Interactive Lecture
Linear and Angular Impulse, Momentum, and Impact
by Dr Arnaldo Delli Carri
The principle of impulse and momentum relates forces acting over time to changes in an object's motion. It is the natural tool for analysing collisions, explosions, and other events where large forces act over short times.
Impulse is the effect of a force acting over time; momentum is a measure of an object's motion. The impulse-momentum theorem states that the impulse applied to an object equals its change in momentum.
Linear momentum is defined as:
\(\vec{p} = m\vec{v}\)
Where:
Linear impulse is the integral of force over time:
\(\vec{J} = \displaystyle \int_{t_1}^{t_2} \vec{F} \, dt\)
For constant force:
\(\vec{J} = \vec{F} \cdot \Delta t\)
Where:
Starting from Newton's Second Law:
\(\vec{F} = m\vec{a} = m\displaystyle\frac{d\vec{v}}{dt}\)
Integrating both sides with respect to time and recalling the definition of linear impulse above:
\(\displaystyle \int_{t_1}^{t_2} \vec{F} \, dt = m \displaystyle\int_{v_1}^{v_2} d\vec{v}\)
\(\vec{J} = m\vec{v}_2 - m\vec{v}_1\)
\(\vec{J} = \Delta\vec{p}\)
Therefore: Impulse = Change in Momentum
Physical Interpretation: The impulse-momentum theorem tells us that to change an object's momentum, we must apply a force over time. A large force over a short time can produce the same momentum change as a small force over a long time.
NOTA BENE: both impulse and momentum are vector quantities, therefore they must be resolved into components when solving problems.
For a system with no external forces:
\(\sum \vec{p}_{\text{initial}} = \sum \vec{p}_{\text{final}}\)
Or for a two-body collision:
\(m_1\vec{v}_{1i} + m_2\vec{v}_{2i} = m_1\vec{v}_{1f} + m_2\vec{v}_{2f}\)
Collisions between objects can be classified based on how kinetic energy is conserved or lost during the impact:
1. Perfectly Elastic Collision (e = 1)
2. Inelastic Collision (0 < e < 1)
3. Perfectly Inelastic Collision (e = 0)
The coefficient of restitution (e) is a measure of how "bouncy" a collision is. It quantifies the elasticity of the collision:
Definition:
\(e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}\)
Or mathematically:
\(e = -\frac{v_{2f} - v_{1f}}{v_{2i} - v_{1i}}\)
Where:
Physical Interpretation:
The coefficient of restitution was first studied by Isaac Newton, who found it depends on the materials of the colliding objects:
Typical Values:
The kinetic energy lost in a collision can be calculated using:
For a collision between two objects:
\(\Delta E_k = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}(v_{1i} - v_{2i})^2(1 - e^2)\)
Note that:
For collisions that are not head-on, the coefficient of restitution applies only along the line of impact (normal direction):
The collision is analysed in two components:
The coefficient of restitution equation applies only to the normal components:
\(e = -\frac{(v_{2n})_f - (v_{1n})_f}{(v_{2n})_i - (v_{1n})_i}\)
Tangential components are unchanged (assuming no friction)
Important Note: In all types of collisions, momentum is ALWAYS conserved (in isolated systems). The coefficient of restitution tells us how kinetic energy behaves, not momentum!
The coefficient of restitution directly shapes engineering design choices:
Impact Force and Time:
For a given momentum change, the average impact force depends on collision time:
\(F_{avg} = \frac{\Delta p}{\Delta t}\)
Design Considerations:
Explore 2D momentum conservation with two billiard balls. The blue ball remains centred whilst you can adjust the white ball's vertical position to create oblique (diagonal) impacts. The animation runs at reduced speed to clearly observe the collision dynamics.
Creating Different Collision Types: The balls only make contact within one diameter of the centre line, so the offset slider is scoped to that window — every setting produces a genuine collision.
Impact Parameter (b): A measure of how "off-centre" the collision is:
Watch how the total momentum remains constant regardless of the impact parameter or coefficient value!
Newton's Cradle is a classic demonstration of sequential elastic collisions. Use the buttons to select how many balls to lift, then press Release to watch the collision cascade. Reduce the coefficient of restitution to see energy gradually dissipate with each swing.
The Dual Conservation Constraint: For a perfectly elastic cradle (e = 1), both momentum and kinetic energy must be conserved simultaneously. Consider 1 ball striking a stationary group of 4:
Scenario B conserves momentum but loses kinetic energy. Only Scenario A satisfies both conservation laws simultaneously — this is why the number of swinging balls is preserved.
Try e < 1: Watch the energy bar decrease with each swing. Energy is lost to heat and sound at each impact, and the cradle gradually comes to rest.
Note: With e < 1, you will notice the middle balls also begin to move. This is a consequence of the simplified sequential pairwise collision model used here: when e < 1, the incoming ball retains a residual velocity after each impact (v′ = ½(1−e)v) and continues pushing into the chain, spreading energy to all balls. In a real Newton's Cradle, the balls are always in contact and the impulse travels as an elastic compression wave through the chain simultaneously — so clean n-in/n-out behaviour is preserved even for e slightly below 1. The pairwise model is a useful simplification for e = 1, but overestimates energy spreading for e < 1.
For a particle:
\(\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}\)
(where \(\times\) denotes the vector product between two vectors)
For a rigid body rotating about a fixed axis it can be proven that:
\(L = I\omega\)
Where:
Angular impulse is the integral of torque/moment over time:
\(\vec{J}_\theta = \displaystyle\int_{t_1}^{t_2} \vec{M} \, dt\)
For constant torque/moment:
\(\vec{J}_\theta = \vec{M} \cdot \Delta t\)
Where:
Starting from the rotational form of Newton's Second Law:
\(\vec{M} = I\vec{\alpha} = I\displaystyle\frac{d\vec{\omega}}{dt}\)
Integrating both sides:
\(\displaystyle\int_{t_1}^{t_2} \vec{M} \, dt = I\int_{\omega_1}^{\omega_2} d\vec{\omega}\)
\(\vec{J}_\theta = I\vec{\omega}_2 - I\vec{\omega}_1 = \Delta\vec{L}\)
Therefore: Angular Impulse = Change in Angular Momentum
When no external torques act on a system:
\(\vec{L}_{\text{initial}} = \vec{L}_{\text{final}}\)
or:
\(I_1\omega_1 = I_2\omega_2\)
Classic Example: A figure skater spins faster when pulling their arms inward. As the moment of inertia \(I\) decreases, the angular velocity \(\omega\) must increase to conserve angular momentum.
Observe how changing the moment of inertia affects rotational velocity. Click the buttons to extend or retract the masses!
Observe: When the masses are extended, the system slows down. When retracted, it speeds up. The angular momentum \(L = I\omega\) remains constant!
For 2D collisions at an angle, momentum is conserved in both x and y directions independently:
X-direction: \(m_1v_{1xi} + m_2v_{2xi} = m_1v_{1xf} + m_2v_{2xf}\)
Y-direction: \(m_1v_{1yi} + m_2v_{2yi} = m_1v_{1yf} + m_2v_{2yf}\)
For systems with changing mass (like rockets):
Rocket equation (Tsiolkovsky):
\(\Delta v = v_e \ln\left(\frac{m_0}{m_f}\right)\)
Where:
When a spinning object experiences a torque perpendicular to its angular momentum, it precesses:
Precession rate: \(\Omega = \frac{M}{L} = \frac{M}{I\omega}\)
where \(M\) is the applied torque — the same quantity denoted \(M\) throughout the angular impulse and momentum sections above.
Impulse and momentum appear throughout engineering — particularly wherever something hits, stops, or changes direction abruptly:
| Concept | Equation | Description |
|---|---|---|
| Linear momentum | \(\vec{p} = m\vec{v}\) | \(m\) = mass; \(\vec{v}\) = velocity |
| Linear impulse | \(\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt\) | \(\vec{F}\) = force; \(t\) = time |
| Impulse-momentum theorem | \(\vec{J} = \Delta\vec{p}\) | Impulse equals change in momentum |
| Conservation of linear momentum | \(\sum \vec{p}_{\text{initial}} = \sum \vec{p}_{\text{final}}\) | For isolated systems (no external forces) |
| Coefficient of restitution | \(e = -\frac{v_{2f} - v_{1f}}{v_{2i} - v_{1i}}\) | \(e = 1\) (elastic); \(e = 0\) (perfectly inelastic) |
| Angular momentum (rigid body) | \(L = I\omega\) | \(I\) = moment of inertia; \(\omega\) = angular velocity |
| Angular impulse | \(\vec{J}_\theta = \int_{t_1}^{t_2} \vec{M} \, dt\) | \(\vec{M}\) = torque/moment |
| Angular impulse-momentum theorem | \(\vec{J}_\theta = \Delta\vec{L}\) | Angular impulse equals change in angular momentum |
| Conservation of angular momentum | \(I_1\omega_1 = I_2\omega_2\) | For systems with no external torques |
Two cars collide head-on. Car A (1500 kg) travels at 20 m/s east, whilst Car B (1200 kg) travels at 15 m/s west. If they stick together after collision, find their final velocity.
Show Solution
Define east as positive direction.
Initial momentum: \(p_i = m_Av_A + m_Bv_B = 1500(20) + 1200(-15) = 30000 - 18000 = 12000 \text{ kg·m/s}\)
After collision (perfectly inelastic): \(p_f = (m_A + m_B)v_f\)
By conservation: \(12000 = (1500 + 1200)v_f = 2700v_f\)
\(v_f = \frac{12000}{2700} = 4.44 \text{ m/s east}\)
Answer: \(v_f = 4.44\,\text{m/s}\) east — the wreckage moves in Car A's original direction.
A figure skater with arms extended has a moment of inertia of 2.5 kg·m² and rotates at 3 rad/s. When pulling arms in, the moment of inertia reduces to 1.0 kg·m². Find the new angular velocity.
Show Solution
By conservation of angular momentum:
\(L_i = L_f\)
\(I_i\omega_i = I_f\omega_f\)
\(2.5 \times 3 = 1.0 \times \omega_f\)
\(\omega_f = 7.5 \text{ rad/s}\)
The skater spins 2.5 times faster!
Answer: \(\omega_f = 7.5\,\text{rad/s}\).
A 0.15 kg cricket ball travelling at 40 m/s is struck by a bat and leaves at 50 m/s in the opposite direction. If contact time is 0.003 s, find the average force on the ball.
Show Solution
Define initial direction as positive:
Initial momentum: \(p_i = 0.15 \times 40 = 6 \text{ kg·m/s}\)
Final momentum: \(p_f = 0.15 \times (-50) = -7.5 \text{ kg·m/s}\)
Change in momentum: \(\Delta p = -7.5 - 6 = -13.5 \text{ kg·m/s}\)
Impulse: \(J = F_{avg} \times \Delta t = \Delta p\)
\(F_{avg} = \frac{-13.5}{0.003} = -4500 \text{ N}\)
The negative sign shows the force opposes the ball's initial motion.
Answer: \(F_{avg} = 4500\,\text{N}\), directed opposite to the ball's initial motion.