Question

Area of muzzle is $A$. Due to ejection of water (density $=\rho$) at a velocity $v$, the spring is compressed as shown in figure. The compression in spring, in 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 $\frac{\rho v^{2}A}{K}$


Mass flow rate of liquid flowing $=\rho vA$
Rate of momentum change of liquid $=\rho v^{2}A$
going out in forward direction.
Rate of momentum given to trolley in backward direction.
$=\rho v^{2}A$
= force on spring (Since force is rate of change of momentum)
$F=-Kx$
$\Rightarrow x=\frac{F}{K}$
Compression in spring, $x=\frac{\rho v^{2}A}{K}$