MechanicsTopic #7 of 35

Momentum and Impulse

Linear momentum, impulse-momentum theorem, and conservation of momentum.

Overview

Momentum is a fundamental quantity in physics that describes the "quantity of motion" of an object. It's particularly useful for analyzing collisions and interactions.

Linear Momentum

Definition

p=mv\vec{p} = m\vec{v}

  • Vector quantity (same direction as velocity)
  • SI unit: kg·m/s

Properties

  • Momentum depends on both mass and velocity
  • Total momentum of a system is the vector sum of individual momenta
  • Can be positive or negative based on direction

Newton's Second Law (Momentum Form)

Fnet=dpdt=d(mv)dt\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}

For constant mass:

Fnet=mdvdt=ma\vec{F}_{\text{net}} = m\frac{d\vec{v}}{dt} = m\vec{a}

Impulse

Definition

Impulse is the change in momentum:

J=Δp=pfpi\vec{J} = \Delta\vec{p} = \vec{p}_f - \vec{p}_i

Impulse-Momentum Theorem

J=Favg×Δt=Δp\vec{J} = \vec{F}_{\text{avg}} \times \Delta t = \Delta\vec{p}

For variable force:

J=Fdt\vec{J} = \int \vec{F} \, dt

Units

  • Same as momentum: kg·m/s or N·s

Conservation of Momentum

In an isolated system (no external forces), total momentum is conserved:

pinitial=pfinal\sum\vec{p}_{\text{initial}} = \sum\vec{p}_{\text{final}}

For two objects:

m1v1i+m2v2i=m1v1f+m2v2fm_1\vec{v}_{1i} + m_2\vec{v}_{2i} = m_1\vec{v}_{1f} + m_2\vec{v}_{2f}

When Momentum is Conserved

  • No external forces acting on the system
  • External forces are negligible compared to internal forces
  • During the brief time of collision

When Momentum is NOT Conserved

  • Friction with ground
  • External applied forces
  • Gravitational effects over long times

Force and Time

The impulse-momentum theorem shows:

FΔt=ΔpF\Delta t = \Delta p

Same change in momentum can result from:

  • Large force over short time (rigid collision)
  • Small force over long time (cushioned collision)

Applications:

  • Airbags increase collision time → reduce force
  • Landing with bent knees increases stopping time
  • Catching a ball by "giving" with it

Examples

Example 1: Impulse Calculation

A 0.15 kg baseball moving at 40 m/s is hit by a bat, leaving at 50 m/s in the opposite direction. Find impulse.

J=Δp=m(vfvi)=0.15(50(40))=0.15×90=13.5 kg\cdotpm/sJ = \Delta p = m(v_f - v_i) = 0.15(50 - (-40)) = 0.15 \times 90 = 13.5 \text{ kg·m/s}

Example 2: Average Force

If the ball in Example 1 is in contact with bat for 0.002 s, find average force.

Favg=JΔt=13.50.002=6750 NF_{\text{avg}} = \frac{J}{\Delta t} = \frac{13.5}{0.002} = 6750 \text{ N}

Example 3: Momentum Conservation

A 5 kg cart moving at 4 m/s collides with a stationary 3 kg cart. They stick together. Find final velocity.

m1v1i+m2v2i=(m1+m2)vfm_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f

5(4)+3(0)=(5+3)vf5(4) + 3(0) = (5 + 3)v_f

vf=208=2.5 m/sv_f = \frac{20}{8} = 2.5 \text{ m/s}

Example 4: Recoil

A 60 kg person throws a 0.5 kg ball at 20 m/s. Find recoil velocity.

Initial momentum = 0 (both at rest)

0=mperson×vperson+mball×vball0 = m_{\text{person}} \times v_{\text{person}} + m_{\text{ball}} \times v_{\text{ball}}

vperson=mball×vballmperson=0.5×2060=0.167 m/sv_{\text{person}} = -\frac{m_{\text{ball}} \times v_{\text{ball}}}{m_{\text{person}}} = -\frac{0.5 \times 20}{60} = -0.167 \text{ m/s}

Example 5: Rocket Propulsion

Momentum is conserved as exhaust is expelled:

Fthrust=dmdt×vexhaustF_{\text{thrust}} = -\frac{dm}{dt} \times v_{\text{exhaust}}

Two-Dimensional Momentum

Conservation applies to each component:

px,initial=px,final\sum p_{x,\text{initial}} = \sum p_{x,\text{final}}

py,initial=py,final\sum p_{y,\text{initial}} = \sum p_{y,\text{final}}

Example: Glancing Collision

Two balls collide at an angle. Solve by treating x and y components separately.

Key Points

  • Momentum is conserved in all isolated collisions
  • Kinetic energy may or may not be conserved
  • Impulse equals change in momentum
  • Reducing impact force requires increasing impact time
  • Vector nature is essential for 2D and 3D problems