---
title: "KE of system of particles; jet on moving plate"
subject: "Applied Mechanics (Dynamics)"
chapter: 3
key: "semester_3/applied_mechanics_dynamics:kinetic-energy-system-water-plate"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/kinetic-energy-system-water-plate
site: BCT Engineering
---

# KE of system of particles; jet on moving plate

Applied Mechanics (Dynamics) · Chapter 3: System of particles

- Asked in: 2081 Baishakh, 4+4 marks
- Video: [Introduction: Impulse-Momentum Principle | Impact of Jet | Hydraulics and Fluid Mechanics](https://www.youtube.com/watch?v=ZHWNFBi2G8w) — The Content Paradox
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/kinetic-energy-system-water-plate
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3

## Question

Derive a expression for kinetic energy of a system of particles. A stream of water of cross-sectional area A and velocity $v_1$ strikes a plate which moves to the right with a velocity $v$. Determine the magnitude of $v$ if $A = 600$ mm$^2$, $v_1 = 30$ m/s and $P = 400$ N. [Figure: horizontal jet with velocity $v_1$ striking a vertical plate moving right at velocity $v$; force $P$ acts on the plate toward the left.]

## Short answer

Kinetic energy of a system is T = ½ m v̄² + ½ Σ m_i v_i'², where v̄ is the mass centre velocity and v_i' are relative velocities. For the moving plate, force P = ρ A (v1 – v)² gives 400 = 1000(600×10⁻⁶)(30 – v)², so v = 4.18 m/s.

## Answer

#### Kinetic energy of a system of particles
Let the particle $i$ of mass $m_i$ have velocity $\vec v_i=\vec{\bar v}+\vec v'_i$, where $\vec{\bar v}$ is the velocity of the mass centre G and $\vec v'_i$ is the velocity relative to G. Then

$$
T=\tfrac12\sum m_iv_i^{2}=\tfrac12\sum m_i(\vec{\bar v}+\vec v'_i)\cdot(\vec{\bar v}+\vec v'_i)
$$

$$
T=\tfrac12\bar v^{2}\sum m_i+\vec{\bar v}\cdot\sum m_i\vec v'_i+\tfrac12\sum m_iv_i'^{2}
$$

The middle term is zero because $\sum m_i\vec v'_i=0$ (momentum relative to G). So

$$
\boxed{T=\tfrac12m\bar v^{2}+\tfrac12\sum m_iv_i'^{2}}
$$

The kinetic energy of the system equals the kinetic energy of the total mass moving with the mass centre plus the kinetic energy due to motion relative to the mass centre.

#### Water jet on a moving plate
**Assumption:** the plate is flat and perpendicular to the jet, and the water leaves along the plate (so all the relative velocity normal to the plate is lost). The plate moves at $v$ in the jet direction, so the mass of water that strikes it per second is based on the relative velocity $v_1-v$:

$$
\dot m=\rho A(v_1-v)
$$

The force on the plate equals the rate of change of the momentum of water:

$$
P=\dot m(v_1-v)=\rho A(v_1-v)^{2}
$$

$$
400=1000(600\times10^{-6})(30-v)^{2}\ \Rightarrow\ (30-v)^{2}=666.7
$$

$$
30-v=25.82\ \Rightarrow\ v=4.18\ \text{m/s}
$$

**Answer: the plate speed is $v\approx4.18$ m/s.**

## Check yourself

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

1. Why is the middle term in the KE expansion zero?
   A. v_bar is zero
   B. Sum of m_i v'_i = 0 (momentum relative to G)
   C. The masses are equal
   D. Velocities are perpendicular
   Answer: B. Momentum measured relative to the mass centre sums to zero, so v_bar . sum(m_i v'_i) = 0. (Key point: Cross term vanishes)

2. What does T of a system equal?
   A. (1/2) m v_bar^2 only
   B. Sum of (1/2) m_i v'_i^2 only
   C. (1/2) m v_bar^2 - (1/2) sum m_i v'_i^2
   D. (1/2) m v_bar^2 + (1/2) sum m_i v'_i^2
   Answer: D. KE of the system = KE of total mass moving with G plus KE due to motion relative to G. (Key point: Result of derivation)

3. What mass flow rate strikes a plate moving at v in the jet direction?
   A. rho A v_1
   B. rho A (v_1 + v)
   C. rho A (v_1 - v)
   D. rho A v
   Answer: C. Only the relative velocity v_1 - v determines how much water reaches the plate per second. (Key point: Mass flow on moving plate)

4. What is v from 400 = 1000(600e-6)(30 - v)^2?
   A. 4.18 m/s
   B. 25.8 m/s
   C. 10 m/s
   D. 0.2 m/s
   Answer: A. (30 - v)^2 = 666.7 gives 30 - v = 25.82, so v = 4.18 m/s. (Key point: Plate speed)

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