A railroad gun of mass M = 3.0 kg fires a shell of mass m=1.0 kg at an angle of θ= 45.0 ∘ with respect to the horizontal as measured relative to the gun. After the firing is complete, the final speed of the projectile relative to the gun (muzzle velocity) is v0=120.0 m/s . The gun recoils with speed vr and the instant the projectile leaves the gun, it makes an angle ϕ with respect to the ground.

Question

A railroad gun of mass M = 3.0 kg fires a shell of mass m=1.0 kg at an angle of θ= 45.0 ∘ with respect to the horizontal as measured relative to the gun. After the firing is complete, the final speed of the projectile relative to the gun (muzzle velocity) is v0=120.0 m/s . The gun recoils with speed vr and the instant the projectile leaves the gun, it makes an angle ϕ with respect to the ground.

What is vp, the speed of the projectile with respect to the ground (in m/s)?

What is ϕ, the angle that the projectile makes with the horizontal with respect to the ground (in degrees)?

Answer 1

To find the speed of the projectile with respect to the ground, we can use the principle of conservation of momentum. Since there are no external forces acting horizontally, the horizontal component of momentum is conserved.

Let's define the positive direction as to the right. Initially, the gun and the projectile have zero horizontal velocity. After firing, the gun recoils with speed vr and the projectile has a horizontal velocity component vpx with respect to the ground.

Conservation of momentum equation:
0 = (M + m) * vpx - M * vr

From this equation, we can solve for vpx:
vpx = M * vr / (M + m)

Now let's find the vertical component of the velocity of the projectile with respect to the ground. Since the firing angle is 45 degrees, the vertical component of the velocity is equal to the horizontal component.

vpy = vpx
vpy = M * vr / (M + m)

The total velocity of the projectile with respect to the ground can be found using the Pythagorean theorem:

vp = √(vpx² + vpy²)
vp = √( (M * vr / (M + m))² + (M * vr / (M + m))² )

To find the angle ϕ that the projectile makes with the horizontal with respect to the ground, we can use trigonometry. The tangent of the angle is given by the ratio of the vertical component to the horizontal component:

tan(ϕ) = vpy / vpx
tan(ϕ) = (M * vr / (M + m)) / (M * vr / (M + m))

Canceling out common terms, we get:

tan(ϕ) = 1
ϕ = arctan(1)
ϕ = 45 degrees

So, the speed of the projectile with respect to the ground is vp = √( (M * vr / (M + m))² + (M * vr / (M + m))² ) m/s, and the angle that the projectile makes with the horizontal with respect to the ground is ϕ = 45 degrees.