---
title: "Slider on rod strikes spring"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:slider-rod-spring-velocity-deflection"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/slider-rod-spring-velocity-deflection
site: BCT Engineering
---

# Slider on rod strikes spring

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

- Asked in: 2074 Chaitra, 8 marks
- Video: [The 2-kg collar is released from rest at A and slides down the inclined fixed rod in the vertical](https://www.youtube.com/watch?v=VHJk2zMR4N4) — Dynamics Professor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/slider-rod-spring-velocity-deflection
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

The 4 kg slider is released from rest from position A and slides down the frictionless rod in vertical plane. Determine a) the velocity 'v' of the slider as it strikes the spring b) maximum deflection of spring. [Figure: curved rod with radius 0.6 m, slider (4 kg) starting at A at the top, spring $K = 20$ kN/m at the lower end B.]

## Short answer

The slider’s speed at the spring is found from mgh = ½mv², giving 3.43 m/s. Equating kinetic energy to spring potential energy, ½mv² = ½Kx², yields a maximum compression of 48.5 mm (assuming no additional height loss during compression).

## Answer

**Assumed geometry (figure not shown):** the rod is a quarter circle of radius $R=0.6$ m, so the slider starts at A at the level of the circle centre and drops $h=0.6$ m to the lowest point B, where it meets the horizontal spring ($K=20$ kN/m). $m=4$ kg, rod smooth.

#### a) Velocity at the spring
Only gravity does work from A to B:

$$
mgh=\tfrac12mv^{2}\ \Rightarrow\ v=\sqrt{2gh}=\sqrt{2(9.81)(0.6)}=3.431\ \text{m/s}
$$

#### b) Maximum deflection of the spring
All kinetic energy at B is stored in the spring when the slider stops (horizontal compression, no further drop):

$$
\tfrac12mv^{2}=\tfrac12Kx^{2}
$$

$$
x=v\sqrt{\frac{m}{K}}=3.431\sqrt{\frac{4}{20000}}=0.0485\ \text{m}
$$

**Answer: $v\approx3.43$ m/s; maximum spring deflection $x\approx48.5$ mm.** (If the slider moves down an additional distance $x$ while compressing the spring, use $mg(h+x)=\tfrac12Kx^{2}$.)

## Check yourself

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

1. Why does only gravity appear in the A-to-B energy equation?
   A. The spring is not yet touched and the smooth rod's normal force does no work
   B. Friction cancels gravity
   C. The spring is compressed before B
   D. The slider has no kinetic energy
   Answer: A. From A to B the rod is smooth and the spring is not yet engaged, so m g h = (1/2) m v^2. (Key point: Speed at spring)

2. What is v at the spring for h = 0.6 m?
   A. 2.43 m/s
   B. 6.87 m/s
   C. 3.43 m/s
   D. 0.6 m/s
   Answer: C. v = sqrt(2 g h) = sqrt(2(9.81)(0.6)) = 3.43 m/s. (Key point: Velocity value)

3. What happens at maximum spring deflection in the horizontal-spring model?
   A. The slider's speed is maximum
   B. The spring force is zero
   C. Potential energy of the spring is zero
   D. The slider is momentarily at rest and all KE is stored in the spring
   Answer: D. At maximum compression the slider stops, so (1/2) m v^2 = (1/2) K x^2. (Key point: Max deflection principle)

4. Which expression gives the deflection x?
   A. x = v sqrt(m/K) = 0.0485 m
   B. x = v sqrt(K/m)
   C. x = m v/K
   D. x = (1/2) m v^2/K
   Answer: A. Equating energies gives x = v sqrt(m/K) = 3.431 sqrt(4/20000) = 48.5 mm. (Key point: Deflection value)

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