Question

Calculate the recoil velocity of a gun of mass $5kg$ which fires a bullet of mass $50g$ with velocity $500m/s$ . Calculate the impulse given to the bullet.

Answer 1

Hint: To solve this question, we first have to assume different variables for the mass of the gun and the bullet as we as their respective velocities and put their values as given in the question. Then, we have to calculate the initial momentum, i.e. momentum before the firing and the final momentum. Now, using the law of conservation of linear momentum we can find the recoil velocity of the gun.

Complete answer:
Let the mass of the gun be $M$ and the mass of the bullet be$m$. The recoil velocity of the gun be ${v_1}$ and the velocity of the bullet be ${v_{2.}}$. Let the combined velocity before firing be $V$. But, here we assume that the whole system initially was at rest, therefore $V = 0m/s$ Now, in the question the given values are:
$M = 5kg$
$m = 50g = 0.05kg$
${v_2} = 500m/s$
The initial momentum of the whole system with the gun and the bullet is given by,
$(M + m)V$
$ \Rightarrow (M + m) \times 0......(1)$
Thus, the initial momentum is zero.
Now, after firing the final momentum of the system will be given by,
$M{v_1} + m{v_2}$
Putting the given values we get,
$ \Rightarrow 5 \times {v_1} + 0.05 \times 500$
$ \Rightarrow 5 \times {v_1} + 25.......(2)$
But, according to the law of conservation of momentum, the initial momentum before firing is equal to the momentum after firing. Therefore equations (1) and (2) we get,
$ \Rightarrow (M + m) \times 0 = 5 \times {v_1} + 25$
$ \Rightarrow 5 \times {v_1} + 25 = 0$
$ \Rightarrow 5 \times {v_1} = - 25$
$ \Rightarrow {v_1} = \dfrac{{ - 25}}{5}$
$ \Rightarrow {v_1} = - 5m/s$
Therefore the recoil velocity of the gun is $5m/s$ in the direction opposite to the direction of the motion of the bullet as indicated by the negative sign.
To find the impulse given to the bullet, we have to use the following formula,
$I = mv - mu......(3)$
Where,
$I$ is the impulse.
$m$ is the mass of the bullet.
$v$ is the final velocity of the bullet.
$u$ is the initial velocity of the bullet.
Putting the values of the initial and final velocity of the bullet in equation (3) we get,
$I = m{v_2} - mV$
$ \Rightarrow I = m({v_2} - V)$
$ \Rightarrow I = 0.05(500 - 0)$
$ \Rightarrow I = 25kgm/s$
Therefore, the impulse given to the bullet is $25kgm/s$.

Note:
Here, we are able to use the law of conservation of energy in this question because there are no external forces acting on the system. If any external force acts on the system, then the law of conservation of linear momentum cannot be used as in such cases the initial and final momentum will not be conserved. As momentum is a vector quantity, the direction also matters in this case. So, if the bullet moves in the forward direction, then to conserve the momentum, the recoil velocity of the gun will be in the opposite direction.