m/s}

Rearrange to get

v=\sqrt{2E_k/m}

. Thus

v=\sqrt{2(32\,\mathrm{J})/(4.0\,\mathrm{kg})}=4.0\,\mathrm{m/s}

.

Key terms

System: The object or group of objects selected for a physics analysis.
Reference frame: The viewpoint relative to which position and motion are described.
Momentum: A vector quantity equal to an object's mass multiplied by its velocity.
Impulse: A vector quantity equal to the change in momentum; for constant net force, it is net force multiplied by elapsed time.
Kinetic energy: The scalar energy associated with motion.
Net force: The overall force on a system after its forces are combined.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation C2.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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C2.5 · Solve momentum, impulse, mass, velocity, and kinetic-energy problems

Ontario Grade 12 Physics

SPH4U C2.5 — solving for momentum, impulse, mass, velocity, and kinetic energy

What you will learn

1. Choose a system and describe motion

2. Momentum and kinetic energy

3. Impulse and change in momentum

4. A reliable solving method

1. Momentum and kinetic energy from velocity

2. Impulse changes momentum

3. Find speed from kinetic energy

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review