Answer:
Explanation:
Let's just have our reference frame travel along with the original un broken mass. This way the original velocity is not relevant.
Each half will have a mass of 770/2 = 385 kg
Each half will have the same magnitude of velocity (conservation of momentum) which will be 2.6 x 10³/2 = 1.30 x 10³ m/s
Now add back the reference frame velocity to get velocity relative to earth.
Section one will have velocity 6.90 x 10³ + 1.30 x 10³ = 8.2 x 10³ m/s
Section two will have velocity 6.90 x 10³ - 1.30 x 10³ = 5.6 x 10³ m/s
In the moving reference frame, each half will have kinetic energy which could only come from the explosion
KE = ½(385)(1.3 x 10³)² = 325,325,000 J
2(325,325,000) = 650,650,000 J released in the explosion.
Rounding to the three significant figures of the problem numerals
E = 6.50 x 10⁸ J or 650 MJ released
