AP Physics 1 · Lesson 2

Conservation of momentum

Know Know it

Start every momentum problem by choosing a system. Forces between objects inside the system are internal: by Newton’s third law they come in equal and opposite pairs, so their impulses cancel and cannot change the system’s total momentum. Only a net external force can.

So if the net external force on a system is zero (or too small to matter during a brief collision), the total momentum is conserved: m1v1i+m2v2i=m1v1f+m2v2fm_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}. The parts trade momentum; the total stays put.

Momentum is a vector, so in two dimensions you write one conservation equation for the x-components and another for the y-components.

Worked example Easy

A 60 kg person sits in a 120 kg rowboat at rest on calm water. The person throws a 4.0 kg anchor horizontally off the front of the boat at 5.4 m/s. Ignore water drag. How fast do the boat and person move just after the throw?

  1. 0.12 m/sAnswer
  2. 0.18 m/s
  3. 0.36 m/s
  4. 5.4 m/s

How to solve it

  1. Before the throw everything is at rest, so the total momentum is zero.
  2. After: the anchor has (4.0)(5.4)=21.6(4.0)(5.4) = 21.6 kg·m/s forward, so boat plus person must have 21.6 kg·m/s backward.
  3. v=21.6/(60+120)=0.12v = 21.6/(60 + 120) = 0.12 m/s, opposite to the throw.

Why each choice is right or wrong

  • A. Correct. The total momentum stays zero, so (180)v=(4.0)(5.4)=21.6(180)v = (4.0)(5.4) = 21.6 kg·m/s and v=0.12v = 0.12 m/s backward.
  • B. Incorrect. 0.18 m/s uses only the boat’s 120 kg, but the person moves with the boat, so the mass is 180 kg.
  • C. Incorrect. 0.36 m/s uses only the person’s 60 kg, but the boat moves with the person, so the mass is 180 kg.
  • D. Incorrect. 5.4 m/s gives the boat the anchor’s speed; equal momenta, not equal speeds, are what conservation requires.

Trap A force inside the system cannot move the total

Your muscles, a spring between two carts or the push of one puck on another are internal forces when both objects are in the system. They change each part’s momentum, but never the total. If the total changes, look for an outside push or pull, such as the floor, friction or gravity.

Explain Explain it

Task verbs: explain, calculate

The center-of-mass velocity is the total momentum divided by the total mass: vcm=(m1v1+m2v2)/(m1+m2)v_{\text{cm}} = (m_1v_1 + m_2v_2)/(m_1 + m_2). When momentum is conserved, vcmv_{\text{cm}} does not change, whatever kind of collision happens. If two objects stick together, they move off at exactly vcmv_{\text{cm}}.

Whether momentum is conserved depends on the system you pick. For one cart in a collision, the other cart’s push is external, so the first cart’s momentum changes. For both carts together, that push is internal, and the total momentum stays the same.

Worked example Medium

On a level air track, a 0.60 kg glider moves at +1.5 m/s and a 0.90 kg glider moves at -0.50 m/s, toward each other. What is the velocity of the center of mass of the two gliders?

  1. -0.30 m/s
  2. +0.30 m/sAnswer
  3. +0.50 m/s
  4. +0.90 m/s

How to solve it

  1. Total momentum: (0.60)(+1.5)+(0.90)(−0.50)=0.90−0.45=+0.45(0.60)(+1.5) + (0.90)(-0.50) = 0.90 - 0.45 = +0.45 kg·m/s.
  2. Total mass: 0.60 + 0.90 = 1.50 kg.
  3. vcm=0.45/1.50=+0.30v_{\text{cm}} = 0.45/1.50 = +0.30 m/s, and it stays +0.30 m/s through the collision.

Why each choice is right or wrong

  • A. Incorrect. -0.30 m/s has the right size but the wrong sign; the total momentum, +0.45 kg·m/s, is positive.
  • B. Correct. vcm=[(0.60)(1.5)+(0.90)(−0.50)]/1.5=0.45/1.5=+0.30v_{\text{cm}} = [(0.60)(1.5) + (0.90)(-0.50)]/1.5 = 0.45/1.5 = +0.30 m/s.
  • C. Incorrect. +0.50 m/s averages the two velocities without weighting them by mass.
  • D. Incorrect. +0.90 m/s adds the sizes of the momenta, 0.90 and 0.45, ignoring that they point opposite ways.

Explain Your turn

A student stands at rest on a skateboard on a smooth, level floor and catches a heavy medicine ball thrown to her horizontally. Ignore friction from the floor. Explain whether the momentum of the student and skateboard alone stays the same during the catch, and whether the momentum of the student, skateboard and ball together stays the same.

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Compare with a model answer

A full answer does this

  • Says the momentum of the student and skateboard alone changes (is not conserved)
  • Names the ball’s push on the student as the external force on that system
  • Says the total momentum of the student, skateboard and ball stays the same

Model answer

The momentum of the student and skateboard alone changes: they start at rest and, after the catch, roll in the direction the ball was moving. That happens because the ball pushes on the student, and the ball is outside that system, so its push is an external force. For the student, skateboard and ball together, the push of the ball on the student and the push of the student on the ball are internal, and no net horizontal outside force acts, so the total momentum of the three stays the same.

Apply Apply it

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