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)?
To solve this problem, we can use the conservation of momentum and the conservation of energy principles.
First, let's calculate the recoil speed of the gun, vr:
Using the conservation of momentum:
Initial momentum of the system (before firing) = Final momentum of the system (after firing)
0 = (M + m) * vr - m * v0 * cos(θ)
Simplifying and solving for vr:
vr = (m * v0 * cos(θ)) / (M + m)
= (1.0 kg * 120.0 m/s * cos(45.0°)) / (3.0 kg + 1.0 kg)
vr = 30.0 m/s
Now let's calculate the speed of the projectile with respect to the ground, vp:
Using the conservation of energy:
Initial total energy of the system (before firing) = Final total energy of the system (after firing)
Initial total energy = (1/2) * (M + m) * vr^2
Final total energy = (1/2) * m * vp^2
Setting them equal and substituting vr with the value we calculated:
(1/2) * (3.0 kg + 1.0 kg) * (30.0 m/s)^2 = (1/2) * 1.0 kg * vp^2
90.0 * 900.0 = 0.5 * vp^2
vp^2 = 90000.0
vp = √(90000.0) ≈ 300.0 m/s
Therefore, vp, the speed of the projectile with respect to the ground, is approximately 300.0 m/s.
Now let's calculate the angle ϕ that the projectile makes with the horizontal with respect to the ground:
Using trigonometry:
sin(ϕ) = (v0 * sin(θ)) / vp
sin(ϕ) = (120.0 m/s * sin(45.0°)) / 300.0 m/s
sin(ϕ) ≈ 0.204
ϕ ≈ arcsin(0.204) ≈ 11.8°
Therefore, ϕ, the angle that the projectile makes with the horizontal with respect to the ground, is approximately 11.8°.
To find the speed of the projectile with respect to the ground (vp), we can use the concept of vector addition. We can break down the velocity vectors into their horizontal and vertical components.
Let's denote the horizontal component of the muzzle velocity as v0x and the vertical component as v0y.
v0x = v0 * cos(θ)
v0y = v0 * sin(θ)
Now, let's analyze the motion of the projectile once it leaves the gun. The only force acting on the projectile is gravity, which causes it to follow a parabolic path. The horizontal component of the velocity remains constant, while the vertical component changes due to the acceleration due to gravity.
For the projectile, the horizontal component of velocity remains the same throughout the motion. So, the horizontal component of the final velocity (vp x) is equal to the horizontal component of the muzzle velocity (v0x).
vp x = v0x
For the vertical component, we need to consider the initial vertical velocity (v0y) and the acceleration due to gravity (g) acting downwards. The height at which the projectile lands is the same as its initial height, which means the time of flight is the same.
Using the equation for displacement in the vertical direction, knowing the initial vertical velocity, acceleration due to gravity, and time of flight, we can find the vertical component of the final velocity (vp y).
vp y = v0y - g * t
Now, to find the time of flight, we can use the vertical equation of motion:
v0y * t + (1/2) * (-g) * t^2 = 0
Solving for t:
t = 2 * (v0y / g)
Substituting t into the equation for vp y:
vp y = v0y - g * (2 * v0y / g)
= v0y - 2 * v0y
= - v0y
Since the vertical component of the final velocity is negative, it means that the projectile is moving downward. The magnitude of the vertical component is equal to the magnitude of the initial vertical component. Hence, the speed of the projectile with respect to the ground (vp) is:
vp = sqrt(vp x^2 + vp y^2)
= sqrt(v0x^2 + (-v0y)^2)
To find the angle that the projectile makes with the horizontal with respect to the ground (ϕ), we can use the equation:
ϕ = tan^(-1)(vp y / vp x)
Now, let's substitute the given values into the equations to find the final answers.