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Why is the total momentum of an isolated system conserved, and how is this used to analyze collisions?

State the law of conservation of momentum, explain it using Newton's third law, and apply it to collisions and explosions where the total momentum before equals the total momentum after.

A Regents Physics answer on conservation of momentum: why total momentum is conserved in an isolated system, how Newton's third law explains it, and how to solve collision and explosion problems with total momentum before equal to total momentum after, with worked examples.

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  1. What this topic is asking
  2. The law of conservation of momentum
  3. Why momentum is conserved: Newton's third law
  4. Applying the law to collisions
  5. Applying the law to explosions and recoils
  6. Reference Tables note
  7. Try this

What this topic is asking

Conservation of momentum is one of the two great conservation laws of Regents mechanics (the other is energy). The Physical Setting/Physics course asks you to state the law, explain it through Newton's third law, and apply it to collisions (objects coming together) and explosions or recoils (objects pushing apart). The recurring exam task is to set the total momentum before an interaction equal to the total momentum after, and solve for an unknown velocity.

The law of conservation of momentum

An "isolated system" is one where outside forces either do not act or cancel, so the only forces are the internal ones between the objects. In practice the Regents treats brief collisions and explosions as isolated, because the internal forces during the brief interaction dwarf any external ones (like friction) acting in that instant.

Why momentum is conserved: Newton's third law

This explanation is often worth a mark in itself. It connects the topic to Newton's third law and momentum and impulse, and it makes clear why the law needs no external force.

Applying the law to collisions

For a collision between two objects, write

m1v1i+m2v2i=m1v1f+m2v2fm_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}

choosing a positive direction and giving each velocity a sign. Two cases are common:

  • Perfectly inelastic: the objects stick together and move off with a common velocity. The right side becomes (m1+m2)vf(m_1 + m_2)v_f.
  • Elastic or general: the objects separate with different velocities; momentum is still conserved, and you solve for the unknown velocity.

In a perfectly inelastic collision, momentum is conserved but kinetic energy is not (some becomes heat, sound or deformation), a distinction explored in energy and its conservation.

Applying the law to explosions and recoils

In an explosion or recoil, objects initially together push apart. If the system starts at rest, the total momentum is zero before, so it must be zero after: the fragments carry equal and opposite momenta. A skater throwing a ball recoils backward; a gun recoils when a bullet is fired; a rocket moves forward as exhaust is expelled. The equation is the same conservation statement, often with a total of zero.

Reference Tables note

Conservation of momentum is not printed as an equation on the Reference Tables; the tables give only p=mvp = mv and FΞ”t=Ξ”pF\Delta t = \Delta p. You must therefore recall the law and write "total momentum before equals total momentum after" yourself. This is one of the few relationships the Regents expects you to supply from memory rather than read from the booklet.

Try this

Q1. State the law of conservation of momentum. [2 points]

  • Cue. In an isolated system, the total momentum before an interaction equals the total momentum after.

Q2. A 1.01.0 kg ball at 5.05.0 m/s strikes a stationary 1.01.0 kg ball and stops; they do not stick. State the velocity of the second ball. [1 point]

  • Cue. By conservation, the second ball moves off at 5.05.0 m/s (the first ball's momentum is transferred entirely).

Exam-style practice questions

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

Regents (style)3 marksPart C (extended response). A 2.02.0 kg cart moving at 3.03.0 m/s collides with a stationary 4.04.0 kg cart, and the two stick together. (a) State the law of conservation of momentum. (b) Calculate the common velocity of the carts after the collision. Show all work.
Show worked answer β†’

A 3-point Part C perfectly inelastic collision.

(a) Law (1 point): in an isolated system (no external net force) the total momentum before an interaction equals the total momentum after.
(b) Velocity after (2 points): total momentum before =(2.0)(3.0)+(4.0)(0)=6.0= (2.0)(3.0) + (4.0)(0) = 6.0 kg m/s. After, the combined 6.06.0 kg moves at vv: 6.0=(6.0)v6.0 = (6.0)v, so v=1.0v = 1.0 m/s in the original direction.

Markers reward writing total momentum before equal to total momentum after and solving for the common velocity. A common error is to conserve velocity or kinetic energy instead of momentum.

Regents (style)2 marksPart B-2 (constructed response). A 60.60. kg skater, initially at rest, throws a 2.02.0 kg ball east at 9.09.0 m/s. Calculate the skater's recoil velocity. Show the equation, substitution and answer.
Show worked answer β†’

A 2-point constructed-response recoil (explosion-type) problem. Total momentum before is zero, so the momenta after must be equal and opposite.

Equation: total momentum before == total momentum after, so 0=mballvball+mskatervskater0 = m_{ball}v_{ball} + m_{skater}v_{skater}.
Substitution: 0=(2.0)(9.0)+(60.)vskater0 = (2.0)(9.0) + (60.)v_{skater}, so (60.)vskater=βˆ’18(60.)v_{skater} = -18.
Answer: vskater=βˆ’0.30v_{skater} = -0.30 m/s, that is 0.300.30 m/s west (opposite to the ball).

Markers reward setting the total momentum to zero before and after and solving for the recoil velocity with its (opposite) direction.

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