Projectile Recoil: Find Gun Horizontal Speed

[verden]

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.

 

[rock.freak667]

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.

 

[verden]

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?

 

[rock.freak667]

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.

 

[rwisz]

I would approach this problem by first breaking it down into its components. The gun and the projectile are two separate objects, so we can analyze their motions separately.

First, let's consider the motion of the projectile. We can use the equations of projectile motion to find the horizontal and vertical components of its velocity. We know that the initial velocity of the projectile is 600 m/s at an angle of 30.0 degrees above the horizontal, so we can use trigonometry to find its horizontal component, which is 600*cos(30) = 519.62 m/s. This is the velocity at which the projectile will travel horizontally after leaving the gun.

Next, let's consider the motion of the gun itself. We know that the gun has a mass of 1200 kg and that it will experience a recoil force when the projectile is fired. This recoil force is equal to the force exerted by the projectile on the gun, which can be calculated using the equation F = ma. The mass of the projectile is 8.00 kg and we can use the vertical component of its velocity (600*sin(30) = 300 m/s) to calculate the force: F = (8.00 kg)*(300 m/s) = 2400 N.

Now, we can use the principle of conservation of momentum to find the horizontal recoil speed of the gun. This principle states that the total momentum before an event is equal to the total momentum after the event. In this case, the total momentum before the projectile is fired is 0, as the gun and projectile are at rest. After the projectile is fired, the gun will have a horizontal recoil velocity (v) and the projectile will have a horizontal velocity of 519.62 m/s. The total momentum after the event is equal to the momentum of the gun (1200 kg * v) plus the momentum of the projectile (8.00 kg * 519.62 m/s). Therefore, we can set up the equation:

0 = (1200 kg * v) + (8.00 kg * 519.62 m/s)

Solving for v, we get v = -3.46 m/s, which is the horizontal recoil speed of the gun. The negative sign indicates that the gun will recoil in the opposite direction of the projectile's motion.

In conclusion, the angle of 30.0 degrees does not have a direct effect on the equation used