Question

A model rocket flies straight up into the air. After the engine burns out, a spring pushes the bottom and top of the rocket apart. The velocity just before the rocket separates is vi, going up. Immediately after the rocket separates, the bottom part has speed vb = 0. If the top has mass mt and the bottom has mass mb, what is the energy provided by the spring? Assume all of the energy stored in the spring was released in the separation process and ignore air resistance and gravity.

Answer 1

Solution Initially mt rocket is moving upward with Velocity vi mb Vp Omi spring mo After seperation top part is moving with velocity = Up (soy) bottom part with 'zero'. y=0 On and on af applying conservation of momentum - Initial momentum final momentum (matmo) vi ть Хо tmput om va тт Now to the 2 my ug = (mytono) vi i Mytmb :)ui come Energy part- Initial energy of Rocket (Ernitial) (My+mo) e (gravitational Energy neglected) → final e energy of Rocket parts Imove 2 mylet EO + тт 2 + 1 2 3 myt mo -ju? MT Effnal = (my+mo 2 left e MT

As whole energy In spring the The would be the stored in spring released on seperation process. extra part in final Energy energy provided by spropong There fore. Energy of sprlong = fioral Energy ve 1 / 2 (my+mo) e Initial Energy Espring ca = loft myth ont lole ģimet moles mitmo mt --] : Esprtony = 1 / 2 (min) (notmy) u? Answer,