A 12.0 kg shell is launched at an angle of 55 degrees above the horizon with an initial velocity of 150 m/s. When at its highest point, the shell explodes into 2 fragments - 1 fragments is 3 times heavier than the other. The 2 fragments reach the ground at the same time, ignore air resistance. If the heavier fragment lands back at the same point from which the shell was launched, where will the 2nd fragment land and how much energy was released by the explosion?
consider the explosion. You know momentum before, so the sum of the momentum after must equal that. That means the heavier fragment had exactly equal and opposite the veloity of the shell just before explosion, then that leaves the other fragement momentum. Some tricky algebra will be required.
I have the same problem and I can't figure it out the answers on these forums are useless
To solve this problem, we can break it down into several steps. First, we need to determine the initial velocity of the shell in the horizontal and vertical directions.
Step 1: Determine the initial velocity components
The horizontal component (Vx) remains constant throughout the trajectory and can be calculated using the initial velocity (Vi) and the launch angle (θ):
Vx = Vi * cos(θ)
In this case, Vi = 150 m/s and θ = 55°:
Vx = 150 * cos(55°)
Vx ≈ 88.24 m/s
The vertical component (Vy) changes due to gravity and can be determined using the formula:
Vy = Vi * sin(θ) - g * t
Where g is the acceleration due to gravity (approximately 9.8 m/s²) and t is the time after the shell is launched.
Step 2: Determine the time to reach the highest point
At the highest point of the shell's trajectory, the vertical velocity (Vy) will be zero. So, we can set Vy = 0 and solve for t:
0 = Vi * sin(θ) - g * t
t = Vi * sin(θ) / g
Step 3: Determine the time to reach the ground for the shell fragments
Since the fragments reach the ground at the same time, we can calculate the time for both fragments using the same formula:
t_total = 2 * t
Step 4: Determine the vertical displacement of the heavier fragment
At the same final time, the heavier fragment will reach the same vertical displacement as the launch point of the shell. The vertical displacement can be calculated using the formula:
d = Vi * sin(θ) * t - (1/2) * g * t^2
Since t_total is the time for both fragments to reach the ground, we can calculate the vertical displacement (d_heavy) for the heavier fragment:
d_heavy = Vi * sin(θ) * t_total - (1/2) * g * t_total^2
Step 5: Determine the vertical displacement of the lighter fragment
Since the lighter fragment is three times lighter, its initial velocity (Vi_light = Vi_heavy) is the same as the heavier fragment. However, its mass (m_light) is one-third of the heavier fragment's mass (m_heavy = 3 * m_light). Therefore, its acceleration due to gravity is three times greater (g_light = 3 * g). Using the same formula as in Step 4, we can calculate the vertical displacement (d_light) for the lighter fragment:
d_light = Vi_light * sin(θ) * t_total - (1/2) * g_light * t_total^2
Step 6: Determine the horizontal displacement of the lighter fragment
The horizontal displacement of the lighter fragment is the same as the shell's horizontal displacement, which can be calculated using:
d_horizon = Vx * t_total
Now, let's calculate the values for each step:
Step 1: Determine the initial velocity components
Vx = 150 * cos(55°) ≈ 88.24 m/s
Step 2: Determine the time to reach the highest point
t = (150 * sin(55°)) / 9.8 ≈ 9.23 s
Step 3: Determine the time to reach the ground for the shell fragments
t_total = 2 * 9.23 ≈ 18.46 s
Step 4: Determine the vertical displacement of the heavier fragment
d_heavy = 150 * sin(55°) * 18.46 - 0.5 * 9.8 * (18.46)^2 ≈ 1278.37 m
Step 5: Determine the vertical displacement of the lighter fragment
d_light = 150 * sin(55°) * 18.46 - 0.5 * (3 * 9.8) * (18.46)^2 ≈ 1061.54 m
Step 6: Determine the horizontal displacement of the lighter fragment
d_horizon = 88.24 * 18.46 ≈ 1626.1 m
Therefore, the lighter fragment will land approximately 1061.54 meters away from the launch point, and the heavier fragment will land back at the same point from which the shell was launched.
Finally, to determine the energy released by the explosion, we need to calculate the difference in kinetic energy before and after the explosion.
The initial kinetic energy of the shell can be calculated using the formula:
KE_initial = (1/2) * m * V^2
m = 12.0 kg (mass of the shell)
V = Vi (initial velocity of the shell)
KE_initial = (1/2) * 12.0 * (150)^2 = 135,000 J
Since the fragments reach the ground at the same time, the total kinetic energy after the explosion will be:
KE_total = KE_heavy + KE_light
KE_total = (1/2) * m_heavy * V_heavy^2 + (1/2) * m_light * V_light^2
Assuming the lighter fragment has a velocity of V_light, we can determine the heavier fragment's velocity based on the conservation of momentum:
m_shell * V_shell = m_heavy * V_heavy + m_light * V_light
12.0 * 150 = 3 * m_light * V_heavy + m_light * V_light
1800 = 4 * V_light + V_heavy -- (Equation 1)
Since both fragments reach the ground at the same time, they will have the same time of flight (t_total) and initial vertical velocity (Vi) but different horizontal velocities (Vx_heavy and Vx_light). We can equate the equations for vertical motion and horizontal motion to find V_light:
Vi * t_total = g_light * t_total
(V_light * sin(θ)) * t_total = (3 * g * t_total)
V_light * sin(θ) = 9.8 * 3
V_light = 29.4 m/s
Using Equation 1, we can substitute the values and solve for V_heavy:
1800 = 4 * 29.4 + V_heavy
1800 - 117.6 = V_heavy
V_heavy = 1682.4 m/s
Now, we can calculate the kinetic energies:
KE_heavy = (1/2) * m_heavy * V_heavy^2 = (1/2) * 3 * (1682.4)^2 ≈ 4,503,972 J
KE_light = (1/2) * m_light * V_light^2 = (1/2) * (29.4)^2 ≈ 254.43 J
Finally, the total energy released by the explosion is:
Energy released = KE_initial - KE_total
Energy released = 135,000 - (4,503,972 + 254.43) ≈ 130,773 J
So, approximately 130,773 joules of energy were released by the explosion.
To find out where the second fragment will land, we can first analyze the motion of the shell before it explodes. We can break down the initial velocity of the shell into its horizontal and vertical components.
The initial velocity in the horizontal direction (Vx) remains constant throughout the motion and can be found using the formula:
Vx = V * cos(theta),
where V is the initial velocity of the shell (150 m/s) and theta is the launch angle (55 degrees).
Vx = 150 * cos(55)
Vx ≈ 88.63 m/s
The initial velocity in the vertical direction (Vy) changes due to the acceleration due to gravity. The equation for the vertical component of the velocity is:
Vy = V * sin(theta) - g * t,
where g is the acceleration due to gravity (9.8 m/s²) and t is the time.
At the highest point, the vertical component of the velocity becomes zero. We can find the time it takes for the shell to reach this point using the equation:
0 = Vy - g * t_max,
0 = V * sin(theta) - g * t_max,
t_max = (V * sin(theta)) / g
t_max ≈ (150 * sin(55)) / 9.8
t_max ≈ 9.83 s
Since we know it takes the same amount of time for both fragments to reach the ground, we can use the time t_max to determine the horizontal distance traveled by the fragments.
For the heavier fragment (Fragment A):
The horizontal distance traveled by Fragment A is given by:
d_A = Vx * t_max,
d_A = 88.63 * 9.83,
d_A ≈ 871.46 m.
Since Fragment A lands back at the same point from which the shell was launched, we conclude that Fragment B will also land at this point.
To find the energy released by the explosion, we can use the principle of conservation of momentum. Since the vertical component of momentum is zero at the highest point, the total momentum before and after the explosion should be conserved.
Let's assume the mass of Fragment B is m_B and the mass of Fragment A is 3 times heavier, which would make it 3m_B. The velocity of Fragment B can be represented as v_B and the velocity of Fragment A as v_A.
Using the principle of conservation of momentum, we have:
(12 kg * Vy) + (0 kg * Vx) = (m_B * v_B) + (3m_B * v_A),
0 = m_B * v_B + 3m_B * v_A.
Since we know velocity (v) is the change in position (d) divided by the time (t), we can rewrite the above equation:
0 = m_B * (d_B / t_max) + 3m_B * (d_A / t_max),
0 = m_B * d_B + 3m_B * d_A.
We know that Fragment A travels a distance of d_A ≈ 871.46 m and that Fragment A is three times heavier than Fragment B. Let's denote the mass of Fragment B as m_B.
So, we have:
0 = m_B * d_B + 3m_B * 871.46.
Simplifying the equation, we find:
3m_B * 871.46 = -m_B * d_B,
2614.38 m_B = -d_B.
Therefore, the second fragment (Fragment B) lands at a horizontal distance (d_B) of -2614.38 meters from the launch point. The negative sign indicates that it lands behind the launch point.
To calculate the energy released by the explosion, we need to use the equation for kinetic energy:
Kinetic Energy = (1/2) * mass * velocity^2.
Let's assume the velocity of Fragment B is v_B.
The kinetic energy released by the explosion is given by:
Energy = (1/2) * m_B * v_B^2.
However, since we don't have the values for mass or velocity, we cannot calculate the exact energy released by the explosion without additional information.