---
title: "Pool game, three balls, v0 = 5 m/s"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:pool-balls-three-elastic-5"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pool-balls-three-elastic-5
site: BCT Engineering
---

# Pool game, three balls, v0 = 5 m/s

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

- Asked in: 2076 Chaitra, 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-5
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

In a game of pool ball A is moving with a velocity $V_0$ of magnitude 5 m/s when it strikes balls B and C, which are at rest and aligned as shown. Knowing that after the collision the three balls move in the directions indicated and assuming frictionless surfaces and perfectly elastic impact, determine the magnitude of the velocities $V_A$, $V_B$ and $V_C$. [Figure: $V_0$ at $45^\circ$ to the horizontal striking A; after impact $V_A$ is vertically upward, $V_B$ is $30^\circ$ below the horizontal, $V_C$ is $30^\circ$ above the horizontal.]

## Short answer

Using momentum and kinetic energy conservation for equal masses and perfectly elastic impact, the velocities are VA = 4.02 m/s, VB = 2.53 m/s, VC = 1.55 m/s.

## Answer

**Setup.** All three balls have equal mass $m$, the table is frictionless and the impact is perfectly elastic, so linear momentum and kinetic energy are both conserved. Axes: $x$ horizontal, $y$ vertical. From the figure: $v_0=5$ m/s at $45^\circ$ above the horizontal, $V_A$ vertically upward, $V_B$ at $30^\circ$ below the horizontal and $V_C$ at $30^\circ$ above the horizontal (B and C on the same side).

**Momentum in $x$:**
$$
m(5\cos45^\circ)=mV_B\cos30^\circ+mV_C\cos30^\circ
$$

$$
V_B+V_C=\frac{3.5355}{0.8660}=4.0825\qquad(1)
$$

**Momentum in $y$:**
$$
m(5\sin45^\circ)=mV_A-mV_B\sin30^\circ+mV_C\sin30^\circ
$$

$$
V_A=3.5355+0.5(V_B-V_C)\qquad(2)
$$

**Energy:**
$$
V_A^{2}+V_B^{2}+V_C^{2}=5^{2}=25\qquad(3)
$$

Let $D=V_B-V_C$. From (1): $V_B^2+V_C^2=\dfrac{(4.0825)^2+D^2}{2}=8.333+0.5D^2$. From (2): $V_A^2=12.5+3.5355D+0.25D^2$. Substituting in (3):

$$
0.75D^{2}+3.5355D-4.1667=0\ \Rightarrow\ D=0.9763
$$

(The other root, $D=-5.69$, is rejected because it makes $V_B$ negative.)

Then

$$
V_B=\frac{4.0825+0.9763}{2}=2.529\ \text{m/s},\qquad V_C=\frac{4.0825-0.9763}{2}=1.553\ \text{m/s}
$$

$$
V_A=3.5355+0.5(0.9763)=4.024\ \text{m/s}
$$

Check: $4.024^2+2.529^2+1.553^2=16.19+6.40+2.41=25.0$.

**Answer: $V_A=4.02$ m/s, $V_B=2.53$ m/s, $V_C=1.55$ m/s.**

## Check yourself

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

1. Which quantities are conserved in this collision?
   A. Momentum in x and y, and kinetic energy
   B. Only kinetic energy
   C. Only x momentum
   D. Velocity of each ball
   Answer: A. Frictionless table, perfectly elastic: both are conserved. (Key point: Conserved quantities)

2. What is the x-momentum equation (equal masses)?
   A. 5 cos45 = V_B cos30 + V_C cos30
   B. 5 sin45 = V_B cos30 + V_C cos30
   C. 5 = V_A + V_B + V_C
   D. 5 cos45 = V_A cos30
   Answer: A. V_A is vertical, so only B and C contribute horizontally: V_B + V_C = 4.0825. (Key point: x-momentum)

3. Which is the y-momentum equation?
   A. 5 sin45 = V_A - V_B sin30 + V_C sin30
   B. 5 sin45 = V_A + V_B sin30 + V_C sin30
   C. 5 sin45 = V_A
   D. 5 cos45 = V_A
   Answer: A. B moves below horizontal and C above, so B's y component is negative. (Key point: y-momentum)

## Close variants set in other exams

- [Pool game, three balls, v0 = 4.57 m/s](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Apool-balls-three-elastic-4-57) — Applied Mechanics (Dynamics), chapter 2, asked 2×