Question

Question #62270

Answer 1

The recoil velocity of the gun is `=-0.25ms^-1`

Explanation:

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The mass of the gun is `M=6kg`

The mass of the bullet is `m=0.005kg`

The velocity of the bullet is `=300ms^-1`

The recoil velocity of the gun is

`v_(gun)=-(0.005/6)*300ms^-1`

`=-0.25ms^-1`

Answer 2

`0.25ms^-1` in the opposite direction

Explanation:

The concept behind this question is the conservation of linear momentum . When there is no external force acting on the system "Initial momentum of the system is equal to the final momentum of the system i.e, `P_i=P_f` ".

`P_i` denotes the initial momentum of the system.
`P_f` denotes the final momentum of the system .

Let's get into the question ,

Mass of the gun `M_g`= `6kg`.
Mass of the bullet `M_b`= `5xx10^-3kg`.

Since initially the system is at rest the initial momentum of the system is zero i.e, `P_i=0`

When the gun shots the bullet moves with a velocity `V_b=300ms^-1`

And we need to find the recoil velocity `V_g` of the gun

The final momentum of the system is the sum of the final momentums of the bullet and the gun

Therefore `P_f=M_b*V_b+M_g*V_g`

Since there is no external force acting on the system the initial momentum is equal to the final momentum

`P_i=P_f`
`0=M_b*V_b+M_g*V_g`
Rewriting this equation
`V_g=-(M_b*V_b)/M_g`
`V_g=-(300xx5xx10^-3)/6`
`V_g=0.25ms^-1`

Hence the recoil velocity of the gun is `0.25ms^-1` and the negative sign indicates that the gun move in the opposite direction of the bullet since velocity is a vector quantity ( quantities which also depend on their direction)