---
title: "40 kg block on spring scale, central impact"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:pan-spring-scale-40kg"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pan-spring-scale-40kg
site: BCT Engineering
---

# 40 kg block on spring scale, central impact

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

- Asked in: 2081 Baishakh, 6+2 marks
- Video: [Dynamics: Lesson 24 - Work of Springs and Friction](https://www.youtube.com/watch?v=TVGUVPoYmfE) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pan-spring-scale-40kg
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

Determine the maximum deflection of the pan as a 40 kg block is dropped from a height of 1.5 m onto the 10 kg pan of a spring scale. Assume the impact to be perfectly plastic. Explain central impact and its types. [Figure: 40 kg block A above 10 kg pan B supported by a spring, $k = 25$ kN/m, drop height 1.5 m.]

## Short answer

Block speed just before impact is sqrt(2gh) = 5.425 m/s. Perfectly plastic impact gives common speed 4.340 m/s. Using spring stiffness and work-energy principle, maximum deflection of the pan below impact position is 0.210 m.

## Answer

#### Central impact and its types
**Central impact** occurs when the centres of mass of the two colliding bodies lie on the line of impact (the common normal at the contact point).

- **Direct central impact:** both velocities are along the line of impact.
- **Oblique central impact:** one or both velocities are inclined to the line of impact; the velocity components along the line of impact follow the impact laws, and the tangential components stay unchanged (smooth bodies).

By the coefficient of restitution $e$:

| Type | $e$ | Behaviour |
|---|---|---|
| Perfectly elastic | $1$ | Kinetic energy conserved; bodies separate |
| Inelastic (plastic) | $0$ | Bodies move together after impact; maximum energy loss |
| Partly elastic | $0<e<1$ | Some energy lost; bodies separate with reduced relative speed |

#### Maximum deflection of the pan
**Data:** $m_A=40$ kg, $m_B=10$ kg, $h=1.5$ m, $k=25$ kN/m, $e=0$ (perfectly plastic), $g=9.81$ m/s$^2$.

**Step 1: block just before impact**
$$
v_A=\sqrt{2gh}=\sqrt{2(9.81)(1.5)}=5.425\ \text{m/s}
$$

**Step 2: plastic impact** (pan initially at rest, momentum conserved):
$$
40(5.425)=(40+10)v'\ \Rightarrow\ v'=4.340\ \text{m/s}
$$

**Step 3: initial spring compression** (pan alone in equilibrium):
$$
x_1=\frac{m_Bg}{k}=\frac{10(9.81)}{25000}=0.003924\ \text{m}
$$

**Step 4: work-energy from just after impact to the lowest point** (total compression $x_2$, mass $M=50$ kg):
$$
\tfrac12 M v'^{2}+\tfrac12kx_1^{2}+Mg(x_2-x_1)=\tfrac12kx_2^{2}
$$

With $\tfrac12Mv'^2=470.88$ J, this gives

$$
12500x_2^{2}-490.5x_2-(470.88+0.1925-1.9247)=0
$$

$$
12500x_2^{2}-490.5x_2-469.15=0\ \Rightarrow\ x_2=0.2143\ \text{m}
$$

The pan therefore moves down $x_2-x_1=0.2143-0.0039=0.2104$ m from its position at impact.

**Answer: maximum deflection of the pan from its position at impact $\approx0.210$ m (total spring compression $\approx0.214$ m).**

## Check yourself

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

1. What defines central impact?
   A. The centres of mass lie on the line of impact and velocities are along it
   B. The bodies collide at the centre of the pan only
   C. Both bodies have equal mass
   D. Kinetic energy is conserved
   Answer: A. If the line of impact passes through both centres of mass, the impact is central. (Key point: Central impact)

2. What happens in perfectly plastic impact (e = 0)?
   A. Bodies separate with equal speeds
   B. Bodies move together after impact and energy loss is maximum
   C. No energy is lost
   D. Momentum is not conserved
   Answer: B. e = 0 means no relative velocity after impact; the bodies stick. (Key point: Plastic impact)

3. How is the common velocity v' found after the 40 kg block hits the pan?
   A. Energy conservation: 40(5.425)^2 = 50 v'^2
   B. Momentum: 40(5.425) = 50 v', giving 4.34 m/s
   C. v' = 5.425 m/s since block does not slow
   D. v' = 5.425 x 10/50
   Answer: B. Momentum is conserved in impact; energy is not (e = 0). (Key point: Speed after impact)

4. What is the correct treatment of the spring in the work-energy equation after impact?
   A. Ignore the initial compression of the spring
   B. Include the pan's initial compression x1 = m_B g / k and gravity work over (x2 - x1)
   C. Use only the kinetic energy
   D. Use block mass only for gravity work
   Answer: B. The spring is already compressed by the pan's weight, and the combined 50 kg moves down x2 - x1 from the impact position. (Key point: Energy method for deflection)

## Close variants set in other exams

- [30 kg block on spring scale, k = 30 kN/m](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Apan-spring-scale-30kg-k30) — Applied Mechanics (Dynamics), chapter 2, asked 3×
- [30 kg block on spring scale, k = 20 kN/m](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Apan-spring-scale-30kg-k20) — Applied Mechanics (Dynamics), chapter 2, asked 3×

## Related in other subjects

- [Castigliano: central deflection of simply supported beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-central-deflection-ss-beam-udl) — Theory of Structures II, chapter 2