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Why is the total momentum of a system unchanged when objects interact, and how is that used to predict their motion?

State the law of conservation of momentum and use it to calculate an unknown velocity after a collision when no external force acts (MA STE Introductory Physics, HS-PS2-2).

A standard-level answer on conservation of momentum for the Massachusetts High School Introductory Physics MCAS: why total momentum is conserved with no external force, how to set up the before-equals-after equation, and how to solve for an unknown velocity.

Generated by Claude Opus 4.813 min answer

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  1. What this topic is asking
  2. The law of conservation of momentum
  3. Setting up the equation
  4. Common situations
  5. Reference-sheet note
  6. Try this

What this topic is asking

Conservation of momentum is one of the most powerful ideas in physics, and the Massachusetts Introductory Physics standard HS-PS2-2 asks you to use it. You must state the law and calculate an unknown velocity after a collision when no external force acts. The skill is setting up the equation total momentum before = total momentum after, then solving. This is the practice of using mathematics, applied to a system, and it rests on the crosscutting concept that systems can be analyzed by tracking what is conserved.

The law of conservation of momentum

Why is momentum conserved? It follows from Newton's third law. In a collision, the two objects push on each other with equal and opposite forces for the same time, so they receive equal and opposite impulses. One object gains exactly the momentum the other loses, so the total does not change. The phrase "no external force" matters: if an outside force (like friction or a wall) acts, it can change the total. In the brief moment of a collision, the internal forces between the objects are usually far larger than friction, so momentum is conserved to a good approximation.

Setting up the equation

Every conservation-of-momentum problem uses the same structure:

pbefore=pafterp_{before} = p_{after}

Written out for two objects:

m1v1+m2v2 (before)=m1v1+m2v2 (after)m_1 v_1 + m_2 v_2 \ (\text{before}) = m_1 v_1' + m_2 v_2' \ (\text{after})

The method:

  1. Choose a positive direction (usually the direction of the first object's motion).
  2. Write the total momentum before, adding mvmv for each object, with signs for direction.
  3. Write the total momentum after, again summing mvmv with signs. If objects stick together, use their combined mass and a single velocity.
  4. Set before equal to after and solve for the unknown.

A negative answer simply means that object moves in the negative direction.

Common situations

  • Objects that stick together (inelastic): after the collision they move as one combined mass with a single velocity. Two carts that couple together are the classic case.
  • Recoil or explosion: the objects start together (or at rest) and push apart, like a skater throwing a ball or a gun firing. The total momentum stays at its starting value (often zero), so the two objects move in opposite directions with equal and opposite momenta.
  • Objects that bounce apart (elastic): they separate with their own velocities; momentum is still conserved, and you set up the same before-equals-after equation.

Reference-sheet note

The reference sheet gives momentum as p=mvp = mv but does not print the conservation law pbefore=pafterp_{before} = p_{after}. You must recall the conservation principle and apply it. This is one of the few relationships on the MCAS that is not handed to you, so it is worth memorizing the before-equals-after structure and practicing the sign bookkeeping for head-on and recoil situations.

Try this

Q1. Two objects collide and stick together. What quantity is the same before and after, assuming no external force? [1]

  • Cue. The total momentum of the system.

Q2. A stationary 4040 kg object explodes into two pieces. One piece moves left with 120120 kg m/s of momentum. State the momentum of the other piece. [2]

  • Cue. 120120 kg m/s to the right, so that the total stays zero (momentum is conserved and started at zero).

Exam-style practice questions

Practice questions written in the style of MA DESE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

MA Physics MCAS (style)3 marksA 2.02.0 kg cart moving at 3.03.0 m/s collides with and sticks to a stationary 1.01.0 kg cart. (a) State the law of conservation of momentum. (b) Calculate the velocity of the combined carts after the collision.
Show worked answer →

A 3-point item applying conservation of momentum to a sticking (inelastic) collision.

(a) 1 point: in the absence of an external force, the total momentum of a system before an interaction equals the total momentum after.
(b) Up to 2 points: before, total momentum =(2.0)(3.0)+(1.0)(0)=6.0= (2.0)(3.0) + (1.0)(0) = 6.0 kg m/s. After, the combined mass is 3.03.0 kg moving at vv: 6.0=3.0v6.0 = 3.0v, so v=2.0v = 2.0 m/s in the original direction.

Markers reward setting total momentum before equal to total momentum after, and using the combined mass after the carts stick.

MA Physics MCAS (style)3 marksA 6060 kg skater, initially at rest, throws a 2.02.0 kg ball forward at 9.09.0 m/s. (a) State why the total momentum of the skater and ball stays zero. (b) Calculate the skater's recoil velocity.
Show worked answer →

A 3-point recoil (explosion-type) problem using conservation of momentum.

(a) 1 point: before the throw the system is at rest, so the total momentum is zero; with no external horizontal force, the total momentum stays zero afterward.
(b) Up to 2 points: total momentum after =0= 0. Ball: (2.0)(9.0)=18(2.0)(9.0) = 18 kg m/s forward. Skater: (60)v(60)v. So 0=18+60v0 = 18 + 60v, giving v=0.30v = -0.30 m/s, meaning 0.300.30 m/s backward.

Markers reward the total being zero and a backward (negative) recoil velocity for the skater.

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