Question

The last stage of a rocket, which is traveling at a speed of 7700 m/s, consist of two parts that are clamped together: a rocket case with a mass of 250.0 kg and a payload capsule with a mass of 100.0 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910.0 m/s. Assume that all velocities are along the same line.What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line. Find the total kinetic energy of the two parts (c) before and (d) after they separate. (e) Account for the difference.

Answer 1

Answer:

Explanation:

Let the velocity of rocket case and payload after the separation be v₁ and v₂ respectively. v₂ will be greater because payload has less mass so it will be fired with greater speed .

v₂ - v₁ = 910

Applying law of conservation of momentum

( 250 + 100 ) x 7700 = 250 v₁ + 100 v₂

2695000 = 250 v₁ + 100 v₂

2695000 = 250 v₁ + 100 ( 910 +v₁ )

v₁ = 7440 m /s

v₂ = 8350 m /s

Total kinetic energy before firing

= 1/2 ( 250 + 100 ) x 7700²

= 1.037575 x 10¹⁰ J

Total kinetic energy after firing

= 1/2 ( 250 x 7440² + 100 x 8350² )

= 1.0405325 x 10¹⁰ J

The kinetic energy has been increased due to addition of energy generated in firing or explosion which separated the parts or due to release of energy from compressed spring.