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.

**Further Reading**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now