Question

any one explain how do we get like thisGiven :
Mass of the bullet,M = 40gm= 0.04kg
Mass of the gun,m = 8kg
Velocity of bullet,u = 800ms⁻¹
To find :
The recoil of the gun(V)

Solution :
Initially both bullet and gun are at rest

⇒Intial momentum before firing = 0

______________________________

As no external force acts on the system

So According to the law of conservation of Momentum .

Total momentum before firing = Total momentum after the firing

⇒0 = mu+ Mv

⇒MV= -mu

⇒8× V= -0.04×800

\implies\:V=\dfrac{-0.04\times 800}{8}⟹V=
8
−0.04×800
​


\implies\:V=\dfrac{-4\times800}{100\times8}⟹V=
100×8
−4×800
​


\implies \: V= - 4\: ms {}^{ - 1}⟹V=−4ms
−1


Therefore;

{\purple{\boxed{\large{\bold{Recoil\:Velocity =-0.4ms{}^{-1}}}}}}
RecoilVelocity=−0.4ms
−1

​


⇒The negative sign shows that V and u are in opposite directions i.e ,the gun gives a kick in the backward direction or gun recoils with velocity v.​

Answer 1

Answer:you already have the solution

Now why do youbwany us to answer your question

Am confused

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

simple mate u have to use some codes u already have done that