9.5 Impulse-Momentum, Impact, and Coefficient of Restitution

9.5 Impulse-Momentum, Impact, and Coefficient of Restitution

Impulse-momentum is the time-domain counterpart of work-energy, and it is the only tool that resolves collisions. NCEES lists impulse and momentum (linear and angular) as a Dynamics sub-topic; the governing idea is that momentum survives a collision even when kinetic energy does not.

Impulse-Momentum and Impact Mechanics

The Impulse-Momentum Principle integrates forces over time and is ideal for problems involving forces applied over specific time durations or sudden impacts.

Linear Impulse and Momentum

$m \mathbf{v}_1 + \int_{t_1}^{t_2} \mathbf{F} dt = m \mathbf{v}_2$

If the net linear impulse $\sum \mathbf{J} = \mathbf{0}$, linear momentum is conserved: $m \mathbf{v}_1 = m \mathbf{v}_2$.

Angular Impulse and Momentum

$I_G \omega_1 + \int_{t_1}^{t_2} M_G dt = I_G \omega_2$

Direct Central Impact and Coefficient of Restitution ($e$)

During a collision between bodies $A$ and $B$ along the line of impact, conservation of total linear momentum dictates:

$m_A v_{A1} + m_B v_{B1} = m_A v_{A2} + m_B v_{B2}$

The coefficient of restitution $e$ defines the ratio of relative separation velocity to relative approach velocity:

$e = \frac{v_{B2} - v_{A2}}{v_{A1} - v_{B1}}$

• Elastic Impact ($e = 1$): Zero kinetic energy loss ($T_1 = T_2$).
• Inelastic Impact ($0 < e < 1$): Partial kinetic energy loss due to deformation/heat.
• Perfectly Plastic Impact ($e = 0$): Maximum kinetic energy loss; bodies stick together ($v_{A2} = v_{B2} = v_2$).

The Elastic-to-Plastic Spectrum

The coefficient of restitution $e$ measures how much relative velocity survives the impact along the line of impact:

$e = \frac{\text{relative separation velocity}}{\text{relative approach velocity}} = \frac{v_B' - v_A'}{v_A - v_B}$

The rule that anchors every collision problem: momentum is always conserved; kinetic energy is not. The internal impact forces are equal and opposite by Newton's third law, so they cancel in the momentum sum regardless of how much deformation, heat, or sound they produce. Any exam option that conserves kinetic energy in a non-elastic collision is wrong.

The Two Equations for Direct Central Impact

Solve these simultaneously for the two unknown final velocities:

$m_A v_A + m_B v_B = m_A v_A' + m_B v_B' \qquad\text{(momentum)}$ $e(v_A - v_B) = v_B' - v_A' \qquad\text{(restitution)}$

For a perfectly plastic collision ($e = 0$), the bodies move together and one equation suffices:

$v' = \frac{m_A v_A + m_B v_B}{m_A + m_B}$

Drop-Height Test for $e$

Dropping a ball from $h_1$ and measuring its rebound $h_2$ gives $e$ directly, since $v = \sqrt{2gh}$:

$e = \frac{v_{\text{rebound}}}{v_{\text{impact}}} = \sqrt{\frac{h_2}{h_1}}$

A ball rebounding to half its drop height has $e = \sqrt{0.5} = 0.707$, not 0.5 — because $e$ is a velocity ratio and height scales with velocity squared. This square-root relationship is a reliable exam distractor.

Worked Example: Plastic Impact and Energy Loss

A 1{,}500 kg car at 20 m/s strikes a stationary 2{,}500 kg truck. The vehicles lock together. Find the common velocity and the fraction of kinetic energy lost.

Momentum conservation:

$v' = \frac{1{,}500(20) + 2{,}500(0)}{1{,}500 + 2{,}500} = \frac{30{,}000}{4{,}000} = 7.5\ \text{m/s}$

Kinetic energy before and after:

$KE_1 = \tfrac{1}{2}(1{,}500)(20)^2 = 300{,}000\ \text{J}$ $KE_2 = \tfrac{1}{2}(4{,}000)(7.5)^2 = \tfrac{1}{2}(4{,}000)(56.25) = 112{,}500\ \text{J}$

$\text{Fraction lost} = \frac{300{,}000 - 112{,}500}{300{,}000} = \frac{187{,}500}{300{,}000} = \boxed{62.5\%}$

Nearly two-thirds of the kinetic energy is destroyed — converted into deformation, heat, and sound — while momentum is conserved exactly. That is the point of crumple-zone design: the energy absorbed in deforming structure is energy not delivered to occupants.

For a perfectly plastic collision with one body initially at rest, the fraction lost has a closed form worth recognizing:

$\frac{\Delta KE}{KE_1} = \frac{m_B}{m_A + m_B} = \frac{2{,}500}{4{,}000} = 0.625$ ✓

So the heavier the struck object relative to the striker, the larger the fraction of energy destroyed.

Conservation of Angular Momentum

$H_O = I\omega, \qquad \sum M_O = \frac{dH_O}{dt}$

With no external moment about $O$, angular momentum is conserved:

$I_1\omega_1 = I_2\omega_2$

A skater pulling in their arms reduces $I$, so $\omega$ rises to compensate. Note that kinetic energy $\tfrac{1}{2}I\omega^2$ increases in the process — supplied by the muscular work of pulling the arms in, not created from nothing.

For a particle at position $\vec{r}$ with momentum $m\vec{v}$, angular momentum about $O$ is $\vec{H}_O = \vec{r}\times m\vec{v}$, with magnitude $mvr\sin\theta$ — or simply $mvd$, where $d$ is the perpendicular distance from $O$ to the velocity's line of action.