Question

A bullet of mass m is fired from a gun of mass M. The recoiling gun compresses a spring of force constant K by a distance d. Then the velocity of the bullet is:
A. $kd\sqrt {\dfrac{M}{m}} $
B. $\dfrac{d}{M}\sqrt {km} $
C. $\dfrac{d}{m}\sqrt {kM} $
D. $\dfrac{{kM}}{m}\sqrt d $

Answer 1

Hint:According to the law of conservation of momentum, initial momentum is always equal to the final momentum. It states that the total energy of an isolated system remains constant and the energy is conserved over the time. Use a formula for the law of conservation to find the velocity of the bullet.

Complete step by step solution:
Let us consider that the recoiling speed of the gun be $ = V$.
The velocity of the bullet be $ = v$.
The mass of the bullet is $ = m$.
The mass of the gun is $ = M$.
By the law of conservation of the momentum-
We get-
$mv - MV = 0$
Make the required velocity of the gun “V” the subject –
$
mv = MV \\
\Rightarrow V = \dfrac{{mv}}{M}{\text{}}......{\text{(a)}} \\ $
Now using the conservation of energy,
We get –
$ 0.5 \times k{d^2} = 0.5 \times M{V^2}$
Take common factor “$0.5$” from both the sides of the equation and remove it.
$ k{d^2} = M{V^2}$
Make the velocity of the gun “V” the subject –
\[
\dfrac{{k{d^2}}}{M} = {V^2} \\
\Rightarrow {V^2} = \dfrac{{k{d^2}}}{M} \\
\]
Place the value of “V” from the equation (a)
\[
{\left( {\dfrac{{mv}}{M}} \right)^2} = \dfrac{{k{d^2}}}{M} \\
\Rightarrow \left( {\dfrac{{{m^2}{v^2}}}{{{M^2}}}} \right) = \dfrac{{k{d^2}}}{M} \\
 \]
Take common factors from both the sides of the equations and remove them.
$ \Rightarrow \left( {\dfrac{{{m^2}{v^2}}}{M}} \right) = k{d^2}$
Do-cross multiplication and make the velocity of the bullet, $v$ the subject –
$ \Rightarrow {v^2} = \dfrac{{k{d^2}M}}{{{m^2}}}$
Take square-root on both the sides of the equations
$ \Rightarrow \sqrt {{v^2}} = \sqrt {\dfrac{{k{d^2}M}}{{{m^2}}}} $
Square and square-root cancel each other on the left hand side of the equation.
\[\therefore v = \dfrac{d}{m}\sqrt {kM} \]

Hence, the option C is the correct answer.

Note: Remember the concept of law of conservation of momentum and its application to solve these types of problems. Refer the basic mathematical concepts for an efficient and accurate solution. Be careful while applying the concepts of square and square-root; always remember that square and square-root cancel each other.