---
title: "Pool game, three balls, v0 = 4.57 m/s"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:pool-balls-three-elastic-4-57"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pool-balls-three-elastic-4-57
site: BCT Engineering
---

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

Applied Mechanics (Dynamics) · Chapter 2: Kinetics of particles: Energy and Momentum Methods

- Asked in: 2080 Baishakh, 2+6 marks
- Video: [Billiard Balls | Linear Momentum Conservation | Dynamics](https://www.youtube.com/watch?v=UqDacB-WZKE) — Engineering Hack
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pool-balls-three-elastic-4-57
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

Explain variable system of particles. In the game of pool, ball A is moving with a velocity $v_0$ of magnitude $v_0 = 4.57$ 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 $V_A$, $V_B$. Take $V_C = 3.42$ m/s. [Figure: ball A moving at $v_0$ at $30^\circ$ below horizontal strikes aligned balls B and C; after impact A moves vertically downward, B moves at $30^\circ$ below the horizontal line, C moves at $30^\circ$ above the horizontal.]

## 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.

## Answer

#### 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 $t$ plus those that join it in $dt$.

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

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

**Momentum along $x'$:**
$$
m v_0=m V_C+m V_B\cos60^\circ\ \Rightarrow\ V_B=\frac{v_0-V_C}{\cos60^\circ}=\frac{4.57-3.42}{0.5}=2.30\ \text{m/s}
$$

**Momentum along $y'$:**
$$
0=m V_B\sin60^\circ-m V_A\ \Rightarrow\ V_A=2.30(0.866)=1.99\ \text{m/s}
$$

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

**Answer: $V_A\approx1.99$ m/s, $V_B\approx2.30$ m/s.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is a variable system of particles?
   A. A system whose mass changes with time as particles enter or leave
   B. A system of balls of different masses
   C. A system with changing velocity
   D. A system without external forces
   Answer: A. Examples are a rocket expelling gas or sand falling on a conveyor belt. (Key point: Variable system)

2. Which laws govern the three-ball elastic collision?
   A. Only conservation of energy
   B. Conservation of momentum (components) and of kinetic energy
   C. Only Newton's second law
   D. Conservation of angular momentum only
   Answer: B. Frictionless and perfectly elastic: momentum is conserved and so is kinetic energy. (Key point: Conservation laws used)

3. Which equation gives V_B?
   A. v0 = V_C + V_B cos60, so V_B = (4.57 - 3.42)/0.5 = 2.30 m/s
   B. v0 = V_C + V_B sin60
   C. v0 = V_B cos60 only
   D. V_B = V_C
   Answer: A. Equal masses cancel; x' momentum before is v0, after is V_C + V_B cos60. (Key point: Momentum along x')

4. How is V_A obtained?
   A. V_A = V_B sin60 = 1.99 m/s
   B. V_A = V_B cos60
   C. V_A = v0 - V_B
   D. V_A = V_C sin60
   Answer: A. Initial y' momentum is zero, so V_A balances V_B's y' component. (Key point: Momentum along y')

## Close variants set in other exams

- [Pool game, three balls, v0 = 5 m/s](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Apool-balls-three-elastic-5) — Applied Mechanics (Dynamics), chapter 2, asked 2×