MechanicsTopic #8 of 35

Collisions

Elastic and inelastic collisions, center of mass, and collision analysis.

Overview

A collision is a brief interaction between two or more objects. Momentum is always conserved in collisions (assuming no external forces), but kinetic energy may or may not be conserved.

Types of Collisions

Elastic Collision

  • Kinetic energy is conserved
  • Momentum is conserved
  • Objects bounce off each other
  • Examples: billiard balls, atomic collisions

Inelastic Collision

  • Kinetic energy is NOT conserved (some is lost)
  • Momentum is conserved
  • Some energy converted to heat, sound, deformation
  • Most real collisions are inelastic

Perfectly Inelastic Collision

  • Maximum kinetic energy loss
  • Objects stick together after collision
  • Momentum is conserved
  • Final velocity is the same for both objects

Conservation Laws for Collisions

Momentum (Always Conserved)

m1v1i+m2v2i=m1v1f+m2v2fm_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

Kinetic Energy (Elastic Only)

12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f2\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2

Elastic Collision Formulas (1D)

For two objects with m1m_1 moving at v1iv_{1i} hitting stationary m2m_2:

v1f=m1m2m1+m2×v1iv_{1f} = \frac{m_1 - m_2}{m_1 + m_2} \times v_{1i}

v2f=2m1m1+m2×v1iv_{2f} = \frac{2m_1}{m_1 + m_2} \times v_{1i}

Special Cases

Equal masses (m1=m2m_1 = m_2):

v1f=0,v2f=v1iv_{1f} = 0, \quad v_{2f} = v_{1i}

Objects exchange velocities!

Heavy hitting light (m1m2m_1 \gg m_2):

v1fv1i,v2f2v1iv_{1f} \approx v_{1i}, \quad v_{2f} \approx 2v_{1i}

Light hitting heavy (m1m2m_1 \ll m_2):

v1fv1i,v2f0v_{1f} \approx -v_{1i}, \quad v_{2f} \approx 0

Perfectly Inelastic Collision

Objects stick together:

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

vf=m1v1i+m2v2im1+m2v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}

Energy Lost

ΔKE=KEfKEi\Delta KE = KE_f - KE_i

ΔKE=12(m1+m2)vf2[12m1v1i2+12m2v2i2]\Delta KE = \frac{1}{2}(m_1 + m_2)v_f^2 - \left[\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2\right]

For one object initially at rest:

Fraction lost=m2m1+m2\text{Fraction lost} = \frac{m_2}{m_1 + m_2}

Coefficient of Restitution

Measures "bounciness" of a collision:

e=v2fv1fv1iv2i=relative speed afterrelative speed beforee = \frac{v_{2f} - v_{1f}}{v_{1i} - v_{2i}} = \frac{\text{relative speed after}}{\text{relative speed before}}

Typeee value
Perfectly elastice=1e = 1
Partially inelastic0<e<10 < e < 1
Perfectly inelastice=0e = 0

Two-Dimensional Collisions

Apply conservation of momentum in each direction:

x-direction:

m1v1xi+m2v2xi=m1v1xf+m2v2xfm_1 v_{1xi} + m_2 v_{2xi} = m_1 v_{1xf} + m_2 v_{2xf}

y-direction:

m1v1yi+m2v2yi=m1v1yf+m2v2yfm_1 v_{1yi} + m_2 v_{2yi} = m_1 v_{1yf} + m_2 v_{2yf}

For elastic 2D collision:

12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f2\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2

Center of Mass

Position of center of mass:

xcm=m1x1+m2x2m1+m2x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}

Velocity of center of mass:

vcm=m1v1+m2v2m1+m2=ptotalmtotalv_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{p_{\text{total}}}{m_{\text{total}}}

Key property: Center of mass velocity is unchanged by collision (in absence of external forces)

Examples

Example 1: Elastic Collision

A 2 kg ball moving at 5 m/s hits a stationary 3 kg ball elastically.

v1f=232+3×5=1 m/sv_{1f} = \frac{2-3}{2+3} \times 5 = -1 \text{ m/s}

v2f=2(2)2+3×5=4 m/sv_{2f} = \frac{2(2)}{2+3} \times 5 = 4 \text{ m/s}

Example 2: Perfectly Inelastic

A 1500 kg car at 20 m/s rear-ends a 1000 kg car at 10 m/s. They lock together.

vf=1500×20+1000×101500+1000=400002500=16 m/sv_f = \frac{1500 \times 20 + 1000 \times 10}{1500 + 1000} = \frac{40000}{2500} = 16 \text{ m/s}

Energy lost:

KEi=12(1500)(20)2+12(1000)(10)2=350,000 JKE_i = \frac{1}{2}(1500)(20)^2 + \frac{1}{2}(1000)(10)^2 = 350{,}000 \text{ J}

KEf=12(2500)(16)2=320,000 JKE_f = \frac{1}{2}(2500)(16)^2 = 320{,}000 \text{ J}

ΔKE=30,000 J (lost to deformation, heat)\Delta KE = -30{,}000 \text{ J (lost to deformation, heat)}

Example 3: Ballistic Pendulum

A bullet (m=10m = 10 g, v=400v = 400 m/s) embeds in a block (M=2M = 2 kg) on a string. Find max height.

Step 1 - Collision (momentum conserved):

mv=(m+M)vfmv = (m + M)v_f

vf=0.01×4000.01+2=1.99 m/sv_f = \frac{0.01 \times 400}{0.01 + 2} = 1.99 \text{ m/s}

Step 2 - Swing up (energy conserved):

12(m+M)vf2=(m+M)gh\frac{1}{2}(m+M)v_f^2 = (m+M)gh

h=vf22g=(1.99)22×9.8=0.20 mh = \frac{v_f^2}{2g} = \frac{(1.99)^2}{2 \times 9.8} = 0.20 \text{ m}

Problem-Solving Strategy

  1. Identify the type of collision
  2. Define the system and check for external forces
  3. Apply conservation of momentum
  4. If elastic, also apply conservation of kinetic energy
  5. Solve the equations simultaneously if needed