Question

Please answer correctly ...very important question ...must solve and answer with explaination

Answer 1

Answer:

option (1)

Explanation:

using conservation of momentum, velocity of the gun v1 is obtained.

mv = Mv1

v1 = mv/M

Now, kinetic energy of the gun after firing

= 1/2 * M * (mv/M)²

= 1/2 * m²v²/M

Using conservation of energy,

1/2 * k*x²= 1/2 * m²v²/M

x² = m²v²/Mk

x = mv/√(Mk)

therefore option (1)

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Answer 2

Answer:-

$\Large\boxed{\sf{(1)\quad\!\dfrac{mv}{\sqrt{kM}}}}$

Solution:-

Let the velocity of the gun and the mount obtained after firing the bullet be $\sf{-V.}$ Negative sign is because motion of gun and mount is opposite to that of the bullet.

Since the bullet, gun and the mount were at rest, the initial linear momentum of the system was zero. Hence by conservation of linear momentum,

$\longrightarrow\sf{mv+M(-V)=0}$

$\longrightarrow\sf{mv-MV=0}$

$\longrightarrow\sf{V=\dfrac{mv}{M}\quad\quad\dots(1)}$

Now consider the case of compression in the spring.

• Before the spring getting compressed, the gun and mount has the velocity (i.e., $\sf{-V}$) so that it has kinetic energy (maximum) but no potential energy due to elasticity of the spring at that moment.

• After the spring getting compressed to the maximum, the gun and mount attain potential energy due to elasticity of the spring (maximum) but no kinetic energy as their velocity at the instant becomes zero.

Let the maximum compression in the spring be x. Hence, by total mechanical energy conservation,

$\longrightarrow\sf{\dfrac{1}{2}kx^2=\dfrac{1}{2}MV^2}$

$\longrightarrow\sf{kx^2=MV^2}$

From (1),

$\longrightarrow\sf{kx^2=M\cdot\dfrac{m^2v^2}{M^2}}$

$\longrightarrow\sf{kx^2=\dfrac{(mv)^2}{M}}$

$\longrightarrow\sf{x^2=\dfrac{(mv)^2}{kM}}$

$\longrightarrow\sf{\underline{\underline{x=\dfrac{mv}{\sqrt{kM}}}}}$

Hence option (1) is the answer.