Explain how the law of conservation of momentum can be used to explain (i) recoil of a gun (ii) motion of a rocket.

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Solution

We have the mass of bullet as $m_{1}$ and the mass of the pistol as $m_{2}$ .

Let the initial velocity of the bullet be $u_{1}$ and pistol is $u_{2}$ = 0, respectively and the final velocity of the bullet be $v_{1}$ 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 momentum after the fire = Total momentum before the fire

Momentum of the bullet and gun after the fire = Momentum of the bullet and gun after the fire

Since in this case initially, the bullet and pistol are at rest, the initial momentum is zero.

Sum of momentum after the fire is $m_{1}$ $v_{1}$ + $m_{2}$ $v_{2}$

According to the law of conservation,

0 = $m_{1}$ $v_{1}$ + $m_{2}$ $v_{2}$

$m_{1}$ $v_{1}$ = - $m_{2}$ $v_{2}$

$v_{2}$ = $\frac{- m_{1} \times v_{1}}{m_{2}}$

This is same for rocket also, instead of bullet velocity we will consider exhaust velocity of gases and instead of gun velocity we will consider 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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Motion of a rocket can be explained on the basis of Newton's third law.

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