Question

ғɪɴᴅ ᴛʜᴇ ʀᴇᴄᴏɪʟ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ ᴀ ɢᴜɴ ʜᴀᴠɪɴɢ ᴍᴀss ᴇϙᴜᴀʟ ᴛᴏ 5 ᴋɢ , ɪғ ᴀ ʙᴜʟʟᴇᴛ ᴏғ 25 ɢᴍ ᴀᴄϙᴜɪʀᴇs ᴛʜᴇ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ 500ᴍs^-1 ᴀғᴛᴇʀ ғɪʀɪɴɢ ғʀᴏᴍ ᴛʜᴇ ɢᴜɴ . (ɪɴ ᴍs^-1) ​

Answer 1

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}$

GIVEN:

→Mass of the bullet is 25g =0.025kg.$(m_{2})$

→Mass of the gun is 5kg.$(m_{1})$

→Final velocity of the bullet is 500m/s $(v_{2})$

→Initial velocity of gun and bullet=0m/s

(since the whole system is at rest)

TO FIND:

→The recoil velocity of the gun. $(v_{1})$

ANSWER:

Here we will be using the concept of conservation of momentum.

By which,

$\large\green{\boxed{m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}}}$

Substituting the values, we have;

=>$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

=>$5×0+0.025×0=5×v_{1} + 0.025× 500$

=>$0 = 5v_{1}+\dfrac{25}{1000}×500$

=>$0 = 5 v_{1}+\dfrac{25}{2}$

=>$5v_{1} = \dfrac{-25}{2}$

=>$v_{1}=\dfrac{-25}{2×5}$

=>$v_{1}=\dfrac{-5}{2}$

=>$v_{1}=-2.5ms^{-1}$

Therefore the recoil velocity of the gun is -2. 5m/s.Negative sign implies that the direction of recoil of gun is in opposite to the direction of the firing of bullet.

$\huge\orange{\boxed{ANSWER:-2.5ms^{-1}}}$