Question

A bullet of mass 0.025 kg is fired from a gun of mass 5 kg with a speed 500 m/s.
Calculate the recoil velocity of gun.

Answer 1

Answer:

2.5 m/s

Explanation:

V(g) = - { M(b) × V(b) } ÷ M(g)

= - {0.025 × 500} ÷ 5

= -2.5 m/s

Answer 2

Answer:

A bullet of mass 0.025 kg is fired from a gun of mass 5 kg with a speed of 500 m/s.  The recoil velocity of the gun is 2.5 m/s.

Explanation:

we know,

momentum = mass $\times$ velocity

that is, p = mv...(1)

we use the conservation of linear momentum to find out the recoil velocity of the gun.

let,

$m_{1}$ and $m_{2}$ be the mass of bullet and gun respectively

$v_{1}$ and$v_{2}$ be the velocity of the bullet and the recoil velocity of the gun respectively.

Using equation (1),

The momentum of bullet = $m_{1}v_{1}$

The momentum of gun    = $m_{2}v_{2}$

According to the conservation of momentum,

The momentum of the bullet = momentum of the gun

That is,

$m_{1}v_{1} = m_{2}v_{2}$

$v_{2} = \frac{m_{1}v_{1}}{m_{2} }$

Given,

$m_{1}$ = 0.025 kg

$m_{2}$ = 5 kg

$v_{1}$   = 500 m/s

substituting,

$v_{2} = \frac{0.025\times500}{5 }$

    = 2.5 m/s

Ans:

The recoil velocity of the gun = 2.5 m/s