Question

bullet of mass 40gm is fired from a gun of mass 8kg with a velocity of 800 m/s, calculate the recoil velocity of gun

(a) 1m/s
(b) -1m/s
(c) 2m/s
(d) -4m/s

​

Answer 1

Given :

• 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

$\implies\:V=\dfrac{-0.04\times 800}{8}$

$\implies\:V=\dfrac{-4\times800}{100\times8}$

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

Therefore;

${\purple{\boxed{\large{\bold{Recoil\:Velocity =-4ms{}^{-1}}}}}}$

⇒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 2

Answer:

Given :

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 =-4ms{}^{-1}}}}}}

RecoilVelocity=−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.