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.
💡 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:
Therefore: constant ✓
Internal forces come in action-reaction pairs that cancel!
One-Dimensional Collisions
For motion along a line (use + and - for direction):
Problem-Solving Steps:
- Choose positive direction (usually direction of motion)
- List known values with proper signs
- Write conservation equation
- Solve for unknown
Two-Dimensional Collisions
Momentum is conserved in each direction independently:
x-direction:
y-direction:
Need to use components: ,
Explosions and Recoil
In an explosion, objects initially at rest fly apart:
Before: (at rest)
After:
Therefore:
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:
If system is isolated, is constant, so 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:
- Object moving left:
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 () and which are final ().
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:
- Identify the system (what objects are involved?)
- Check if isolated (are external forces negligible?)
- Choose coordinate system (which direction is positive?)
- List knowns and unknowns
- Before collision:
- After collision: (unknowns?)
- Apply conservation of momentum
- 1D: One equation
- 2D: Two equations (x and y)
- Solve algebraically
- Check your answer (reasonable magnitude? correct direction?)
Applications
Rocket Propulsion
Momentum of rocket + exhaust is conserved:
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) | | Isolated system | | Conservation (general) | | Isolated system | | Explosion/recoil | | Initially at rest | | Center of mass velocity | | 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: kg
- Car initial velocity: m/s
- Truck mass: kg
- Truck initial velocity: m/s (at rest)
- After collision: stuck together (same final velocity )
Find: Final velocity
Step 1: Identify the system
System = car + truck (isolated during collision)
Step 2: Calculate initial momentum
Step 3: Calculate final momentum
After sticking together, combined mass moves with velocity :
Step 4: Apply conservation of momentum
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: kg
- Wrench mass: kg
- Initial state: both at rest
- Wrench final velocity: m/s (choose this as positive direction)
Find: Astronaut's recoil velocity
Step 1: Calculate initial momentum
Both at rest initially:
Step 2: Calculate final momentum
Step 3: Apply conservation of momentum
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: kg·m/s = ✓
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: kg
- Before: m/s in +x direction
- After: m/s in +y direction
- Ball 2: kg
- Before: m/s at 60° above +x axis
- After: (unknown magnitude and direction)
Find: (magnitude and direction)
Step 1: Set up initial velocities in components
Ball 1 (before):
- m/s
- m/s
Ball 2 (before):
- m/s
- m/s
Step 2: Set up final velocities in components
Ball 1 (after):
- m/s
- m/s
Ball 2 (after):
- (unknown)
- (unknown)
Step 3: Apply conservation of momentum in x-direction
Step 4: Apply conservation of momentum in y-direction
Step 5: Find magnitude of final velocity
Step 6: Find direction
(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).
Practice with Flashcards
Review key concepts with our flashcard system
Browse All Topics
Explore other calculus topics