Explain how the law of conservtion of momentum can be used to explain
(i) recoil of a gun
(ii) motion of a rocket. [3 MARKS]

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Solution

Solution: 1.5 Marks each

Law of Conservation to explain the recoil of a gun:
We have the mass of bullet as $m_{1}$ and the mass of the pistol, $m_{2}$
Let the initial velocities of the bullet be $(u_{1})$ and pistol be $(u_{2}) = 0$ , respectively and the final velocity of the bullet, $v_{1}$ and the recoil velocity of the gun be $v_{2}$ . The direction of bullet is taken as positive and the direction of recoil of the gun is taken as negative.
According to the law of conservation of momentum:
Total momenta after the fire = Total momenta before the fire
Momenta of the bullet and gun after the fire = Momenta of the bullet and gun after the fire.
Since in this case initially, the bullet and pistol are at rest, the initial momenta is zero.
Sum of momenta after the fire is $m_{1} \times v_{1} + m_{1} \times v_{2}$

according to the law of conservation:
$0 = m_{1} \times v_{1} + m_{2} \times v_{2}$
$m_{1} v_{1} = - m_{2} v_{2}$
$v_{2} = \frac{- m_{1} \times v_{1}}{m_{2}}$

It is the similar in the case of rocket too.Replace bullet velocity, with exhaust velocity of gases and instead of gun velocity, substitute it witih rocket velocity $v_{r o c k e t} = \frac{m_{G a s} \times v_{G a s}}{m_{r o c k e t}}$

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