Question

A gun of mass ${m_1}$ fires a bullet of mass ${m_2}$ with a horizontal speed ${v_o}$. The gun is fitted with a concave mirror of focal length $f$ facing toward a receding bullet. Find the speed of separations of the bullet and the image just after the gun was fired.

Answer 1

- Hint: In this question, you should have prior knowledge about the conservation of momentum and also remember that in this case the velocity of the bullet will be equal to the velocity of the image, use this information to approach towards the solution of the question.

Complete step-by-step answer:
According to the given information we know that bullet of ${m_2}$ with a speed of ${v_o}$ which is fired by the gun of mass ${m_1}$
As we know that by the conservation of momentum ${m_1}{v_1} = {m_2}{v_2}$
So in this situation we can say that the momentum of bullet and mirror will be conserved and will be given by ${m_1}{v_1} = {m_2}{v_o}$
So the velocity of mirror will be $\dfrac{{{m_2}{v_o}}}{{{m_1}}}$ with respect to the ground
We know that the velocity ${v_o}$ is the speed of bullet with respect to mirror
Therefore the velocity of bullet will be ${v_o} + \dfrac{{{m_2}{v_o}}}{{{m_1}}}$
Also we know that the speed of image and the speed of bullet will be same therefore
Velocity of image will be ${v_o} + \dfrac{{{m_2}{v_o}}}{{{m_1}}}$
And we know that speed of separation is equal to the sum of velocity of bullet and velocity of image
So the speed of separation of the bullet and the image = ${v_o} + \dfrac{{{m_2}{v_o}}}{{{m_1}}}$ + ${v_o} + \dfrac{{{m_2}{v_o}}}{{{m_1}}}$ = 2(${v_o} + \dfrac{{{m_2}{v_o}}}{{{m_1}}}$)
Therefore the speed of separation of the bullet and the image will be 2(${v_o} + \dfrac{{{m_2}{v_o}}}{{{m_1}}}$).

Note: In the above solution we used the concept of conservation of momentum to find the force exerted on wall the concept of conversion of momentum according to which when two bodies of collide having some initial velocities in an isolated system thus the total momentum of system remain constant i.e. total momentum before collision is equal to the total momentum after collision which can be given by the equation ${m_1}{v_1} + {m_1}{v_1} = {m_1}{v_2} + {m_2}{v_1}$ Here $m_1$ and $m_2$ are the masses of the objects colliding and $v_1$ and $v_2$ are the velocities of the colliding objects the example of the conservation of momentum is the firing of bullets from a gun.