CLASSICAL MECHANICS & THERMODYNAMICS - Chapter 4, Exercise 1 Solution
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Conservation of Momentum: Two Skaters Pushing Apart

PROBLEM
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A 70 kg ice skater at rest pushes off a stationary 50 kg skater. If
the 70 kg skater moves backward at 1.5 m/s, use conservation of
momentum to find the velocity of the 50 kg skater.

SOLUTION
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Before pushing off, both skaters are at rest, so total momentum is
zero. Momentum must still total zero immediately after the push,
since no external horizontal force acts on the two-skater system
(ignoring friction with the ice).

    m_A v_A + m_B v_B = 0

Let the 70 kg skater be A (moving backward, so a negative velocity)
and the 50 kg skater be B.

    (70)(-1.5) + (50)(v_B) = 0
    -105 + 50 v_B = 0
    50 v_B = 105
    v_B = 105 / 50
    v_B = 2.1 m/s

ANSWER: The 50 kg skater moves forward (opposite direction to the
70 kg skater) at 2.1 m/s.

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WHY THIS WORKS AS AN ANSWER

Before the push, the two-skater system's total momentum is exactly
zero, since neither skater is moving. Because the push is an internal
force (each skater pushes on the other, an action-reaction pair per
Chapter 2's Third Law), it cannot change the SYSTEM's total momentum
- only redistribute it between the two skaters. The lighter skater
(50 kg) ends up moving faster (2.1 m/s) than the heavier skater
(70 kg, at 1.5 m/s) in the opposite direction, since a smaller mass
needs a larger velocity to carry the same magnitude of momentum -
exactly the same principle behind the cannon-recoil example in the
chapter, where the far lighter cannonball moves at a much higher speed
than the heavy cannon's own recoil.