Momentum and Impulse

The Concept of Momentum

Momentum is a Vector quantity representing the 'quantity of motion' an object possesses, determined by its mass and velocity. The SI unit is kg·m/s, which is equivalent to N·s.

$$p = mv$$

Momentum is the product of how much stuff is moving and how fast it is going.

$p$=Linear Momentum(kg·m/s)

$m$=Mass(kg)

$v$=Velocity(m/s)

$v = 0$

→

The object has zero momentum regardless of its mass.

$m$ is changing (e.g., rocket)

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Use $F = \frac{dp}{dt}$ instead of $F = ma$, since mass is not constant.

Directionality: Because velocity is a vector, momentum always points in the same direction as the motion. Direction matters in all momentum problems — assign +/− signs carefully.

Mass vs Velocity: A slow truck and a fast bullet can have equal momentum if their $m \times v$ products are the same.

Newton's Second Law Form: Force equals the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$. This is actually the more general form of $F = ma$.

Impulse: Changing Momentum

Impulse is the product of the Average Force applied and the time interval of contact, resulting in a Change in Momentum. When force varies during impact (like a bat hitting a ball), impulse captures the total effect.

$$J = F_{avg} \Delta t = \Delta p$$

Impulse links the force applied over time to the resulting change in an object's motion.

$J$=Impulse(N·s)

$F_{avg}$=Average impact force(N)

$\Delta t$=Contact time(s)

$\Delta p$=Change in momentum(kg·m/s)

Object bounces back at the same speed

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$\Delta v = 2v$ (not zero!), so impulse is $2mv$ — double that of simply stopping.

$\Delta t \to 0$ (instantaneous hit)

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Force becomes infinitely large — physically impossible, so real collisions always have some contact time.

Unit Equivalence: $1 \text{ N·s}$ is mathematically identical to $1 \text{ kg·m/s}$. Both measure impulse/momentum.

Graph Area: On a Force-Time graph, the total impulse is the area under the curve — even if the force varies.

Bouncing vs Stopping: A rubber bullet that bounces back delivers more impulse than a lead bullet that embeds — the reversal adds extra momentum change ($\Delta p = 2mv$ vs $mv$).

The Impulse-Momentum Theorem

This theorem states that the impulse applied to an object is exactly equal to its change in momentum, highlighting the trade-off between force and time.

$$F \Delta t = m(v_f - v_i)$$

Relates the external force acting on a mass to its velocity change over time.

$v_i$=Initial Velocity(m/s)

$v_f$=Final Velocity(m/s)

$\Delta t$ is very large

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Force becomes negligible — like a feather slowly drifting to rest.

Object comes to rest ($v_f = 0$)

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Simplifies to $F \Delta t = mv_i$.

Safety Design: Airbags and Crumple Zones increase Contact Time to reduce the average impact force for the same change in momentum.

The Soft Landing: Catching a ball by 'pulling back' your hands increases $\Delta t$, making the required $F$ smaller.

Limiting Case: As $\Delta t \to 0$ (rigid collision), the force approaches infinity — this is why hitting concrete hurts more than hitting a pillow.

Conservation of Linear Momentum

In a closed system where no external forces act, the total momentum before an interaction equals the total momentum after the interaction. This applies to all collisions and explosions.

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$

The sum of momenta in a system remains constant during collisions or explosions.

$u_1, u_2$=Initial velocities of objects 1 and 2(m/s)

$v_1, v_2$=Final velocities of objects 1 and 2(m/s)

One object is initially at rest ($u_2 = 0$)

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Simplifies to $m_1 u_1 = m_1 v_1 + m_2 v_2$

Both initially at rest (explosion: $u_1 = u_2 = 0$)

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Total momentum is zero, so $m_1 v_1 = -m_2 v_2$ — objects move in opposite directions

Internal Forces: Forces between colliding objects are internal and cancel out (Newton's Third Law), leaving the total system momentum unchanged.

Sign Convention: Always define a positive direction first. Objects moving in opposite directions get opposite signs.

Proportionality Shortcut: In explosions from rest, $\frac{v_1}{v_2} = -\frac{m_2}{m_1}$ — the lighter piece always moves faster.

2D Explosions: If a bomb splits into three fragments, conserve momentum separately in $x$ and $y$ directions, then use Pythagoras for the resultant.

Types of Collisions

In an elastic collision, both momentum AND kinetic energy are conserved. The objects bounce off each other with no permanent deformation or heat generation. The Coefficient of Restitution equals 1.

$$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$

Total kinetic energy before equals total kinetic energy after in elastic collisions.

$e$=Coefficient of restitution

Equal masses, one at rest ($m_1 = m_2$, $u_2 = 0$)

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The first object stops completely and the second moves with the initial velocity of the first — like a Newton's cradle.

A light object hits a very heavy stationary object ($m_1 \ll m_2$)

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The light object bounces back at nearly the same speed; the heavy object barely moves.

Coefficient of Restitution: $e = \frac{\text{relative speed after}}{\text{relative speed before}} = \frac{v_2 - v_1}{u_1 - u_2}$. For elastic: $e = 1$.

Heavy Hits Light ($m_1 \gg m_2$, $u_2 = 0$): The heavy object barely slows down; the light object flies off at nearly $2u_1$.

Real-world: Perfectly elastic collisions only occur between atoms/molecules. Billiard balls are approximately elastic.

In an inelastic collision, momentum is conserved but kinetic energy is NOT. Some kinetic energy converts to heat, sound, or deformation. In a perfectly inelastic collision, the objects stick together ($e = 0$) and move with a common final velocity.

$$m_1u_1 + m_2u_2 = (m_1 + m_2)v_f$$

When objects stick together after collision, use combined mass for the final state.

$v_f$=Common final velocity(m/s)

Maximum KE Loss: Perfectly inelastic collisions lose the maximum possible kinetic energy while still conserving momentum.

Final Velocity Shortcut: For target at rest: $v_f = \frac{m_1}{m_1 + m_2} u_1$ — always less than $u_1$.

KE Lost: $\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2}(u_1 - u_2)^2$ — depends on relative velocity and reduced mass.

Explosive Forces and Recoil

When an object at rest breaks apart or ejects mass (explosions, gunfire, rocket propulsion), the total momentum remains zero. The fragments must move in opposite directions to conserve momentum. This produces recoil.

$$m_1 v_1 + m_2 v_2 = 0$$

Starting from rest, the momenta of the fragments are equal in magnitude but opposite in direction.

$m_1 v_1$=Momentum of fragment 1 (e.g., bullet)(kg·m/s)

$m_2 v_2$=Momentum of fragment 2 (e.g., gun)(kg·m/s)

Recoil Velocity: $v_{\text{gun}} = -\frac{m_{\text{bullet}}}{m_{\text{gun}}} v_{\text{bullet}}$ — the gun recoils slowly because it is much heavier.

Rocket Propulsion: A rocket ejects gas backward at high velocity. Thrust = (mass ejected per second) × (exhaust velocity): $F_{\text{thrust}} = \frac{\Delta m}{\Delta t} v_{\text{exhaust}}$

Increasing Acceleration: As a rocket burns fuel, its mass $M$ decreases while thrust stays roughly constant, so acceleration $a = F/M$ increases over time.

Force from Continuous Flow

When a continuous stream of matter (like water from a hose) strikes a surface and stops, it delivers a steady force. The force equals the rate of momentum delivery.

$$F = \frac{\Delta m}{\Delta t} \times v$$

Force from a stream equals mass flow rate times velocity change.

$\frac{\Delta m}{\Delta t}$=Mass striking per second (mass flow rate)(kg/s)

$v$=Velocity of the stream(m/s)

Application: This formula applies to water jets, sand falling on conveyor belts, and any scenario where mass arrives continuously.

If it bounces back: If the stream bounces instead of stopping, the velocity change is $2v$ (stopping + reversing), so the force doubles.

Dimensional Check: $[\frac{\Delta m}{\Delta t}] \times [v] = \frac{\text{kg}}{\text{s}} \times \frac{\text{m}}{\text{s}} = \frac{\text{kg·m}}{\text{s}^2} = \text{N}$ ✓