Recoil and Conservation of Momentum: Choose the Whole System

When the net external impulse on a system is negligible, its total momentum is unchanged. If two objects start at rest and separate, their final momenta add to zero, so their velocities have opposite signs.

Draw the system boundary first

Include both separating objects when applying momentum conservation. The interaction between them is then internal to that system. Their individual momenta can change substantially while the sum remains constant. If an important external impulse acts, the simple conserved-total equation must be modified.

Use an idealised horizontal pair of carts with negligible external impulse, rather than assuming that every real push or collision conserves the momentum of whatever object you happen to examine. Defining the system is part of solving the question.

Worked two-cart recoil

Two model carts initially rest together. One has mass 2 kg and the other 3 kg. After an internal release, the 2 kg cart moves at +1.5 m/s. Initial system momentum was zero, so 2(1.5) + 3v = 0.

Solving gives 3v = −3 and v = −1 m/s for the 3 kg cart. Check the final momenta: +3 kg·m/s and −3 kg·m/s sum to zero. The heavier cart has a smaller speed, but the momentum magnitudes are equal in this initially stationary two-object example.

Momentum does not determine all energy changes

The final kinetic energies here are (1/2) × 2 × 1.5² = 2.25 J and (1/2) × 3 × 1² = 1.5 J. Their sum is 3.75 J. It need not equal the initial kinetic energy of zero; an internal energy store supplies the increase.

Conserving momentum does not mean kinetic energy is always conserved. Elastic collision questions impose an additional kinetic-energy condition. In a recoil calculation with a given final velocity, momentum alone can determine the partner velocity without specifying every energy transfer.

Common mistake

Do not set the two speeds equal unless the masses justify it. Keep a sign for each velocity, and check the total momentum before and after. If the pair already had an initial velocity, the initial total momentum is not zero; calculate it before writing the final equation.

System dono carts ka lo; momentum ka total check karo, speed ka nahi.

Three practice questions with worked answers

1. Two 1 kg carts start at rest. One moves at +2 m/s. Find the other velocity.

   Zero total momentum gives 1(2) + 1v = 0, so v = −2 m/s.

2. A stationary pair has masses 1 kg and 4 kg. The 1 kg cart leaves at +3 m/s. Find the other velocity.

   1(3) + 4v = 0 gives v = −0.75 m/s, assuming negligible external impulse.

3. Carts of 2 kg and 3 kg initially share +2 m/s. Later the 2 kg cart has +5 m/s. Find the 3 kg cart velocity.

   Initial momentum is (2 + 3)2 = 10 kg·m/s. Thus 2(5) + 3v = 10, giving v = 0 m/s.

Related study guides

• Force and Laws of Motion Class 9 Notes with Numericals
• Newton Second Law Numericals

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Conservation of Momentum. No textbook or exam-board affiliation is claimed.