This answer key provides solutions to common conservation of momentum problems. Remember, the law of conservation of momentum states that in a closed system, the total momentum before a collision equals the total momentum after the collision. Momentum (p) is calculated as mass (m) times velocity (v): p = mv.
Section 1: Basic Momentum Calculations
Problem 1: A 2 kg ball rolls at 3 m/s towards a stationary 1 kg ball. After the collision, the 2 kg ball moves at 1 m/s. What is the velocity of the 1 kg ball after the collision?
Answer:
- Momentum before: (2 kg)(3 m/s) + (1 kg)(0 m/s) = 6 kg⋅m/s
- Momentum after: (2 kg)(1 m/s) + (1 kg)(v) = 2 kg⋅m/s + v⋅kg⋅m/s
- Conservation of momentum: 6 kg⋅m/s = 2 kg⋅m/s + v⋅kg⋅m/s
- Solving for v: v = 4 m/s
The 1 kg ball moves at 4 m/s after the collision.
Problem 2: A 5 kg object moving at 4 m/s collides with a 3 kg object moving in the opposite direction at 6 m/s. If they stick together after the collision, what is their final velocity?
Answer:
- Momentum before: (5 kg)(4 m/s) + (3 kg)(-6 m/s) = 20 kg⋅m/s - 18 kg⋅m/s = 2 kg⋅m/s (Note the negative sign for the 3 kg object's velocity since it's moving in the opposite direction).
- Momentum after: (5 kg + 3 kg)(v) = 8 kg * v
- Conservation of momentum: 2 kg⋅m/s = 8 kg * v
- Solving for v: v = 0.25 m/s
The combined objects move at 0.25 m/s in the direction of the initially faster object.
Section 2: More Complex Scenarios
Problem 3: A 1000 kg car traveling at 20 m/s collides with a stationary 1500 kg truck. After the collision, the car's velocity is reduced to 5 m/s. Assuming the collision is perfectly elastic (kinetic energy is conserved, a simplification for problem-solving purposes), calculate the truck's velocity after the collision. (Note: This problem requires more advanced understanding of collisions; a simpler, inelastic collision would assume the objects stick together).
Answer: This problem involves both momentum and kinetic energy conservation, making the calculation more complex. The detailed solution would involve simultaneous equations solving, which is beyond the scope of a concise answer key. Consult a physics textbook or online resources for guidance on solving elastic collision problems.
Problem 4: A bullet with a mass of 0.01 kg is fired from a 5 kg rifle with a velocity of 800 m/s. What is the recoil velocity of the rifle?
Answer:
- Momentum before: 0 kg⋅m/s (both rifle and bullet are initially at rest)
- Momentum after: (0.01 kg)(800 m/s) + (5 kg)(v) = 8 kg⋅m/s + 5v kg⋅m/s
- Conservation of momentum: 0 kg⋅m/s = 8 kg⋅m/s + 5v kg⋅m/s
- Solving for v: v = -1.6 m/s
The rifle recoils with a velocity of -1.6 m/s (the negative sign indicates the opposite direction of the bullet).
Section 3: Real-World Application
Problem 5 (Open-ended): Explain how the principle of conservation of momentum applies to a rocket launching into space.
Answer: A rocket expels hot gas downwards at high velocity. This downwards momentum of the expelled gas is equal and opposite to the upwards momentum gained by the rocket. The total momentum of the system (rocket and expelled gas) remains constant, propelling the rocket upwards.
This answer key provides solutions and explanations. Remember that understanding the underlying concepts is crucial. Further practice problems and consultations with physics resources are recommended for a thorough grasp of the subject.
