---
title: "Two balls impact e = 0.84, energy lost"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:two-balls-e084-ke-lost"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/two-balls-e084-ke-lost
site: BCT Engineering
---

# Two balls impact e = 0.84, energy lost

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

- Asked in: 2075 Asoj, 8 marks
- Video: [Impact: Coefficient of Restitution (learn to solve any problem)](https://www.youtube.com/watch?v=uzfeib3MkF8) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/two-balls-e084-ke-lost
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

The magnitude and direction of the velocities of two balls A and B having masses 1.2 kg and 1.8 kg respectively before they strike each other are shown as in figure below. Assuming $e = 0.84$, determine the velocity of each ball after the impact. How much K.E. will be lost due to the impact? [Figure: ball A with $v_A = 20$ m/s directed along $30^\circ$ to the x-axis; ball B with $v_B = 25$ m/s directed along $45^\circ$ to the x-axis.]

## Short answer

Tangential velocities unchanged; normal from momentum and e=0.84 gives v'_A=23.55 m/s at 25.1° above –x, v'_B=19.44 m/s at 65.4° below +x. Kinetic energy lost = 129.8 J.

## Answer

**Assumptions (figure not shown):** the x-axis is the line of impact. Ball A ($m_A=1.2$ kg, $20$ m/s) moves to the right, $30^\circ$ above the x-axis; ball B ($m_B=1.8$ kg, $25$ m/s) moves to the left, $45^\circ$ below the x-axis. The balls are smooth, $e=0.84$.

#### Components
| Ball | $v_x$ (along line of impact) | $v_y$ (tangential) |
|---|---|---|
| A | $20\cos30^\circ=17.32$ | $20\sin30^\circ=10.00$ |
| B | $-25\cos45^\circ=-17.68$ | $-25\sin45^\circ=-17.68$ |

The $y$ components are unchanged by the impact: $v'_{Ay}=10.00$ m/s, $v'_{By}=-17.68$ m/s.

#### Along the line of impact
Momentum:

$$
1.2(17.32)+1.8(-17.68)=1.2v'_{Ax}+1.8v'_{Bx}\ \Rightarrow\ 1.2v'_{Ax}+1.8v'_{Bx}=-11.04
$$

Restitution:

$$
v'_{Bx}-v'_{Ax}=e(v_{Ax}-v_{Bx})=0.84(17.32+17.68)=29.40
$$

Solving: $v'_{Ax}=-21.32$ m/s, $v'_{Bx}=8.08$ m/s.

#### Final velocities
$$
v'_A=\sqrt{21.32^{2}+10^{2}}=23.55\ \text{m/s},\qquad \text{direction }25.1^\circ\text{ above the negative x-axis}
$$

$$
v'_B=\sqrt{8.08^{2}+17.68^{2}}=19.44\ \text{m/s},\qquad \text{direction }65.4^\circ\text{ below the positive x-axis}
$$

#### Kinetic energy lost
$$
\begin{aligned}
T_1&=\tfrac12(1.2)(20)^{2}+\tfrac12(1.8)(25)^{2}=240+562.5=802.5\ \text{J}\\
T_2&=\tfrac12(1.2)(23.55)^{2}+\tfrac12(1.8)(19.44)^{2}=672.7\ \text{J}\\
\Delta T&=802.5-672.7=129.8\ \text{J}
\end{aligned}
$$

**Answer: $v'_A=23.55$ m/s, $v'_B=19.44$ m/s; kinetic energy lost $\approx129.8$ J.**

## Check yourself

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

1. Which components are unchanged by the impact?
   A. Those along the line of impact (x)
   B. Those perpendicular to the line of impact (y)
   C. None
   D. Both
   Answer: B. No tangential impulse on smooth balls: v'_Ay = 10, v'_By = -17.68 m/s. (Key point: Unchanged components)

2. What is the restitution equation here?
   A. v'_Bx - v'_Ax = 0.84 (17.32 + 17.68) = 29.40
   B. v'_Bx + v'_Ax = 0.84 (17.32 + 17.68)
   C. v'_Bx - v'_Ax = 17.32 - 17.68
   D. v'_Ax = 0.84 (17.32)
   Answer: A. Approach speed along x is 17.32 - (-17.68) = 35.0 m/s, multiplied by e. (Key point: Restitution equation)

3. What are v'_Ax and v'_Bx?
   A. -21.32 m/s and 8.08 m/s
   B. 17.32 m/s and -17.68 m/s
   C. 8.08 m/s and -21.32 m/s
   D. -11.04 m/s and 29.4 m/s
   Answer: A. Solving momentum 1.2v'_Ax + 1.8v'_Bx = -11.04 with restitution gives these values. (Key point: Final x velocities)

4. How is kinetic energy lost computed?
   A. T1 - T2 using total speeds, about 129.8 J
   B. Only the change in x components
   C. Zero, as e is nearly 1
   D. Momentum change
   Answer: A. T1 = 802.5 J, T2 = 672.7 J. Since e < 1 the impact is not perfectly elastic. (Key point: Kinetic energy loss)

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