A railroad gun of mass M = 2.0 kg fires a shell of mass m =1.0 kg at an angle of theta= 35.0 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 v_0 =100.0 m/s . The gun recoils with speed v_r and the instant the projectile leaves the gun, it makes an angle phi with respect to the ground.
a)What is v_p , the speed of the projectile with respect to the ground?
b)What is phi , the angle that the projectile makes with the horizontal with respect to the ground (in degrees)?
To solve this problem, we can use the principles of conservation of momentum and vector addition.
a) First, let's calculate the recoil speed of the gun (v_r). According to the conservation of momentum, the total momentum before firing is equal to the total momentum after firing. The initial momentum is zero since the gun is at rest. The final momentum is the sum of the momenta of the gun and the projectile:
0 = M * v_r + m * v_p
To find v_p, we need to express v_r in terms of v_p. We know that the final speed of the projectile relative to the gun (muzzle velocity) is given by v_0 = 100.0 m/s. Since v_0 is the relative speed between the projectile and the gun, we can write:
v_0 = v_p - v_r
Rearranging the equation, we get:
v_r = v_p - v_0
Substituting this value for v_r in the momentum conservation equation, we have:
0 = M * (v_p - v_0) + m * v_p
Expanding and rearranging the equation, we find:
(M + m) * v_p = M * v_0
Solving for v_p, we find:
v_p = (M * v_0) / (M + m)
Substituting the given values M = 2.0 kg, m = 1.0 kg, and v_0 = 100.0 m/s, we can calculate v_p:
v_p = (2.0 kg * 100.0 m/s) / (2.0 kg + 1.0 kg) = 200.0 m/s / 3.0 kg = 66.67 m/s
Therefore, the speed of the projectile with respect to the ground (v_p) is 66.67 m/s.
b) To find the angle phi, we can use vector addition. The horizontal component of the projectile's velocity remains unchanged, which is v_p*cos(theta), where theta = 35.0°.
The vertical component of the projectile's velocity changes due to the recoil of the gun. We can find this component by using the conservation of angular momentum.
The initial angular momentum is given by:
L_initial = m * v_0 * sin(theta)
The final angular momentum is given by:
L_final = m * v_p * sin(phi)
Since angular momentum is conserved during projectile motion, we can set these two expressions equal to each other:
m * v_0 * sin(theta) = m * v_p * sin(phi)
Dividing both sides of the equation by m and rearranging, we find:
v_0 * sin(theta) = v_p * sin(phi)
Now we can solve for phi by isolating sin(phi):
sin(phi) = (v_0 * sin(theta)) / v_p
Substituting the given values v_0 = 100.0 m/s, theta = 35.0°, and v_p = 66.67 m/s, we can calculate sin(phi):
sin(phi) = (100.0 m/s * sin(35.0°)) / 66.67 m/s
Using a calculator, we find:
sin(phi) ≈ 0.514
Finally, we can find phi by taking the inverse sine of this value:
phi ≈ sin^(-1)(0.514) ≈ 30.87°
Therefore, the angle that the projectile makes with the horizontal with respect to the ground (phi) is approximately 30.87°.
To solve this problem, we need to apply the principle of conservation of momentum and the laws of motion.
Let's start by finding the recoil speed of the gun, v_r. We know that the total momentum before firing is zero, as there is no external force acting on the system:
M * v_r = (M + m) * v_0
Substituting the given values, we have:
2.0 kg * v_r = (2.0 kg + 1.0 kg) * 100.0 m/s
2.0 kg * v_r = 300.0 kg m/s
v_r = 150.0 m/s
Now, let's find the speed of the projectile with respect to the ground, v_p. We need to take into account the horizontal and vertical components separately.
The horizontal component of the projectile's speed (v_px) remains constant throughout the motion. It is given by:
v_px = v_0 * cos(theta)
v_px = 100.0 m/s * cos(35.0 degrees)
v_px = 81.68 m/s
The vertical component of the projectile's speed (v_py) changes due to the acceleration due to gravity. We can use the equations of motion to find the time of flight (t) and the vertical displacement (h) of the projectile:
- h = (1/2) * g * t^2
- v_py = g * t
Since the projectile is fired at an angle of 35.0 degrees, the initial vertical velocity is:
v_py0 = v_0 * sin(theta)
v_py0 = 100.0 m/s * sin(35.0 degrees)
v_py0 = 57.55 m/s
Using the second equation of motion, we can find the time of flight:
v_py = v_py0 - g * t
0 = 57.55 m/s - 9.8 m/s^2 * t
t = 5.88 s
Now we can find the vertical displacement:
h = (1/2) * g * t^2
h = (1/2) * 9.8 m/s^2 * (5.88 s)^2
h = 169.92 m
Finally, we can find the speed of the projectile with respect to the ground using the Pythagorean theorem:
v_p = sqrt(v_px^2 + v_py^2)
v_p = sqrt((81.68 m/s)^2 + (0 m/s - 9.8 m/s^2 * 5.88 s)^2)
v_p = sqrt(81.68^2 + (-57.55)^2) m/s
v_p = sqrt(6657.23 + 3313.10) m/s
v_p ≈ 91.42 m/s
Therefore, the speed of the projectile with respect to the ground is approximately 91.42 m/s.
To find the angle phi that the projectile makes with the horizontal with respect to the ground, we can use the inverse tangent function:
tan(phi) = v_py / v_px
phi = arctan(v_py / v_px)
phi = arctan((-57.55 m/s) / (81.68 m/s))
phi ≈ -37.38 degrees
However, phi is the angle with respect to the horizontal, so we need to take the absolute value:
phi ≈ 37.38 degrees
Therefore, the angle that the projectile makes with the horizontal with respect to the ground is approximately 37.38 degrees.