Question

A gun of mass $m_1$ fires a bullet of mass $m_2$ with a horizontal speed $v_0$. The gun is fitted with a concave mirror of focal length $f$ facing towards a receding bullet. Find the speed of separations of the bullet and the image just after the gun was fired.

• A.
  $2(1+\frac{m_1}{m_2})v_o$
• B.
  $2(1+m_1{m}_2)v_o$
• C.
  $2(1+\frac{m_2}{m_1})v_o$
• D.
  $2\left(\frac{m_2}{m_1}\right)v_o$

Answer 1

The correct option is C

$2(1+\frac{m_2}{m_1})v_o$
Given,
Mass of gun$=m_1$
Mass of bullet$=m_2$
Speed of bullet$=v_o$

Let $v_1$ be the speed of gun (or mirror) just after the firing of bullet. From conservation of linear momentum, as the bullet and gun is at rest before firing,

$m_2{v}_0=m_1{v}_1$

$\Rightarrow v_1=\frac{m_2{v}_0}{m_1}...\left(1\right)$

Let, $u$ is object distnace and $v$ is image distance.

Now, $\frac{du}{dt}$ is the rate which distance between mirror and bullet is increasing $=v_1+v_0...\left(2\right)$

We know that,
$\frac{dv}{dt}$=$\left(\frac{v^2}{u^2}\right)$ $\frac{du}{dt}$

Here, $\frac{v^2}{u^2}=m^2=1$ (as at the time of firing, bullet is at pole)

$\therefore \frac{dv}{dt}=\frac{du}{dt}=v_1+v_0...\left(3\right)$


Here, $\frac{dv}{dt}$ is the rate at which distance between image (of bullet) and mirror is increasing.

So, if $v_2$ is the absolute velocity of image (towards right), then according to the relative velocity between mirror and image,

$\frac{dv}{dt}=v_1+v_0=v_2-v_1$

Therefore, speed of separation of bullet and image will be

$v_r=v_2+v_0$

$\Rightarrow v_r=2v_1+v_o+v_0$

$\Rightarrow v_r=2v_1+2v_o$

Subsituting value of $v_1$ from equation $\left(1\right)$, we have

$v_r=2(1+\frac{m_2}{m_1})v_0$

Hence, option (c) is correct answer.

Why this question ? This is a multi-conceptual question. This question requires the concept of speed of object and image in a spherical mirror, and its interpretation using relative motion.