A 810kg two-stage rocket is traveling at a speed of 5.15×103m/s with respect to Earth...?
A 810kg two-stage rocket is traveling at a speed of 5.15×103m/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 2.50×103m/s relative to each other along the original line of motion.
a)What is the speed of each section (relative to Earth) after the explosion?
b)What are the direction of each section (relative to Earth) after the explosion?
c)How much energy was supplied by the explosion? [Hint: What is the change in KE as a result of the explosion?]
More Questions From This User See All
Answers & Comments
Verified answer
When the explosion occurs, a force causes the two parts to accelerate away from each other. If we assume that the mass of each part of the rocket is the same, the magnitude of the acceleration of each part is the same. Since they move at a speed of 2.50×103m/s relative to each other, the speed of the front part of the rocket increases by ½ of the relative speed; and the speed of the rear part decreases by ½ of the relative speed.
½ * 2.5 * 10^3 = 1.25 * 10^3
Speed of front part relative to the earth = 5.15 * 10^3 + 1.25 * 10^3 = 6.4 * 10^3
Speed of rear part relative to the earth = 5.15 * 10^3 – 1.25 * 10^3 = 3.9 * 10^3
b) Since the speed of both parts is positive, both parts are still moving forward.
c)How much energy was supplied by the explosion? [Hint: What is the change in KE as a result of the explosion?]
Initial KE = ½ * 810 * (5.15 * 10^3)^2
KE of front part = ½ * 810 * (6.4 * 10^3)^2
KE of rear part = ½ * 810 * (3.9 * 10^3)^2
Total KE after the explosion = ½ * 810 * (10.3 * 10^3)^2
Increase of KE = ½ * 810 * [(10.3 * 10^3)^2 – (5.15 * 10^3)]
The increase of the kinetic energy of each part is approximately 3.22 *10^10 Joules!
I hope this helps you understand how to solve this type of problem!