Question

A gun of mass $M$, frees a shell of mass $m$ horizontally and the energy of explosion is such as would be sufficient to project the shell vertically to a height '$h$'. The recoil velocity of the gun is:

• A. ${\left(\frac{2m^{2}gh}{M(m+M)}\right)}^{\frac{1}{2}}$
• B. ${\left(\frac{2m^{2}gh}{M(m-M)}\right)}^{\frac{1}{2}}$
• C. ${\left(\frac{2m^{2}gh}{2M(m-M)}\right)}^{\frac{1}{2}}$
• D. ${\left(\frac{2m^{2}gh}{2M(m+M)}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is A ${\left(\frac{2m^{2}gh}{M(m+M)}\right)}^{\frac{1}{2}}$

Applying linear momentum conservation,

$p_i=p_f$

$\Rightarrow 0=m{V}_g=m{V}_b$

$\Rightarrow V_b=\frac{m{V}_g}{m}$......(1)

According to question, total explosion energy is equal to $mgh$

$\Rightarrow \frac{1}{2}m{V}_g^2+\frac{1}{2}m{V}_b^2=mgh$

$m{v}_g^2+m{\left(\frac{M}{m}\right)}^{2}V62g=2mgh$ ( By (1))

$v_g^2(M+\frac{M^2}{m})=2mgh$

$v_g^2=2mgh\times \frac{m}{m(m+M)}$

$v_g=\sqrt{\frac{2m^{2}gh}{M(m+M)}}$ (A)