---
title: "Elastic impact, final positions with friction"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:elastic-impact-two-particles-friction"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/elastic-impact-two-particles-friction
site: BCT Engineering
---

# Elastic impact, final positions with friction

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

- Asked in: 2078 Kartik, 6 marks
- Video: [Central impact example #2](https://www.youtube.com/watch?v=Mr2Y91NoYDo) — Engineer4Free
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/elastic-impact-two-particles-friction
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

A particle having mass 0.5 kg is released from rest and strikes the stationary particle of mass 0.4 kg as shown in figure. Assume the impact is direct and elastic. If the horizontal surface has a kinetic coefficient of friction $\mu_k = 0.3$. Locate the final position of each mass from the origin of x-axis. [Figure: 0.5 kg particle released from rest at A on a circular quarter arc of radius $R = 0.25$ m (A at $30^\circ$, height $0.25\sin 30^\circ$ above the level of the lowest point B); 0.4 kg particle at rest at B on a horizontal surface with $\mu = 0.3$; x-axis horizontal from the origin.]

## Short answer

The 0.5 kg mass reaches 1.57 m/s before impact. After elastic collision, speeds are 0.174 m/s and 1.74 m/s. With kinetic friction μ_k=0.3, stopping distances are 5.1 mm and 0.514 m from the origin, respectively.

## Answer

**Assumptions:** the arc is smooth; the origin of the x-axis is the point B (the lowest point of the arc, where the 0.4 kg particle rests); $\mu_k=0.3$ acts on the horizontal surface; $g=9.81\ \text{m/s}^2$.

#### Speed of the 0.5 kg particle at B
Drop in height $h=0.25\sin30^\circ=0.125$ m.

$$
\tfrac12m_1u^{2}=m_1gh\ \Rightarrow\ u=\sqrt{2(9.81)(0.125)}=1.566\ \text{m/s}
$$

#### Elastic direct impact ($e=1$)
$m_1=0.5$ kg, $m_2=0.4$ kg (initially at rest). Momentum and restitution:

$$
0.5(1.566)=0.5v_1+0.4v_2,\qquad v_2-v_1=1.566
$$

$$
v_1=\frac{m_1-m_2}{m_1+m_2}u=\frac{0.1}{0.9}(1.566)=0.1740\ \text{m/s}
$$

$$
v_2=\frac{2m_1}{m_1+m_2}u=\frac{1.0}{0.9}(1.566)=1.740\ \text{m/s}
$$

Both move in the +x direction, and $v_2>v_1$, so they do not collide again.

#### Sliding distances (work-energy with friction)
$$
\tfrac12mv^{2}=\mu_kmg\,d\ \Rightarrow\ d=\frac{v^{2}}{2\mu_kg}
$$

$$
d_1=\frac{0.1740^{2}}{2(0.3)(9.81)}=0.00514\ \text{m},\qquad d_2=\frac{1.740^{2}}{2(0.3)(9.81)}=0.514\ \text{m}
$$

**Answer: the 0.5 kg particle stops $0.0051$ m (5.1 mm) from the origin and the 0.4 kg particle stops $0.514$ m from the origin, both on the positive x-side.**

## Check yourself

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

1. How is the 0.5 kg particle's speed at B found?
   A. From the impulse of gravity over time
   B. From restitution with the 0.4 kg mass
   C. By energy conservation: m g h = (1/2) m u^2 with h = 0.25 sin 30
   D. From friction work on the arc
   Answer: C. The arc is smooth, so u = sqrt(2 g h) = sqrt(2(9.81)(0.125)) = 1.566 m/s. (Key point: Speed before impact)

2. Which two relations solve for the post-impact velocities?
   A. Energy lost equals zero and friction equals zero
   B. Conservation of momentum and v2 - v1 = u (e = 1)
   C. Only conservation of momentum
   D. Only e = 1 with equal masses
   Answer: B. Momentum is conserved and for e = 1 the relative speed of separation equals the speed of approach. (Key point: Elastic impact equations)

3. What are the velocities after impact?
   A. v1 = 0.174 m/s, v2 = 1.740 m/s, both in +x
   B. v1 = -0.174 m/s, v2 = 1.740 m/s
   C. v1 = 0, v2 = 1.566 m/s
   D. v1 = 1.740 m/s, v2 = 0.174 m/s
   Answer: A. v1 = (0.1/0.9)u = 0.174 m/s and v2 = (1.0/0.9)u = 1.740 m/s, so the heavier particle continues forward slowly. (Key point: Post-impact velocities)

4. How is the stopping distance on the rough surface found?
   A. d = v^2/(2 mu_k g), from (1/2) m v^2 = mu_k m g d
   B. d = v/(mu_k g)
   C. d = mu_k v^2/(2g)
   D. d = v^2/(mu_k g)
   Answer: A. Friction work mu_k m g d removes the kinetic energy, so d = v^2/(2 mu_k g): 0.0051 m and 0.514 m. (Key point: Sliding distance)

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