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Pool game, three balls, v0 = 4.57 m/s

Short answer

Variable system of particles: mass changes with time (e.g., rocket); impulse-momentum applied to particles in system plus those entering/leaving. Conserving momentum along and perpendicular to V0, with VC=3.42 m/s, gives VB=2.30 m/s, VA=1.99 m/s.

Explain variable system of particles. In the game of pool, ball A is moving with a velocity v0v_0 of magnitude v0=4.57v_0 = 4.57 m/s. When it strikes balls B and C, which are at rest and aligned as shown. After the collision, the three balls move as indicated and assuming frictionless surfaces and perfectly elastic impact, determine VAV_A, VBV_B. Take VC=3.42V_C = 3.42 m/s. [Figure: ball A moving at v0v_0 at 30∘30^\circ below horizontal strikes aligned balls B and C; after impact A moves vertically downward, B moves at 30∘30^\circ below the horizontal line, C moves at 30∘30^\circ above the horizontal.]

Variable system of particles

A variable system of particles is a system whose mass changes with time: particles continuously enter (are gained by) or leave (are lost from) the system, for example a rocket expelling gas, a jet of water striking a vane, or a conveyor belt receiving sand. The principles of impulse-momentum are applied to the particles that are in the system at time tt plus those that join it in dtdt.

Pool balls

Three equal balls (equal mass mm). Ball A with v0=4.57v_0=4.57 m/s strikes B and C at rest. Reading of the figure (chosen to be consistent with the given VC=3.42V_C=3.42 m/s): use axes x′x' along v0v_0 and y′y' perpendicular to it. After impact C moves along x′x' (the same line as v0v_0), A moves along −y′-y' (perpendicular to v0v_0) and B moves at 60∘60^\circ to x′x' on the side opposite to A.

        y'
        |   B (60 deg to x')
        |  /
        | /
  v0 -> +------------> C   (x')
        |
        A (along -y')

Momentum along x′x':

mv0=mVC+mVBcos⁡60∘ ⇒ VB=v0−VCcos⁡60∘=4.57−3.420.5=2.30 m/sm v_0=m V_C+m V_B\cos60^\circ\ \Rightarrow\ V_B=\frac{v_0-V_C}{\cos60^\circ}=\frac{4.57-3.42}{0.5}=2.30\ \text{m/s}

Momentum along y′y':

0=mVBsin⁡60∘−mVA ⇒ VA=2.30(0.866)=1.99 m/s0=m V_B\sin60^\circ-m V_A\ \Rightarrow\ V_A=2.30(0.866)=1.99\ \text{m/s}

Check with energy (perfectly elastic): initial v02=20.88v_0^{2}=20.88; final VA2+VB2+VC2=3.97+5.29+11.70=20.96V_A^{2}+V_B^{2}+V_C^{2}=3.97+5.29+11.70=20.96. These agree within rounding of the given VCV_C, so the interpretation is consistent.

Answer: VA≈1.99V_A\approx1.99 m/s, VB≈2.30V_B\approx2.30 m/s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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