Momentum Revision Notes

Momentum (symbol p) is defined as the product of an object's mass (m) and its velocity (v). It quantifies the quantity of motion an object possesses.

The formula for momentum is: p = mv, where: p is momentum in kilogram metres per second (kg·m/s), m is mass in kilograms (kg), and v is velocity in metres per second (m/s).

Since velocity is a vector quantity (it has both magnitude and direction), momentum is also a vector quantity, meaning it has both magnitude and direction.

• Momentum (p) = mass (m) × velocity (v)
• Units: kg·m/s
• Momentum is a vector quantity

Mass is measured in kilograms (kg), velocity in metres per second (m/s), so momentum has units of kilogram metres per second (kg·m/s).

Because velocity has direction, momentum also has direction. For example, an object moving east with a certain speed has momentum directed east.

When representing momentum, include both magnitude and direction, e.g., 10 kg·m/s east.

• Mass unit: kilogram (kg)
• Velocity unit: metres per second (m/s)
• Momentum unit: kilogram metre per second (kg·m/s)
• Momentum direction is the same as velocity direction

To calculate momentum, multiply the mass of the object by its velocity.

Example: A car of mass 800 kg is moving at 20 m/s east. Calculate its momentum.

Using p = mv: p = 800 kg × 20 m/s = 16,000 kg·m/s east.

• Step 1: Write down the formula p = mv
• Step 2: Substitute values: p = 800 × 20
• Step 3: Calculate: p = 16,000 kg·m/s
• Step 4: Include direction: east

A frequent mistake is to confuse mass and weight; remember mass is in kilograms, weight is a force in newtons.

Another error is ignoring the direction of velocity, thus ignoring momentum's vector nature.

Also, ensure units are consistent: mass in kg, velocity in m/s, so momentum is in kg·m/s.

• Do not use weight (N) instead of mass (kg) in calculations.
• Always include direction when stating momentum.
• Check units before calculating to avoid errors.

To draw a momentum vector diagram for an object: start with a straight line representing the object's path.

Draw an arrow along this line pointing in the direction of the velocity.

Label the arrow with the momentum value and units, e.g., 'p = 16,000 kg·m/s east'.

• Draw object's path as a straight line
• Draw arrow along path indicating direction
• Label arrow with momentum magnitude and units

Impulse is a vector quantity defined as the product of the average force (F) applied to an object and the time interval (Δt) over which the force acts.

Mathematically, impulse (J) is expressed as: J = F Δt.

The SI unit of impulse is newton second (N·s), derived from force in newtons (N) and time in seconds (s).

• Impulse has both magnitude and direction, matching the direction of the applied force.
• Impulse represents the effect of a force acting over a finite time, rather than an instantaneous force.

Momentum (p) is the product of an object's mass (m) and velocity (v): p = mv.

When a force acts on an object, it changes the object's momentum.

Impulse equals the change in momentum: J = Δp = p_final − p_initial = m v_final − m v_initial.

• This relationship links force, time, and motion changes.
• It is a direct consequence of Newton's second law in its momentum form.

The impulse-momentum equation is: F Δt = Δp = m v_final − m v_initial.

This equation can be rearranged to find any unknown quantity when the others are known.

It is especially useful in analysing collisions, where forces act over short time intervals to change velocities.

• Example calculation:
• A 2.0 kg ball initially moving at 3.0 m/s is brought to rest by a force applied for 0.5 s. Calculate the average force.
• Solution: Δp = m(v_final − v_initial) = 2.0 × (0 − 3.0) = −6.0 kg·m/s.
• Impulse = F Δt = Δp → F = Δp / Δt = (−6.0) / 0.5 = −12 N.
• The negative sign indicates the force acts opposite to the initial motion direction.

A force-time graph can represent impulse as the area under the force versus time curve.

The x-axis is time (s), and the y-axis is force (N).

The area under the curve between two times gives the impulse delivered during that interval.

• If force is constant, the graph is a rectangle; area = force × time.
• If force varies, the area can be found by counting squares or using integration for precise calculation.

Impulse is important in safety devices like airbags and helmets, which increase the time over which a force acts, reducing the force magnitude and injury risk.

In experiments, measuring force and time accurately requires sensors such as force sensors and timers.

Repeated measurements improve accuracy and help estimate uncertainties.

• Ensure force sensors are calibrated before use.
• Use protective equipment when testing collisions to prevent injury.
• Record multiple trials and calculate average impulse for reliability.

Momentum (symbol p) is a vector quantity defined as the product of an object's mass (m) and its velocity (v):

\[ p = m v \] where p is momentum in kilogram metres per second (kg·m/s), m is mass in kilograms (kg), and v is velocity in metres per second (m/s).

Because velocity is a vector, momentum has both magnitude and direction. The direction of momentum is the same as the direction of the velocity. When solving problems, it is essential to assign a positive or negative sign to velocities to indicate direction consistently.

• Momentum is measured in kg·m/s.
• Momentum direction depends on velocity direction.
• Use a consistent sign convention for directions.

In a closed system where no external forces act, the total momentum before an event (such as a collision or explosion) is equal to the total momentum after the event. This is the principle of conservation of momentum:

\[ \text{Total momentum before} = \text{Total momentum after} \]

Mathematically, for two objects interacting in one dimension:

• \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]
• where subscripts i and f denote initial and final velocities respectively.
• This equation assumes no external forces and that the system is isolated.

To solve problems using conservation of momentum: 1. Identify the system and ensure it is closed and isolated (no external forces). 2. Assign a positive direction and apply it consistently to all velocities. 3. Write down the conservation of momentum equation for the system. 4. Substitute known values of masses and velocities. 5. Rearrange the equation to find the unknown quantity (velocity or mass). 6. Calculate the answer with correct units and direction.

Example: Two objects collide and stick together. Object 1 has mass 2.0 kg and velocity 3.0 m/s to the right; object 2 has mass 3.0 kg and velocity 0 m/s. Find their common velocity after collision.

Solution: Using right as positive, initial momentum: \( p_i = 2.0 \times 3.0 + 3.0 \times 0 = 6.0 \, \text{kg·m/s} \). After collision, combined mass = 5.0 kg, velocity = v_f. Conservation of momentum: \( 6.0 = 5.0 \times v_f \) so \( v_f = \frac{6.0}{5.0} = 1.2 \, \text{m/s} \) to the right.

• Define positive direction and assign signs to velocities.
• Write conservation of momentum equation.
• Substitute known values and solve for unknown.
• Include units and direction in final answer.

Forgetting that momentum is a vector and neglecting direction leads to incorrect answers. Always assign positive and negative signs based on chosen direction.

Applying conservation of momentum when external forces act significantly invalidates the principle. Confirm the system is isolated.

Mixing units or forgetting to include units in answers reduces clarity and can cause mistakes. Always include units in calculations and final answers.

• Use consistent sign conventions for direction.
• Check for external forces before applying conservation.
• Always include units in calculations and answers.

To draw a clear diagram for a one-dimensional collision problem: 1. Draw a horizontal line representing the path of motion. 2. Represent each object as a labelled box or circle on the line.

3. Use arrows to show initial velocities with length proportional to speed and direction indicated by arrow direction.

4. After collision, draw arrows for final velocities similarly, showing changes in speed and direction if any. 5. Label masses and velocities clearly with units. This visual helps in assigning directions and understanding momentum changes.

• Draw horizontal line for motion axis.
• Represent objects as labelled shapes.
• Use arrows for velocity direction and magnitude.
• Label masses and velocities with units.

Momentum (\( p \)) is a vector quantity defined as the product of an object's mass (\( m \)) and its velocity (\( v \)): \( p = mv \). It has both magnitude and direction, measured in kilogram metres per second (\( \mathrm{kg\,m/s} \)).

Resultant force (\( F \)) is the single force that has the same effect as all the forces acting on an object combined. It causes a change in the object's motion, specifically a change in its momentum.

The key relationship between force and momentum is that the resultant force acting on an object equals the rate of change of its momentum with respect to time. This is expressed mathematically as: \[ F = \frac{\Delta p}{\Delta t} \] where \( \Delta p \) is the change in momentum and \( \Delta t \) is the time interval over which this change occurs.

• Momentum \( p = mv \) (vector, units \( \mathrm{kg\,m/s} \))
• Resultant force causes change in momentum
• Resultant force \( F = \frac{\Delta p}{\Delta t} \) (units \( \mathrm{N} = \mathrm{kg\,m/s^2} \))

Starting from Newton's second law in its original form, the force acting on an object is equal to the rate of change of its momentum. If an object's momentum changes from \( p_1 \) to \( p_2 \) over a time interval \( \Delta t = t_2 - t_1 \), then:

\[ F = \frac{p_2 - p_1}{t_2 - t_1} = \frac{\Delta p}{\Delta t} \]

This formula applies to any situation where the momentum changes, including changes in speed or direction, since momentum is a vector. The force and momentum change must be considered as vectors, but in this objective, focus is on magnitude and scalar calculations unless otherwise specified.

• Force equals rate of change of momentum
• Applies to changes in speed or direction
• Force and momentum are vector quantities

Momentum has units of kilogram metres per second (\( \mathrm{kg\,m/s} \)).

Time is measured in seconds (\( \mathrm{s} \)). Therefore, the rate of change of momentum \( \frac{\Delta p}{\Delta t} \) has units:

\[ \frac{\mathrm{kg\,m/s}}{\mathrm{s}} = \mathrm{kg\,m/s^2} = \mathrm{N} \] which are the units of force (newtons). This confirms the consistency of the formula.

• Momentum units: \( \mathrm{kg\,m/s} \)
• Time units: seconds (\( \mathrm{s} \))
• Force units: newtons (\( \mathrm{N} = \mathrm{kg\,m/s^2} \))

An object of mass 2.0 kg has its velocity changed from 3.0 m/s to 7.0 m/s in 2.0 seconds. Calculate the resultant force acting on the object.

Step 1: Calculate initial momentum \( p_1 \): \[ p_1 = m v_1 = 2.0 \times 3.0 = 6.0 \ \mathrm{kg\,m/s} \]

Step 2: Calculate final momentum \( p_2 \): \[ p_2 = m v_2 = 2.0 \times 7.0 = 14.0 \ \mathrm{kg\,m/s} \] Step 3: Calculate change in momentum \( \Delta p \): \[ \Delta p = p_2 - p_1 = 14.0 - 6.0 = 8.0 \ \mathrm{kg\,m/s} \] Step 4: Calculate time interval \( \Delta t = 2.0 \ \mathrm{s} \) Step 5: Calculate resultant force \[ F = \frac{\Delta p}{\Delta t} = \frac{8.0}{2.0} = 4.0 \ \mathrm{N} \] Answer: The resultant force acting on the object is 4.0 newtons in the direction of the velocity change.

• Calculate initial and final momentum
• Find change in momentum \( \Delta p \)
• Divide by time interval \( \Delta t \)
• Resultant force \( F = \frac{\Delta p}{\Delta t} \)

1. \textit{Force equals momentum, not rate of change of momentum.} Force is not the momentum itself but how quickly momentum changes.

2. \textit{Momentum change must be considered as a vector.} Direction matters; a change in direction means a change in momentum even if speed is constant.

3. \textit{Resultant force is the net force, not any single force.} Only the combined effect of all forces causes momentum change.

• Force is rate of change of momentum, not momentum itself
• Momentum is a vector; direction changes count
• Resultant force is the net force causing momentum change