A 915 kg two-stage rocket is traveling at a speed of 5.50×10^3 m/s with respect to Earth when a pre-designed explosion separates the rocket into two sections of equal mass that then move at a speed of 4.00×10^3 m/s relative to each other along the original line of motion. What is the speed of each section (relative to Earth) after the explosion? AND How much energy was supplied by the explosion? [Hint: What is the change in KE as a result of the explosion?] Please help step by step. I've tried this problem so many times but cant get the correct answer
For any future students (likely SBU like myself), the line "4000m/s relative to each other along the original line of motion" translates to Va=Vb+4000m/s
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy. Let's break it down step by step:
Step 1: Calculate the total momentum before the explosion.
The momentum (p) of an object can be calculated by multiplying its mass (m) by its velocity (v). In this case, the mass (m) of the two-stage rocket is 915 kg, and its velocity (v) is 5.50×10^3 m/s.
Therefore, the total momentum before the explosion is:
Total momentum = 915 kg × 5.50×10^3 m/s
Total momentum = 5.0325×10^6 kg·m/s
Step 2: Apply the conservation of momentum.
According to the conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion. Since the two sections of the rocket have equal mass after the explosion, we can assume they have the same momentum.
Let the velocity of one section be v1 and the velocity of the other section be v2.
The momentum of the first section is:
Momentum of section 1 = (mass of section 1) × (velocity of section 1)
Momentum of section 1 = (915 kg/2) × v1
The momentum of the second section is:
Momentum of section 2 = (mass of section 2) × (velocity of section 2)
Momentum of section 2 = (915 kg/2) × v2
As mentioned earlier, the total momentum after the explosion is equal to the total momentum before the explosion:
Momentum of section 1 + Momentum of section 2 = Total momentum
Substituting the values:
(915 kg/2) × v1 + (915 kg/2) × v2 = 5.0325×10^6 kg·m/s
Step 3: Calculate the velocities after the explosion.
From the given information, we know that the velocity of one section relative to the other section is 4.00×10^3 m/s. Therefore, we can express the velocity of one section in terms of the other section's velocity:
v1 = v2 + 4.00×10^3 m/s
Now, we can substitute this expression into the equation from Step 2:
(915 kg/2) × (v2 + 4.00×10^3 m/s) + (915 kg/2) × v2 = 5.0325×10^6 kg·m/s
Simplifying the equation:
915 kg × v2 + 4.00×10^3 kg·m/s + 915 kg × v2 = 10.065×10^6 kg·m/s
1830 kg × v2 = 10.065×10^6 kg·m/s - 4.00×10^3 kg·m/s
1830 kg × v2 = 10.061×10^6 kg·m/s
Dividing both sides by 1830 kg:
v2 = (10.061×10^6 kg·m/s) / (1830 kg)
Calculating v2:
v2 ≈ 5497.82 m/s
Since v1 = v2 + 4.00×10^3 m/s, we can calculate v1:
v1 ≈ 5497.82 m/s + 4.00×10^3 m/s
v1 ≈ 9497.82 m/s
Step 4: Calculate the change in kinetic energy.
The change in kinetic energy can be calculated by subtracting the initial kinetic energy from the final kinetic energy.
The initial kinetic energy of the two-stage rocket is:
Initial kinetic energy = (1/2) × (mass of the rocket) × (velocity of the rocket)^2
Initial kinetic energy = (1/2) × 915 kg × (5.50×10^3 m/s)^2
The final kinetic energy of each section can be calculated using their mass and velocity:
Final kinetic energy = (1/2) × (mass of each section) × (velocity of each section)^2
Final kinetic energy = (1/2) × (915 kg/2) × (5497.82 m/s)^2
The change in kinetic energy is then:
Change in kinetic energy = (Final kinetic energy) - (Initial kinetic energy)
Plug in the given values and calculate to find the answer.
I hope this step-by-step explanation helps you understand the problem better and assists you in finding the correct answer.