Conservation of Linear Momentum - AP Physics C Mechanics

  1. Determine the impulse for a 3 kg object accelerating from 2 m/s to 5 m/s.
    J=9 N⋅sJ = 9 \, \text{N·s}. J=mΔv=3(5−2)=9J = m\Delta v = 3(5-2) = 9 N·s.
  2. What is impulse in relation to momentum?
    Impulse is change in momentum. Impulse equals the change in momentum.
  3. What is the momentum of a 7 kg object moving at 5 m/s?
    p=35 kg⋅m/sp = 35 \, \text{kg·m/s}. Using p=mv=7×5=35p = mv = 7 \times 5 = 35 kg·m/s.
  4. In momentum conservation, what is the role of internal forces?
    Internal forces do not change total momentum. Internal forces cancel by Newton's third law.
  5. What is the effect of longer collision time on force experienced?
    Force decreases as collision time increases. Impulse J=FΔtJ = F\Delta t remains constant.
  6. Calculate the average force if a 5 kg object changes speed by 4 m/s in 2 s.
    F=10 NF = 10 \, \text{N}. From F=mΔvΔt=5×42=10F = \frac{m\Delta v}{\Delta t} = \frac{5 \times 4}{2} = 10 N.
  7. Which type of collision conserves only momentum?
    Inelastic collision. Kinetic energy is lost but momentum conserved.
  8. Calculate the momentum of a 2 kg object moving at 3 m/s.
    p=6 kg⋅m/sp = 6 \, \text{kg·m/s}. Using p=mvp = mv: 2×3=6 kg⋅m/s2 \times 3 = 6 \, \text{kg·m/s}
  9. Find the velocity of a 4 kg object with 20 kg·m/s momentum.
    v=5 m/sv = 5 \, \text{m/s}. From p=mvp = mv: v=p/m=20/4=5v = p/m = 20/4 = 5 m/s.
  10. State the formula relating impulse and momentum.
    J=△p=F△tJ = \triangle p = F \triangle t. Impulse-momentum theorem relating force and time.
  11. What is the relationship between force and time in changing momentum?
    Impulse equals force times time. Impulse-momentum theorem: J=FΔtJ = F\Delta t.
  12. What is the momentum of a 10 kg mass moving with 0 m/s velocity?
    Zero, since velocity is zero. Zero velocity results in zero momentum.
  13. What is the principle of conservation of momentum in a closed system?
    Momentum is conserved in a closed system with no external forces. External forces must be absent for conservation.
  14. Find the final momentum of a 5 kg object initially 10 kg·m/s with 2 m/s velocity change.
    pf=20 kg⋅m/sp_f = 20 \, \text{kg·m/s}. pf=pi+mΔv=10+5(2)=20 kg⋅m/sp_f = p_i + m\Delta v = 10 + 5(2) = 20 \, \text{kg·m/s}
  15. Calculate final speed of two 3 kg objects colliding with v1=2v_1=2 m/s, v2=0v_2=0 m/s.
    vf=1 m/sv_f = 1 \, \text{m/s}. Perfectly inelastic: (3)(2)+(3)(0)=(6)vf(3)(2) + (3)(0) = (6)v_f.
  16. What condition is necessary for momentum conservation in a collision?
    No external net forces. Essential requirement for momentum conservation.
  17. If two objects have equal but opposite momenta, what is their total momentum?
    Zero. Equal magnitudes cancel each other out.
  18. Determine the impulse for a force of 10 N over 2 seconds.
    J=20 N⋅sJ = 20 \, \text{N·s}. Using J=FΔt=10×2=20J = F\Delta t = 10 \times 2 = 20 N·s.
  19. What factor does not affect total momentum in a closed system?
    Internal forces. Internal forces cancel by Newton's third law.
  20. Identify the SI unit for linear momentum.
    Kilogram meter per second (kg·m/s). Derived from mass (kg) times velocity (m/s).
  21. How is momentum affected by external forces?
    Momentum changes due to external forces. External forces create momentum change.
  22. Define an isolated system in terms of momentum.
    A system with no external forces acting on it. Required condition for momentum conservation.
  23. State the Law of Conservation of Linear Momentum.
    Total momentum before collision = total momentum after collision. Applies to isolated systems with no external forces.
  24. What is the formula for linear momentum?
    p=mvp = mv. Mass times velocity defines momentum.
  25. Find the final momentum of a 5 kg5 \, \text{kg} object initially 10 kg⋅m/s10 \, \text{kg·m/s} with 2 m/s2 \, \text{m/s} velocity change.
    pf=20 kg⋅m/sp_f = 20 \, \text{kg·m/s}. pf=pi+mΔv=10+5(2)=20 kg⋅m/sp_f = p_i + m\Delta v = 10 + 5(2) = 20 \, \text{kg·m/s}
  26. What is conserved in an isolated system of colliding objects?
    Momentum. Fundamental conservation law for isolated systems.
  27. Calculate the momentum of a 3 kg object at rest.
    p=0 kg⋅m/sp = 0 \, \text{kg·m/s}. Zero velocity means zero momentum.
  28. Describe inelastic collision in terms of momentum.
    Total momentum is conserved, kinetic energy is not. Only momentum is conserved, energy is lost.
  29. What is the effect of an external net force on a system's momentum?
    Changes the system's momentum. External net force creates momentum change.
  30. How does conservation of momentum apply to rocket propulsion?
    Momentum of expelled gas equals momentum change of rocket. Action-reaction pairs conserve total momentum.
  31. What is the momentum of a stationary object?
    Zero, since velocity is zero. Momentum equals mass times velocity.
  32. Calculate the velocity of two 5 kg objects sticking together post-collision, v1=3v_1=3 m/s, v2=0v_2=0 m/s.
    vf=1.5 m/sv_f = 1.5 \, \text{m/s}. Perfectly inelastic: (5)(3)+(5)(0)=(10)vf(5)(3) + (5)(0) = (10)v_f.
  33. If momentum is conserved, what is true about net external forces?
    Net external forces are zero. Required condition for momentum conservation.
  34. Describe elastic collision in terms of momentum.
    Total momentum and kinetic energy are conserved. Both momentum and energy remain unchanged.
  35. In a two-object system, how does momentum transfer during collision?
    Momentum lost by one is gained by the other. Newton's third law applied to momentum.
  36. What happens to total momentum when two objects collide in isolation?
    Total momentum remains constant. No external forces means momentum is conserved.
  37. Which type of collision conserves both momentum and kinetic energy?
    Elastic collision. Kinetic energy is also conserved in elastic collisions.
  38. What determines the change in momentum in a system?
    Impulse, or force applied over time. Impulse creates the momentum change.
  39. Calculate the change in momentum for a 2 kg object accelerating from 0 to 4 m/s.
    Δp=8 kg⋅m/s\Delta p = 8 \, \text{kg·m/s}. Δp=mΔv=2(4−0)=8 kg⋅m/s\Delta p = m\Delta v = 2(4-0) = 8 \, \text{kg·m/s}

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