Collisions

Elastic and inelastic collisions, coefficient of restitution

💥 Collisions

Types of Collisions

Collisions are classified based on what happens to kinetic energy:

| Type | Momentum Conserved? | Kinetic Energy Conserved? | What Happens | |------|-------------------|------------------------|--------------| | Elastic | ✅ Yes | ✅ Yes | Objects bounce off, no deformation | | Inelastic | ✅ Yes | ❌ No | Some KE lost to heat, sound, deformation | | Perfectly Inelastic | ✅ Yes | ❌ No (max loss) | Objects stick together |

💡 Key Point: Momentum is ALWAYS conserved in collisions (if system is isolated). Kinetic energy is only conserved in elastic collisions.


Elastic Collisions

In an elastic collision:

  • Momentum is conserved
  • Kinetic energy is conserved
  • Objects bounce apart
  • No energy lost to heat, sound, or deformation

Examples:

  • Colliding billiard balls (nearly elastic)
  • Molecules in ideal gas
  • Atomic particles (fundamental particles)

Conservation Equations (1D)

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

Kinetic Energy: 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

Two equations, two unknowns (if we know initial velocities and masses)


Special Case: Equal Masses (Elastic)

If m1=m2m_1 = m_2 and object 2 is initially at rest:

Before: Object 1 moving at vv, object 2 at rest

After:

  • Object 1 comes to rest (v1f=0v_{1f} = 0)
  • Object 2 moves with original velocity (v2f=vv_{2f} = v)

They exchange velocities!


Special Case: One Object Much Heavier

Light object hits heavy object at rest:

  • Light object bounces back (reverses direction)
  • Heavy object barely moves

Heavy object hits light object at rest:

  • Heavy object continues forward (barely slows)
  • Light object shoots forward at high speed

Inelastic Collisions

In an inelastic collision:

  • Momentum is conserved ✓
  • Kinetic energy is NOT conserved ✗
  • Some KE converted to other forms (heat, sound, deformation)
  • Objects may or may not stick together

Examples:

  • Car crash (crumple zones absorb energy)
  • Dropped ball (loses energy on each bounce)
  • Most real-world collisions

Energy Loss

KElost=KEiKEfKE_{lost} = KE_i - KE_f

KElost=(12m1v1i2+12m2v2i2)(12m1v1f2+12m2v2f2)KE_{lost} = \left(\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2\right) - \left(\frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2\right)

This energy goes into:

  • Heat (friction)
  • Sound
  • Deformation (crushing, bending)
  • Internal energy

Perfectly Inelastic Collisions

In a perfectly inelastic collision:

  • Objects stick together after collision
  • Move with same final velocity
  • Maximum possible kinetic energy loss (for given initial conditions)
  • Momentum still conserved!

Examples:

  • Clay balls colliding
  • Train cars coupling
  • Tackle in football
  • Meteor impact creating crater

Simplified Equation

Since objects stick together: v1f=v2f=vfv_{1f} = v_{2f} = v_f

Momentum conservation: 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}

Only one unknown! Much simpler than elastic collisions.


Coefficient of Restitution

The coefficient of restitution ee measures "bounciness":

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}}

Values:

  • e=1e = 1: Perfectly elastic (ideal bounce)
  • 0<e<10 < e < 1: Inelastic (some energy lost)
  • e=0e = 0: Perfectly inelastic (stick together, no bounce)

For a ball bouncing: e=hfhie = \sqrt{\frac{h_f}{h_i}}

where hih_i is drop height, hfh_f is bounce height.


Problem-Solving Strategy

For Elastic Collisions:

  1. Write momentum conservation: m1v1i+m2v2i=m1v1f+m2v2fm_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}
  2. Write energy conservation: 12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f2\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2
  3. Solve system of equations (algebra intensive!)
  4. Check: KE should be equal before and after

For Perfectly Inelastic Collisions:

  1. Use simplified equation: m1v1i+m2v2i=(m1+m2)vfm_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f
  2. Solve for vfv_f
  3. Optional: Calculate energy lost

Much simpler because objects stick together!


⚠️ Common Mistakes

Mistake 1: Assuming KE is Always Conserved

Wrong: KE is conserved in all collisions ✅ Right: KE is only conserved in ELASTIC collisions

Mistake 2: Perfectly Inelastic = No Motion

Wrong: Perfectly inelastic means objects stop moving ✅ Right: Perfectly inelastic means objects stick together (they can still move!)

Mistake 3: Sign Errors in Momentum

Remember: velocities in opposite directions have opposite signs!

Mistake 4: Confusing Speed and Velocity

In coefficient of restitution, use relative speeds (magnitudes), but in momentum conservation use velocities (with signs).


2D Collisions

For collisions in two dimensions:

Conserve momentum in EACH direction:

x-direction: m1v1ix+m2v2ix=m1v1fx+m2v2fxm_1 v_{1ix} + m_2 v_{2ix} = m_1 v_{1fx} + m_2 v_{2fx}

y-direction: m1v1iy+m2v2iy=m1v1fy+m2v2fym_1 v_{1iy} + m_2 v_{2iy} = m_1 v_{1fy} + m_2 v_{2fy}

For elastic collisions, also conserve energy: 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

Need to use components: vx=vcosθv_x = v\cos\theta, vy=vsinθv_y = v\sin\theta


Real-World Applications

Car Safety

Modern cars designed for inelastic collisions:

  • Crumple zones deform (absorb energy)
  • Reduces force on passengers
  • Energy goes into crushing metal, not hurting people

Sports

Elastic collisions:

  • Golf ball and club (nearly elastic for maximum energy transfer)
  • Billiards (designed to be elastic)

Inelastic collisions:

  • Football tackle (maximum energy dissipation)
  • Boxing glove (padding increases collision time, decreases force)

Particle Physics

Particle accelerators create collisions:

  • Analyze momentum and energy
  • Discover new particles
  • Test fundamental physics laws

Comparing Collision Types

Same Initial Conditions

Consider object 1 (mass mm, velocity vv) hitting object 2 (mass mm, at rest):

Elastic:

  • Object 1 stops: v1f=0v_{1f} = 0
  • Object 2 moves: v2f=vv_{2f} = v
  • KEi=KEf=12mv2KE_i = KE_f = \frac{1}{2}mv^2
  • No energy lost

Perfectly Inelastic:

  • Both move together: vf=v2v_f = \frac{v}{2}
  • KEi=12mv2KE_i = \frac{1}{2}mv^2
  • KEf=12(2m)(v2)2=14mv2KE_f = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 = \frac{1}{4}mv^2
  • Half the kinetic energy lost!

Explosions (Reverse Collisions)

An explosion is like a collision in reverse:

  • Start: objects together (or at rest)
  • End: objects flying apart

Still conserve momentum!

If explosion from rest: m1v1+m2v2=0m_1\vec{v}_1 + m_2\vec{v}_2 = 0

Energy is ADDED to system (from chemical energy, spring, etc.)


Key Formulas Summary

| Type | Momentum | Kinetic Energy | Formula | |------|----------|----------------|---------| | Elastic | Conserved | Conserved | Two conservation equations | | Inelastic | Conserved | Not conserved | Momentum equation only | | Perfectly Inelastic | Conserved | Not conserved (max loss) | (m1+m2)vf=m1v1i+m2v2i(m_1 + m_2)v_f = m_1v_{1i} + m_2v_{2i} | | Coefficient of restitution | - | - | e=vseparationvapproache = \frac{v_{separation}}{v_{approach}} |

📚 Practice Problems

1Problem 1easy

Question:

Two identical 2 kg balls collide elastically. Ball 1 is moving at 6 m/s and ball 2 is at rest. Find the final velocities of both balls.

💡 Show Solution

Given Information:

  • Ball 1: m1=2m_1 = 2 kg, v1i=6v_{1i} = 6 m/s
  • Ball 2: m2=2m_2 = 2 kg, v2i=0v_{2i} = 0 m/s (at rest)
  • Elastic collision (momentum AND energy conserved)

Find: v1fv_{1f} and v2fv_{2f}


Step 1: Apply conservation of momentum

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

(2)(6)+(2)(0)=(2)v1f+(2)v2f(2)(6) + (2)(0) = (2)v_{1f} + (2)v_{2f}

12=2v1f+2v2f12 = 2v_{1f} + 2v_{2f}

6=v1f+v2f(Equation 1)6 = v_{1f} + v_{2f} \quad \text{(Equation 1)}


Step 2: Apply conservation of kinetic energy

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

12(2)(6)2+0=12(2)v1f2+12(2)v2f2\frac{1}{2}(2)(6)^2 + 0 = \frac{1}{2}(2)v_{1f}^2 + \frac{1}{2}(2)v_{2f}^2

36=v1f2+v2f2(Equation 2)36 = v_{1f}^2 + v_{2f}^2 \quad \text{(Equation 2)}


Step 3: Use special case result

For elastic collision with equal masses where one is at rest:

The velocities are exchanged!

v1f=0 m/sv_{1f} = 0 \text{ m/s} v2f=6 m/sv_{2f} = 6 \text{ m/s}


Step 4: Verify with equations

Check Equation 1: v1f+v2f=0+6=6v_{1f} + v_{2f} = 0 + 6 = 6 \quad ✓

Check Equation 2: v1f2+v2f2=02+62=36v_{1f}^2 + v_{2f}^2 = 0^2 + 6^2 = 36 \quad ✓


Answers:

  • Ball 1 final velocity: 0 m/s (comes to rest)
  • Ball 2 final velocity: 6 m/s (takes on ball 1's velocity)

Physical Interpretation: Ball 1 transfers all its momentum and energy to ball 2. This is what you see in Newton's cradle!

2Problem 2medium

Question:

A 3.0 kg ball moving at 8.0 m/s collides head-on with a 5.0 kg ball at rest. The collision is elastic. (a) Find the final velocity of each ball. (b) Verify that kinetic energy is conserved.

💡 Show Solution

Solution:

Given: m₁ = 3.0 kg, v₁ᵢ = 8.0 m/s, m₂ = 5.0 kg, v₂ᵢ = 0

For elastic collision with one object initially at rest:

v₁f = ((m₁ - m₂)/(m₁ + m₂))v₁ᵢ = ((3.0 - 5.0)/(3.0 + 5.0))(8.0) v₁f = (-2.0/8.0)(8.0) = -2.0 m/s (bounces back)

v₂f = (2m₁/(m₁ + m₂))v₁ᵢ = (2(3.0)/(8.0))(8.0) v₂f = (6.0/8.0)(8.0) = 6.0 m/s

(b) Check KE conservation: KE_i = ½m₁v₁ᵢ² = ½(3.0)(8.0)² = 96 J

KE_f = ½m₁v₁f² + ½m₂v₂f² KE_f = ½(3.0)(2.0)² + ½(5.0)(6.0)² KE_f = 6.0 + 90 = 96 J ✓

Kinetic energy is conserved!

3Problem 3medium

Question:

A 1500 kg car traveling at 20 m/s rear-ends a 1000 kg car traveling at 15 m/s in the same direction. The cars stick together after the collision. (a) Find their common velocity after the collision. (b) How much kinetic energy is lost?

💡 Show Solution

Given Information:

  • Car 1: m1=1500m_1 = 1500 kg, v1i=20v_{1i} = 20 m/s
  • Car 2: m2=1000m_2 = 1000 kg, v2i=15v_{2i} = 15 m/s (same direction)
  • Perfectly inelastic collision (stick together)

(a) Find common velocity after collision


Step 1: Apply conservation of momentum

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

(1500)(20)+(1000)(15)=(1500+1000)vf(1500)(20) + (1000)(15) = (1500 + 1000)v_f

30,000+15,000=2500vf30,000 + 15,000 = 2500v_f

45,000=2500vf45,000 = 2500v_f

vf=45,0002500=18 m/sv_f = \frac{45,000}{2500} = 18 \text{ m/s}


Answer (a): Common velocity after collision is 18 m/s


(b) Find kinetic energy lost


Step 2: Calculate initial kinetic energy

KEi=12m1v1i2+12m2v2i2KE_i = \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2

KEi=12(1500)(20)2+12(1000)(15)2KE_i = \frac{1}{2}(1500)(20)^2 + \frac{1}{2}(1000)(15)^2

KEi=12(1500)(400)+12(1000)(225)KE_i = \frac{1}{2}(1500)(400) + \frac{1}{2}(1000)(225)

KEi=300,000+112,500KE_i = 300,000 + 112,500

KEi=412,500 JKE_i = 412,500 \text{ J}


Step 3: Calculate final kinetic energy

KEf=12(m1+m2)vf2KE_f = \frac{1}{2}(m_1 + m_2)v_f^2

KEf=12(2500)(18)2KE_f = \frac{1}{2}(2500)(18)^2

KEf=12(2500)(324)KE_f = \frac{1}{2}(2500)(324)

KEf=405,000 JKE_f = 405,000 \text{ J}


Step 4: Calculate energy lost

KElost=KEiKEfKE_{lost} = KE_i - KE_f

KElost=412,500405,000KE_{lost} = 412,500 - 405,000

KElost=7,500 JKE_{lost} = 7,500 \text{ J}


Answer (b): Kinetic energy lost = 7,500 J

Percentage lost: 7,500412,500×100%1.8%\frac{7,500}{412,500} \times 100\% \approx 1.8\%

Note: This energy goes into crumpling metal, heat, and sound. Even though both cars were moving in the same direction and the speed difference was only 5 m/s, significant energy was dissipated!

4Problem 4hard

Question:

A 0.20 kg ball moving at 5.0 m/s collides with a 0.30 kg ball initially at rest. After the collision, the 0.20 kg ball moves at 2.0 m/s at 60° to its original direction. (a) Find the velocity of the 0.30 kg ball after collision. (b) Is this collision elastic or inelastic?

💡 Show Solution

Solution:

Given: m₁ = 0.20 kg, v₁ᵢ = 5.0 m/s, m₂ = 0.30 kg, v₂ᵢ = 0 After: v₁f = 2.0 m/s at 60°

(a) Velocity of m₂: Conservation of momentum (x-direction): m₁v₁ᵢ = m₁v₁f cos 60° + m₂v₂f,x 0.20(5.0) = 0.20(2.0)(0.5) + 0.30v₂f,x 1.0 = 0.20 + 0.30v₂f,x v₂f,x = 2.67 m/s

Conservation of momentum (y-direction): 0 = m₁v₁f sin 60° + m₂v₂f,y 0 = 0.20(2.0)(0.866) + 0.30v₂f,y v₂f,y = -1.15 m/s

Magnitude: v₂f = √(2.67² + 1.15²) = 2.9 m/s Direction: θ = tan⁻¹(1.15/2.67) = 23° below original direction

(b) Elastic or inelastic? KE_i = ½(0.20)(5.0)² = 2.5 J KE_f = ½(0.20)(2.0)² + ½(0.30)(2.9)² = 0.40 + 1.26 = 1.66 J

KE lost = 2.5 - 1.66 = 0.84 J Inelastic collision (KE not conserved)

5Problem 5hard

Question:

A 0.2 kg ball is dropped from a height of 2 m and bounces to a height of 1.2 m. (a) Find the coefficient of restitution. (b) If the ball was in contact with the ground for 0.01 s, what average force did the ground exert on the ball?

💡 Show Solution

Given Information:

  • Mass: m=0.2m = 0.2 kg
  • Drop height: hi=2h_i = 2 m
  • Bounce height: hf=1.2h_f = 1.2 m
  • Contact time: Δt=0.01\Delta t = 0.01 s

(a) Find coefficient of restitution


Step 1: Calculate velocity just before hitting ground

Using energy conservation: mghi=12mvbefore2mgh_i = \frac{1}{2}mv_{before}^2

vbefore=2ghi=2(9.8)(2)v_{before} = \sqrt{2gh_i} = \sqrt{2(9.8)(2)}

vbefore=39.2=6.26 m/s (downward)v_{before} = \sqrt{39.2} = 6.26 \text{ m/s (downward)}


Step 2: Calculate velocity just after leaving ground

Using energy conservation: 12mvafter2=mghf\frac{1}{2}mv_{after}^2 = mgh_f

vafter=2ghf=2(9.8)(1.2)v_{after} = \sqrt{2gh_f} = \sqrt{2(9.8)(1.2)}

vafter=23.52=4.85 m/s (upward)v_{after} = \sqrt{23.52} = 4.85 \text{ m/s (upward)}


Step 3: Calculate coefficient of restitution

e=vseparationvapproach=vaftervbeforee = \frac{v_{separation}}{v_{approach}} = \frac{v_{after}}{v_{before}}

e=4.856.26=0.775e = \frac{4.85}{6.26} = 0.775

Alternative formula using heights:

e=hfhi=1.22=0.6=0.775e = \sqrt{\frac{h_f}{h_i}} = \sqrt{\frac{1.2}{2}} = \sqrt{0.6} = 0.775


Answer (a): Coefficient of restitution = 0.775

This means the collision is inelastic (not perfectly elastic where e=1e = 1).


(b) Find average force from ground


Step 4: Set up velocity directions

Taking upward as positive:

  • Velocity before collision: vi=6.26v_i = -6.26 m/s (downward)
  • Velocity after collision: vf=+4.85v_f = +4.85 m/s (upward)

Step 5: Calculate change in momentum

Δp=m(vfvi)\Delta p = m(v_f - v_i)

Δp=(0.2)(4.85(6.26))\Delta p = (0.2)(4.85 - (-6.26))

Δp=(0.2)(4.85+6.26)\Delta p = (0.2)(4.85 + 6.26)

Δp=(0.2)(11.11)\Delta p = (0.2)(11.11)

Δp=2.22 kg\cdotpm/s (upward)\Delta p = 2.22 \text{ kg·m/s (upward)}


Step 6: Apply impulse-momentum theorem

FnetΔt=ΔpF_{net}\Delta t = \Delta p

Fnet(0.01)=2.22F_{net}(0.01) = 2.22

Fnet=222 N (upward)F_{net} = 222 \text{ N (upward)}


Step 7: Find force from ground

Net force = Normal force from ground - Weight

Fnet=NmgF_{net} = N - mg

222=N(0.2)(9.8)222 = N - (0.2)(9.8)

222=N1.96222 = N - 1.96

N=223.96 NN = 223.96 \text{ N}


Answer (b): Average force from ground = 224 N

Note: This is about 114 times the ball's weight! The ground must exert a large force because:

  1. It must stop the ball's downward motion
  2. It must accelerate the ball upward
  3. All in a very short time (0.01 s)

If the contact time were longer (softer surface), the force would be smaller.