KE of system of particles; jet on moving plate

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.