Question

A bullet of mass $20g$ is fired horizontally with a velocity of $150m/s$ from a pistol of mass $2Kg$. Find the recoil velocity of the pistol.

Answer 1

Step 1: Given

Mass of the bullet, $m_1=20g=0.02Kg$

Mass of the pistol, $m_2=2Kg$

The initial velocity of the bullet and the pistol, $u_1=u_2=0$

The final velocity of the bullet $v_1=150m/s$

Let the recoil velocity of the pistol be $v_2$

Step 2: Find the total momentums

According to the Law of conservation of momentum, the initial momentum will be equal to the final momentum.

The total momentum before the bullet is fired ($M_i$) will be zero as both the bullet and the pistol are at rest.

The total momentum after the bullet is fired ($M_f$) will be,

$$\begin{gathered} M_f=m_1{v}_1+m_2{v}_2 \\ =0.02\left(150\right)+2\left(v_2\right) \\ =3+2v_2 \end{gathered}$$

Step 3: Equate the momentums

The Law of conservation of momentum states that the initial momentum will be equal to the final momentum.

$\Rightarrow M_f=M_i$

$$\begin{gathered} 3+2v_2=0 \\ {v}_2=\frac{-3}{2} \\ {v}_2=-1.5m/s \end{gathered}$$

The value is negative as it is a recoil velocity. The direction of recoil velocity is opposite to the direction of motion.

Hence, the recoil velocity of the pistol is $1.5m/s$.