From a rifle of mass $ 4 \mathrm{~kg} $, a bullet of mass $ 50 \mathrm{~g} $ is fired with an initial velocity of $ 35 \mathrm{~m} \mathrm{~s}^{-1} $. Calculate the initial recoil velocity of the rifle.

Given:

Mass of the rifle, $m_1$ = 4 kg

Mass of the bullet, $m_2$ = 50 g =  0.05 kg    

Initial velocity of the bullet, $v_2$ = 35 m/s

To find: Initial recoil velocity of the rifle, $v_1$.

Solution:

Let the recoil velocity of the rifle = $v_1$

Before firing-

The bullet, the rifle and bullet were at rest, thus, the initial velocity $(v)$ will be zero.

$\therefore$ The total initial momentum of the rifle and bullet system is-

$=(m1+m2)\times v$               $(\because p=m\times v,\ where\ p=momentum,\ m=mass,\ v=velocity)$

$=(m1+m2)\times 0=0$         $(\because v=0)$

After firing-

Momentum of bullet = $m_1v_1$

Momentum of rifle = $m_2v_2$

Thus, the total momentum of the rifle and bullet system after firing is-

$=m_1v_1+m_2v_2$

Putting the given values in the above expression we get-

$=4\times {v_1}+0.05\times {35}$

$=4{v_1}+1.75$

Now,

According to the law of conservation of momentum

Total momentum after the firing = Total momentum before the firing

$4{v_1}+1.75=0$

$v_1=\frac {-1.75}{4}$

$v_1=\frac {-1.75}{4}$

$v_1=–0.4375m/s$

Thus, the initial recoil velocity of the rifle, $v_1$ is 0.4375m/s, the negative sign indicates that the rifle recoils backward.

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Updated on: 10-Oct-2022

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