Projectile Recoil: Find Gun Horizontal Speed

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To find the horizontal recoil speed of a 1200 kg gun firing an 8 kg projectile at a muzzle velocity of 600 m/s at a 30-degree angle, conservation of momentum is applied. The horizontal component of the projectile's velocity is calculated using 600 cos(30), which is approximately 519.6 m/s. The momentum of the gun before firing is zero, so the equation 1200 kg * V = 8 kg * 519.6 m/s is used to find the recoil speed. Solving this gives a recoil speed of approximately 3.46 m/s for the gun. Understanding the angle is crucial as it affects the horizontal velocity component used in the momentum calculation.
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Homework Statement



A 1200 kg gun mounted on wheels shoots an 8.00 kg projectile with a muzzle velocity of 600 m/s at an angle of 30.0 degrees above the horizontal. Find the horizontal recoil speed of the gun.

Homework Equations



F = ma

The Attempt at a Solution



I can not find anything in the book that resembles this equation. so do not know where to start. The answer in the book is 3.46 m/s

Also does the degrees have anything to do with the equation because it asks for the horizontal speed.
 
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What is happening is that something is being fired and hence losing mass.

So it some initial momentum and then it loses some mass (at some velocity and hence momentum).

Hence you need to apply conservation of momentum here.
 
After checking some of the other threads figured that

1200kg x V = 8 kg x 600 m/s but that puts V at 4 m/s

am I going in the right direction

Also is this 4 m/s in relation with the 600 m/s what if I needed to solve for the bullet and the gun in relation to the Earth instead f the gun?
 
verden said:
After checking some of the other threads figured that

1200kg x V = 8 kg x 600 m/s but that puts V at 4 m/s

am I going in the right direction

Also is this 4 m/s in relation with the 600 m/s what if I needed to solve for the bullet and the gun in relation to the Earth instead f the gun?

You have the general idea down, but the velocity is at an angle of 30 degrees, so the horizontal component of velocity would be 600cos(30) m/s.
 
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