# A 1200kg gun mounted on wheels shoots an 8.0kg projectile with a muzzle velocity of 1800m/s at an angle of 30 degrees above the horizontal. Find the horizontal recoil velocity of the gun?

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### 1 Answer

On the horizontal (x) axis there is no force acting. It means the total horizontal momentum of the gun+projectile is conserved before and after shooting.

On the vertical (y) axis, at the moment of firing the projectile, the vertical reaction from the ground will increase accordingly to keep the vertical speed of the gun uncharged (zero). There is no total momentum conservation on the y axis.

Total horizontal momentum of the system gun+projectile **before firing** is zero (their initial speeds are zero).

`Px_("initial") =0`

The horizontal component of the projectile speed after firing is

`V1_x = V1*cos(theta) =1800*cos(30) =1558.85 m/s`

The horizontal component of the gun speed after firing is `V2_x`

Total horizontal momentum of the system gun+projectile **after firing** is

`Px_("final") = m*V1_x + M*V2_x`

Law of conservation of horizontal momentum is

`0 = m*V1_x +M*V2_x`

`V2_x = -V1_x * m/M = -1558.85*8/1200 =-10.39 m/s`

The negative sign in front of gun speed shows that the recoil is opposite to the direction of the shot.

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