Conservation of Momentum

Conservation law for isolated systems and applications

🎯 Conservation of Momentum

The Law of Conservation of Momentum

In an isolated system (no external forces), the total momentum remains constant.

ptotal,i=ptotal,f\vec{p}_{total,i} = \vec{p}_{total,f}

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

💡 Fundamental Law: This is one of the most important conservation laws in physics! It applies to all interactions, from colliding billiard balls to exploding stars.


Isolated System

A system is isolated when:

  • No external forces act on the system
  • OR external forces sum to zero
  • OR external forces are negligible compared to internal forces

Examples of isolated systems:

  • Two ice skaters pushing apart (on frictionless ice)
  • Collision between two cars (during brief collision, friction negligible)
  • Explosion of fireworks (no external forces during explosion)

Not isolated:

  • Ball rolling on rough surface (friction is external force)
  • Rocket launching (thrust is internal, but gravity/air resistance are external)

Why is Momentum Conserved?

Derivation from Newton's Laws

For two objects interacting:

By Newton's 3rd Law: F12=F21\vec{F}_{12} = -\vec{F}_{21}

dp1dt=dp2dt\frac{d\vec{p}_1}{dt} = -\frac{d\vec{p}_2}{dt}

dp1dt+dp2dt=0\frac{d\vec{p}_1}{dt} + \frac{d\vec{p}_2}{dt} = 0

d(p1+p2)dt=0\frac{d(\vec{p}_1 + \vec{p}_2)}{dt} = 0

Therefore: p1+p2=\vec{p}_1 + \vec{p}_2 = constant ✓

Internal forces come in action-reaction pairs that cancel!


One-Dimensional Collisions

For motion along a line (use + and - for direction):

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

Problem-Solving Steps:

  1. Choose positive direction (usually direction of motion)
  2. List known values with proper signs
  3. Write conservation equation
  4. Solve for unknown

Two-Dimensional Collisions

Momentum is conserved in each direction independently:

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}

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


Explosions and Recoil

In an explosion, objects initially at rest fly apart:

Before: pi=0p_i = 0 (at rest)

After: pf=m1v1+m2v2=0p_f = m_1\vec{v}_1 + m_2\vec{v}_2 = 0

Therefore: m1v1=m2v2m_1\vec{v}_1 = -m_2\vec{v}_2

Objects move in opposite directions with momenta equal in magnitude.

Examples:

  • Rifle recoil: Bullet goes forward, rifle goes backward
  • Rocket propulsion: Exhaust goes backward, rocket goes forward
  • Ice skaters pushing apart: Equal and opposite momenta

Center of Mass

The center of mass of an isolated system moves at constant velocity:

vcm=m1v1+m2v2m1+m2=ptotalmtotal\vec{v}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2} = \frac{\vec{p}_{total}}{m_{total}}

If system is isolated, ptotal\vec{p}_{total} is constant, so vcm\vec{v}_{cm} is constant!

Example: Two skaters pushing apart

  • Each skater accelerates
  • But center of mass continues at same velocity (or stays at rest)

When is Momentum NOT Conserved?

Momentum is NOT conserved when:

  • External forces act on system
  • Friction is significant
  • System is not isolated

However: Even with external forces, conservation can apply during brief interactions where internal forces dominate.

Example: Car collision

  • Friction acts on system (external)
  • But during 0.1 s collision, collision forces >> friction
  • Momentum approximately conserved during collision

⚠️ Common Mistakes

Mistake 1: Forgetting Vector Nature

In 2D problems, must conserve momentum in EACH direction separately.

Mistake 2: Wrong Signs

Velocities in opposite directions have opposite signs!

  • Object moving right: v>0v > 0
  • Object moving left: v<0v < 0

Mistake 3: Including External Forces

Don't apply conservation if significant external forces act on system!

Mistake 4: Confusing Before and After

Make sure you clearly identify which velocities are initial (viv_i) and which are final (vfv_f).


Momentum vs. Energy Conservation

| Conservation Law | Always Valid? | Remarks | |------------------|---------------|---------| | Momentum | Yes (if isolated) | Vector, conserved in all collisions | | Energy | Yes | But can convert forms (not always mechanical) | | Kinetic Energy | No | Only conserved in elastic collisions |

In inelastic collisions:

  • Momentum IS conserved ✓
  • Kinetic energy is NOT conserved (some converts to heat, sound, deformation)

Problem-Solving Strategy

For Collision Problems:

  1. Identify the system (what objects are involved?)
  2. Check if isolated (are external forces negligible?)
  3. Choose coordinate system (which direction is positive?)
  4. List knowns and unknowns
    • Before collision: m1,v1i,m2,v2im_1, v_{1i}, m_2, v_{2i}
    • After collision: v1f,v2fv_{1f}, v_{2f} (unknowns?)
  5. Apply conservation of momentum
    • 1D: One equation
    • 2D: Two equations (x and y)
  6. Solve algebraically
  7. Check your answer (reasonable magnitude? correct direction?)

Applications

Rocket Propulsion

Momentum of rocket + exhaust is conserved: mrocketvrocket+mexhaustvexhaust=0m_{rocket}v_{rocket} + m_{exhaust}v_{exhaust} = 0

Exhaust goes backward → rocket goes forward

Particle Physics

When particles collide or decay, total momentum is always conserved. Used to detect invisible particles!

Asteroid Defense

To deflect asteroid:

  • Explosion changes asteroid's momentum
  • Conservation tells us required impulse

Traffic Accidents

Forensics use conservation of momentum to determine pre-collision speeds from post-collision debris patterns.


Key Formulas Summary

| Concept | Formula | Condition | |---------|---------|-----------| | Conservation (1D) | m1v1i+m2v2i=m1v1f+m2v2fm_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} | Isolated system | | Conservation (general) | ptotal,i=ptotal,f\vec{p}_{total,i} = \vec{p}_{total,f} | Isolated system | | Explosion/recoil | m1v1+m2v2=0m_1\vec{v}_1 + m_2\vec{v}_2 = 0 | Initially at rest | | Center of mass velocity | vcm=ptotalmtotal\vec{v}_{cm} = \frac{\vec{p}_{total}}{m_{total}} | Always |

📚 Practice Problems

1Problem 1easy

Question:

A 1000 kg car traveling at 20 m/s collides with a 2000 kg truck at rest. After the collision, the two vehicles stick together. What is their velocity immediately after the collision?

💡 Show Solution

Given Information:

  • Car mass: m1=1000m_1 = 1000 kg
  • Car initial velocity: v1i=20v_{1i} = 20 m/s
  • Truck mass: m2=2000m_2 = 2000 kg
  • Truck initial velocity: v2i=0v_{2i} = 0 m/s (at rest)
  • After collision: stuck together (same final velocity vfv_f)

Find: Final velocity vfv_f


Step 1: Identify the system

System = car + truck (isolated during collision)


Step 2: Calculate initial momentum

pi=m1v1i+m2v2ip_i = m_1 v_{1i} + m_2 v_{2i}

pi=(1000)(20)+(2000)(0)p_i = (1000)(20) + (2000)(0)

pi=20,000 kg\cdotpm/sp_i = 20,000 \text{ kg·m/s}


Step 3: Calculate final momentum

After sticking together, combined mass moves with velocity vfv_f:

pf=(m1+m2)vfp_f = (m_1 + m_2)v_f

pf=(1000+2000)vf=3000vfp_f = (1000 + 2000)v_f = 3000v_f


Step 4: Apply conservation of momentum

pi=pfp_i = p_f

20,000=3000vf20,000 = 3000v_f

vf=20,0003000=6.67 m/sv_f = \frac{20,000}{3000} = 6.67 \text{ m/s}


Answer: The velocity immediately after collision is 6.67 m/s in the direction the car was traveling.

Check: Final velocity (6.67 m/s) is less than initial car velocity (20 m/s), which makes sense since the car must slow down when it hits the truck. ✓

2Problem 2medium

Question:

A 1500 kg car traveling east at 20 m/s collides with a 2000 kg truck traveling west at 10 m/s. They stick together after the collision. (a) What is the total momentum before the collision? (b) What is their velocity after the collision?

💡 Show Solution

Solution:

Given: m₁ = 1500 kg, v₁ = +20 m/s (east), m₂ = 2000 kg, v₂ = -10 m/s (west)

(a) Total momentum before: p_total = m₁v₁ + m₂v₂ p_total = 1500(20) + 2000(-10) p_total = 30,000 - 20,000 p_total = 10,000 kg·m/s east

(b) Velocity after collision: Conservation of momentum: p_before = p_after 10,000 = (m₁ + m₂)v_f 10,000 = (1500 + 2000)v_f 10,000 = 3500v_f v_f = 2.86 m/s east (or 2.9 m/s east)

The combined wreckage moves east at 2.9 m/s.

3Problem 3easy

Question:

An astronaut (mass 80 kg) floating at rest in space throws a 2 kg wrench away at 10 m/s. What is the astronaut's recoil velocity?

💡 Show Solution

Given Information:

  • Astronaut mass: ma=80m_a = 80 kg
  • Wrench mass: mw=2m_w = 2 kg
  • Initial state: both at rest
  • Wrench final velocity: vw=+10v_{w} = +10 m/s (choose this as positive direction)

Find: Astronaut's recoil velocity vav_a


Step 1: Calculate initial momentum

Both at rest initially:

pi=0p_i = 0


Step 2: Calculate final momentum

pf=mava+mwvwp_f = m_a v_a + m_w v_w

pf=(80)va+(2)(10)p_f = (80)v_a + (2)(10)

pf=80va+20p_f = 80v_a + 20


Step 3: Apply conservation of momentum

pi=pfp_i = p_f

0=80va+200 = 80v_a + 20

80va=2080v_a = -20

va=0.25 m/sv_a = -0.25 \text{ m/s}


Answer: The astronaut recoils at 0.25 m/s in the opposite direction to the wrench.

Interpretation:

  • Negative sign means opposite direction to wrench
  • Astronaut moves backward when wrench is thrown forward
  • Smaller velocity because astronaut has much larger mass
  • Momentum magnitudes: pa=80(0.25)=20|p_a| = 80(0.25) = 20 kg·m/s = pw|p_w|

4Problem 4hard

Question:

A 60 kg astronaut floating in space throws a 2.0 kg tool at 15 m/s. (a) What is the astronaut's recoil velocity? (b) If the astronaut catches the tool coming back at the same speed, what is the final velocity of the astronaut+tool system?

💡 Show Solution

Solution:

(a) Recoil velocity: Initial momentum = 0 (both at rest) Conservation of momentum: 0 = m_astronaut × v_a + m_tool × v_t 0 = 60v_a + 2.0(15) 0 = 60v_a + 30 v_a = -30/60 = -0.50 m/s

The astronaut moves backward at 0.50 m/s.

(b) Final velocity after catching: Before catch:

  • Astronaut: 60 kg at -0.50 m/s → p = -30 kg·m/s
  • Tool: 2.0 kg at -15 m/s (coming back) → p = -30 kg·m/s
  • Total: p = -60 kg·m/s

After catch (inelastic collision): p_total = (m_a + m_t)v_f -60 = (60 + 2.0)v_f v_f = -60/62 = -0.97 m/s

Both move backward together at 0.97 m/s.

5Problem 5hard

Question:

A 3 kg ball moving at 4 m/s in the +x direction collides with a 2 kg ball moving at 3 m/s at an angle of 60° above the +x axis. After the collision, the 3 kg ball moves at 2 m/s in the +y direction. Find the velocity (magnitude and direction) of the 2 kg ball after the collision.

💡 Show Solution

Given Information:

  • Ball 1: m1=3m_1 = 3 kg
    • Before: v1i=4v_{1i} = 4 m/s in +x direction
    • After: v1f=2v_{1f} = 2 m/s in +y direction
  • Ball 2: m2=2m_2 = 2 kg
    • Before: v2i=3v_{2i} = 3 m/s at 60° above +x axis
    • After: v2f=?v_{2f} = ? (unknown magnitude and direction)

Find: v2fv_{2f} (magnitude and direction)


Step 1: Set up initial velocities in components

Ball 1 (before):

  • v1ix=4v_{1ix} = 4 m/s
  • v1iy=0v_{1iy} = 0 m/s

Ball 2 (before):

  • v2ix=3cos(60°)=3(0.5)=1.5v_{2ix} = 3\cos(60°) = 3(0.5) = 1.5 m/s
  • v2iy=3sin(60°)=3(0.866)=2.60v_{2iy} = 3\sin(60°) = 3(0.866) = 2.60 m/s

Step 2: Set up final velocities in components

Ball 1 (after):

  • v1fx=0v_{1fx} = 0 m/s
  • v1fy=2v_{1fy} = 2 m/s

Ball 2 (after):

  • v2fx=?v_{2fx} = ? (unknown)
  • v2fy=?v_{2fy} = ? (unknown)

Step 3: Apply conservation of momentum in x-direction

m1v1ix+m2v2ix=m1v1fx+m2v2fxm_1 v_{1ix} + m_2 v_{2ix} = m_1 v_{1fx} + m_2 v_{2fx}

(3)(4)+(2)(1.5)=(3)(0)+(2)v2fx(3)(4) + (2)(1.5) = (3)(0) + (2)v_{2fx}

12+3=0+2v2fx12 + 3 = 0 + 2v_{2fx}

15=2v2fx15 = 2v_{2fx}

v2fx=7.5 m/sv_{2fx} = 7.5 \text{ m/s}


Step 4: Apply conservation of momentum in y-direction

m1v1iy+m2v2iy=m1v1fy+m2v2fym_1 v_{1iy} + m_2 v_{2iy} = m_1 v_{1fy} + m_2 v_{2fy}

(3)(0)+(2)(2.60)=(3)(2)+(2)v2fy(3)(0) + (2)(2.60) = (3)(2) + (2)v_{2fy}

0+5.2=6+2v2fy0 + 5.2 = 6 + 2v_{2fy}

5.2=6+2v2fy5.2 = 6 + 2v_{2fy}

2v2fy=0.82v_{2fy} = -0.8

v2fy=0.4 m/sv_{2fy} = -0.4 \text{ m/s}


Step 5: Find magnitude of final velocity

v2f=v2fx2+v2fy2v_{2f} = \sqrt{v_{2fx}^2 + v_{2fy}^2}

v2f=(7.5)2+(0.4)2v_{2f} = \sqrt{(7.5)^2 + (-0.4)^2}

v2f=56.25+0.16v_{2f} = \sqrt{56.25 + 0.16}

v2f=56.417.51 m/sv_{2f} = \sqrt{56.41} \approx 7.51 \text{ m/s}


Step 6: Find direction

θ=tan1(v2fyv2fx)\theta = \tan^{-1}\left(\frac{v_{2fy}}{v_{2fx}}\right)

θ=tan1(0.47.5)\theta = \tan^{-1}\left(\frac{-0.4}{7.5}\right)

θ=tan1(0.0533)\theta = \tan^{-1}(-0.0533)

θ3.05°\theta \approx -3.05°

(Negative means below the +x axis)


Answers:

  • Magnitude: 7.51 m/s
  • Direction: 3.05° below the +x axis (or 356.95° measured counterclockwise from +x axis)

Summary: After the collision, the 2 kg ball moves at approximately 7.5 m/s almost horizontally (just slightly downward).