Question

Area of muzzle is A. Due to ejection of water (density = r) at a velocity v, the compression in spring, at equilibrium is:

• A. $\frac{\rho v^{2}A}{2K}$
• B. $\frac{\rho v^{2}A}{K}$
• C. $\frac{2\rho v^{2}A}{K}$
• D. $\frac{4\rho v^{2}A}{K}$

Answer 1

The correct option is B $\frac{\rho v^{2}A}{K}$

The force exerted on the horizontal direction will be

$F_x=\text{rate of change in the momentum in the direction of force}$

$F_x=\frac{\text{initial momentum}-\text{final momentum}}{\text{time}}$

$F_x=\rho A{V}^2$

now spring force in x direction will be $Kx$

so we get the following equation $F_x-Kx$

$x=\frac{\rho v^{2}A}{K}$