Question

The recoil velocity of a gun of $5kg$, if a bullet of $25g$ acquires a velocity of $500m{s}^{-1}$ after firing from the gun, is .

• A. $5m{s}^{-1}$
• B. $2.5m{s}^{-1}$
• C. $2m{s}^{-1}$

Answer 1

The correct option is B $2.5m{s}^{-1}$

Let: $m_1$ = $25g$ = $0.025kg$, $u_1$ = 0,
$v_1=500m{s}^{-1}$, $m_2$ = $5kg$, $u_2$ = 0
Let velocity of gun after firing be $v_2$
From law of conservation of momentum, $m_1$$u_1$+$m_2$$u_2$ = $m_1$$v_1$+$m_2$$v_2$
$\Rightarrow$ $(0.025\times 0)+(5\times 0)=(0.025\times 500)+(5\times v_2)$
$\Rightarrow$$0=12.5+(5\times v_2)$
$\Rightarrow$$5\times v_2=-12.5$
$\Rightarrow$ $v_2=\frac{-12.5}{5}=-2.5m{s}^{-1}$

Thus, recoil velocity of gun is equal to $2.5m{s}^{-1}$. Here, the negative(- ve) sign shows that gun moves in the opposite direction of bullet.