Question

A bullet of mass $40g$ is horizontally fired with a velocity $100m{s}^{-1}$ from a pistol of mass $2kg$. What is the recoil velocity of the pistol ?

• A. 2 ms^-1
• B. 20 ms^-1
• C. -1 ms^-1
• D. 15 ms^-1

Answer 1

The correct option is A

2 ms^-1

Given, mass of bullet $m_b=40g=\frac{40}{1000}=0.04kg$, mass of pistol, $m_p=2kg$
Initial velocity of bullet, $u_b$ and pistol, $u_p$ is zero. (as initially both objects were at rest)
Final velocity of the bullet, $v_b=100m{s}^{-1}$
Let recoil velocity of the pistol be $v_p$
From law of conservation of momentum, total initial momentum of the system before firing = total final momentum of the system after firing.
$\therefore$$(m_b\times u_b)+(m_p\times u_p)=(m_b\times v_b)+(m_p\times v_p)$
$\Rightarrow$$0=(0.04\times 100)+(2\times v_p)$
$\Rightarrow$$0=4+(2\times v_p)$
$\Rightarrow v_p=\frac{-4}{2}=-2m{s}^{-1}$ (negative sign shows that the pistol recoils after firing)
$\therefore$Recoil velocity of the pistol is $2m{s}^{-1}$