We have the mass of bullet = $m_{1} = 20 g (= 0.02 k g)$ and the mass of the pistol, $m_{2} = 2 k g$ ; initial velocities of the bullet $(u_{1})$ and pistol $(u_{2}) = 0$ , respectively. The final velocity of the bullet, $v_{1} = + 150 m s^{- 1}$ . The direction of bullet is taken from left to right (positive, by convention.) Let v be the recoil velocity of the pistol.

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Solution

Since the momentum has to be conserved:
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ where,
$m_{1}$ is mass of bullet.
$m_{2}$ is mass of pistol.
$u_{1}$ is initial velocity of bullet.
$u_{2}$ is initial velocity of pistol.
$v_{1}$ is final velocity of bullet.
$v_{2}$ is final velocity of pistol.
$0 = (0.02 \times 150) + 2 v_{2}$
$v_{2} = \frac{- 3}{2} = - 1.5 m / s$ =recoil velocity of pistol

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