Momentum, Energy, and Simple Systems Revision Notes for HSC SSCE Physics

Momentum and Energy in Inelastic Collisions

Conservation principles in collisions

Momentum is conserved in all interactions, including all collisions. Energy is also a conserved quantity, so energy is conserved in all interactions. However, unlike momentum, energy can change forms. In most collisions, some kinetic energy is transformed into other forms including internal energy and sound.

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While both momentum and energy are conserved quantities, their behavior in collisions differs significantly. Momentum always remains as momentum in the system, but energy can transform between different forms such as kinetic energy, thermal energy, sound, and potential energy stored in deformation.

Elastic collisions review

In a perfectly elastic collision, the total kinetic energy of the objects is the same before and after the collision. In reality, perfectly elastic collisions only happen when the collision is due to a field force and the objects don't actually come into contact, such as in interactions of subatomic particles. The elastic collision model is useful for collisions where friction forces are small.

Inelastic collisions

Most real collisions are inelastic collisions. Inelastic collisions involve a loss of kinetic energy of the colliding objects. If sound or heat is produced, that energy must have come from somewhere - the decrease in the objects' kinetic energy. Energy can also be converted to potential energy in a collision, for example, due to deformation of the objects.

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Key Characteristics of Inelastic Collisions

The defining features that distinguish inelastic collisions are:

Momentum is conserved
Total energy is conserved, but kinetic energy is not
Some kinetic energy is transformed into other forms (heat, sound, deformation)
The colliding objects may separate after collision

Perfectly inelastic collisions

A perfectly inelastic collision is when the colliding objects stick together after the collision to form a single object. This represents the maximum possible loss of kinetic energy while still conserving momentum.

Common examples of perfectly inelastic collisions include:

Vehicle collisions where the vehicles become entangled
Meteorite impacts with Earth
A person catching a ball
A neutron being captured by a nucleus

Worked example: Car-truck collision

Consider a collision between a $1.5 \times 10^3$ kg car moving to the right at $20$ m s $^{-1}$ and a $5.0 \times 10^3$ kg truck moving to the left at $10$ m s $^{-1}$ . The vehicles collide and stick together.

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Worked Example: Finding the velocity after collision

Given information:

$m_{\text{car}} = 1500$ kg, $u_{\text{car}} = +20$ m s $^{-1}$
$m_{\text{truck}} = 5000$ kg, $u_{\text{truck}} = -10$ m s $^{-1}$
After collision: combined mass $m_{\text{car + truck}} = 6500$ kg

Solution:

Using conservation of momentum:

$\sum p_{\text{initial}} = \sum p_{\text{final}}$

$m_{\text{car}}u_{\text{car}} + m_{\text{truck}}u_{\text{truck}} = m_{\text{car + truck}}v_{\text{car + truck}}$

Rearranging for the final velocity:

$v_{\text{car + truck}} = \frac{m_{\text{car}}u_{\text{car}} + m_{\text{truck}}u_{\text{truck}}}{m_{\text{car + truck}}}$

$v_{\text{car + truck}} = \frac{(1500 \text{ kg})(20 \text{ m s}^{-1}) + (5000 \text{ kg})(-10 \text{ m s}^{-1})}{1500 \text{ kg} + 5000 \text{ kg}}$

$v_{\text{car + truck}} = -3.08 \text{ m s}^{-1} \approx -3.1 \text{ m s}^{-1}$

Result: The negative sign indicates the combined wreckage moves to the left.

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Worked Example: Calculating energy loss

To find the fractional change in kinetic energy:

$\frac{\Delta E_k}{E_{k,\text{before}}} = \frac{E_{k,\text{before}} - E_{k,\text{after}}}{E_{k,\text{before}}}$

Before the collision:

$E_{k,\text{before}} = \frac{1}{2}m_{\text{car}}u_{\text{car}}^2 + \frac{1}{2}m_{\text{truck}}u_{\text{truck}}^2$

$E_{k,\text{before}} = \frac{1}{2}(1500 \text{ kg})(20 \text{ m s}^{-1})^2 + \frac{1}{2}(5000 \text{ kg})(10 \text{ m s}^{-1})^2$

$E_{k,\text{before}} = 5.5 \times 10^5 \text{ J}$

After the collision:

$E_{k,\text{after}} = \frac{1}{2}m_{\text{car + truck}}v_{\text{car + truck}}^2$

$E_{k,\text{after}} = \frac{1}{2}(6500 \text{ kg})(3.08 \text{ m s}^{-1})^2$

$E_{k,\text{after}} = 3.08 \times 10^4 \text{ J}$

Fractional change:

$\frac{\Delta E_k}{E_{k,\text{before}}} = \frac{5.5 \times 10^5 \text{ J} - 3.08 \times 10^4 \text{ J}}{5.5 \times 10^5 \text{ J}} = 0.944$

Result: Therefore, 94% of the kinetic energy was transformed to other forms (heat, sound, deformation).

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Important Note About Approximations

The calculation asks about "immediately after" the collision because the car and truck cannot be modelled as an isolated system for long. The road exerts significant forces on them, particularly after the collision, changing the momentum of the system.

However, during the very small time that the collision is actually taking place, the forces that the car and truck exert on each other are much greater than the forces due to anything else. Hence, just for those moments, we can reasonably make the approximation that the only significant interactions are between the car and truck.

Vehicle collisions typically show very large fractional energy losses. This is because initial speeds can be quite high but the final speed is typically much lower. Vehicle collisions are always inelastic, and sometimes perfectly inelastic.

Worked example: Neutron collision in nuclear reactor

Subatomic particles can have both perfectly elastic and perfectly inelastic collisions. For example, a neutron can be captured by a nucleus in a perfectly inelastic collision, or be scattered in an elastic collision.

Consider a neutron in the OPAL (Open Pool Australian Light water) reactor in Sydney. The neutron is emitted with a speed of $1.5 \times 10^5$ m s $^{-1}$ and collides with a hydrogen nucleus (proton) in the water, which is initially at rest.

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Scenario 1: Perfectly elastic collision

Given:

$u_n = 1.5 \times 10^5$ m s $^{-1}$ , $u_p = 0$
$m_p = m_n = 1.67 \times 10^{-27}$ kg (approximation)

For elastic collisions, we use two equations:

Conservation of momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

Elastic collision relation: $u_1 + v_1 = v_2 + u_2$

Simplifying for equal masses and $u_p = 0$ :

From momentum: $u_n = v_p + v_n$ , so $v_p = u_n - v_n$

From elastic relation: $u_n + v_n = v_p$ , so $v_n = v_p - u_n$

Substituting:

$v_n = (u_n - v_n) - u_n = -v_n$

Therefore: $v_n = 0$

And: $v_p = u_n = 1.5 \times 10^5$ m s $^{-1}$

Result: In a perfectly elastic collision between equal masses where one is initially at rest, the velocities are exchanged. The neutron stops, and the proton moves off with the neutron's initial velocity.

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Scenario 2: Perfectly inelastic collision

In this case, the neutron is captured and forms a deuterium nucleus.

Using conservation of momentum:

$m_nu_n = (m_n + m_p)v_{p+n} = 2m_nv_{p+n}$

$v_{p+n} = \frac{1}{2}u_n = \frac{1}{2}(1.5 \times 10^5 \text{ m s}^{-1}) = 7.5 \times 10^4 \text{ m s}^{-1}$

Result: The deuterium nucleus moves off at half the speed of the original neutron.

Investigation: Comparing elastic and inelastic collisions

This investigation compares the collisions of different balls with Earth to identify which are closest to elastic or perfectly inelastic behaviour.

Aim

To compare the collisions of different balls with Earth and identify which are closest to elastic or perfectly inelastic.

Materials

Tape measure or metre ruler
Several different types of ball of approximately the same size (including super-ball and plasticine or blu-tack ball)
Scales
Webcam or video camera

Method

Attach the ruler or tape measure to the wall with zero at the floor
Set up the camera so it has a clear view of the ruler and you can read the scale on the recording
Weigh each ball and record its mass
Start recording
Hold a ball just in front of the 1 m mark on the ruler. Drop the ball (don't throw it)
View the recording and note the maximum height to which the ball bounced. Estimate the uncertainty
Repeat for each remaining ball

Data collection

Record measurements in a table with columns for:

Ball type
Mass $m$ (kg)
Maximum bounce height $h_{\text{max}}$ (m)
Fractional kinetic energy change $\frac{\Delta E_k}{\Delta E_{k,\text{before}}}$
Momentum before collision $p_{\text{before}}$ (kg m s $^{-1}$ )
Momentum after collision $p_{\text{after}}$ (kg m s $^{-1}$ )
Change in momentum $\Delta p$ (kg m s $^{-1}$ )

Analysis

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Key Assumptions:

The loss of energy due to air resistance is small compared with the loss due to friction forces during the collision
All balls are dropped from the same height (1 m)

Calculating fractional energy loss:

Using energy conservation, the kinetic energy just before collision equals the gravitational potential energy at the drop height. Similarly, kinetic energy just after collision equals the potential energy at maximum bounce height.

$\frac{\Delta E_k}{E_{k,\text{before}}} = \frac{1 \text{ m} - h_{\text{max}}}{1 \text{ m}}$

Calculating momentum:

Momentum before collision: $p_{\text{before}} = m\sqrt{2gh_{\text{drop}}}$ (downward)

Momentum after collision: $p_{\text{after}} = m\sqrt{2gh_{\text{max}}}$ (upward)

Since momentum is a vector, the change in momentum accounts for the direction change.

Discussion points

Most elastic: The ball with the smallest fractional energy loss (highest bounce) is closest to elastic
Most inelastic: The ball with the largest fractional energy loss (lowest or no bounce) is closest to perfectly inelastic
Conservation: While momentum of the ball changes (due to Earth's force), the total momentum of the ball-Earth system is conserved. Energy is conserved overall, with kinetic energy transforming to heat, sound, and deformation

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Understanding System Boundaries

While the momentum of the ball alone changes during the collision, this doesn't violate conservation of momentum. The ball-Earth system as a whole conserves momentum - the Earth gains an equal and opposite momentum change, though this is imperceptible due to Earth's enormous mass.

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Key Points to Remember:

Momentum is conserved in all collisions, including both elastic and inelastic collisions
Most real collisions are inelastic, involving a loss of kinetic energy which is transformed into other forms (heat, sound, deformation)
In perfectly inelastic collisions, the objects stick together after collision, representing the maximum possible loss of kinetic energy
Energy is always conserved overall, but kinetic energy can transform into other forms during collisions
When solving collision problems, use conservation of momentum as your primary tool, then calculate energy changes to determine the type of collision

Momentum, Energy, and Simple Systems (HSC SSCE Physics): Revision Notes

Momentum and Energy in Inelastic Collisions

Conservation principles in collisions

Elastic collisions review

Inelastic collisions

Perfectly inelastic collisions

Worked example: Car-truck collision

Worked Example: Finding the velocity after collision

Worked Example: Calculating energy loss

Worked example: Neutron collision in nuclear reactor

Scenario 1: Perfectly elastic collision

Scenario 2: Perfectly inelastic collision

Investigation: Comparing elastic and inelastic collisions

Aim

Materials

Method

Data collection

Analysis

Discussion points

Explore HSC SSCE Physics Model Answers by Topics

Forces

Forces, Acceleration, and Energy