Studyledge

Dynamics: Question 9

Syllabus 3.1, 3.2, 3.3

Structured AS 8 marks

A space probe of total mass 900 kg900\text{ kg} is at rest in deep space, far from any other body, so no external resultant force acts on it. An internal explosive charge separates the probe into two parts: a lander of mass 200 kg200\text{ kg} and a service module of mass 700 kg700\text{ kg}. Immediately after separation, the lander moves away at 12 m s−112\text{ m s}^{-1}.

(a) State the total momentum of the probe immediately before separation, and explain why the total momentum of the two parts must be the same immediately after separation. [2]

(b) Calculate the velocity of the service module immediately after separation, stating its direction relative to the motion of the lander. [3]

(c) Calculate the total kinetic energy of the two parts immediately after separation, and state the origin of this energy. [3]

Show worked solution Hide worked solution

Worked solution

Part (a): Momentum before and after separation

Before separation, the probe is at rest, so its total momentum is 0 kg m s−10\text{ kg m s}^{-1}.

The probe is isolated in deep space, so no external resultant force acts on it. The force of the explosion is internal to the system, by Newton’s third law it acts equally and oppositely on the two parts, so it cannot change the total momentum of the system. With no external resultant force, momentum is conserved, so the total momentum of the two parts immediately after separation must also be 0 kg m s−10\text{ kg m s}^{-1}.

Part (b): Velocity of the service module

Take the lander’s direction of motion as positive. By conservation of momentum: 0=mlandervlander+mservicevservice0 = m_{\text{lander}}v_{\text{lander}} + m_{\text{service}}v_{\text{service}} 0=(200)(12)+(700) vservice0 = (200)(12) + (700)\,v_{\text{service}} 700 vservice=−2400700\,v_{\text{service}} = -2400 vservice=−3.428...≈−3.43 m s−1v_{\text{service}} = -3.428...\approx -3.43\text{ m s}^{-1}

The magnitude of the service module’s velocity is 3.43 m s−13.43\text{ m s}^{-1} (3 s.f.). The negative sign shows it moves in the opposite direction to the lander, as required, since the two momenta must cancel to give a total of zero.

Part (c): Kinetic energy after separation

Ek,lander=12(200)(12)2=12(200)(144)=14400 JE_{k,\text{lander}} = \tfrac12(200)(12)^2 = \tfrac12(200)(144) = 14400\text{ J} Ek,service=12(700)(3.428...)2≈12(700)(11.76)≈4114 JE_{k,\text{service}} = \tfrac12(700)(3.428...)^2 \approx \tfrac12(700)(11.76) \approx 4114\text{ J} Ek,total=14400+4114≈18500 J (3 s.f.)E_{k,\text{total}} = 14400+4114 \approx 18500\text{ J (3 s.f.)}

Before separation the probe was at rest, so its total kinetic energy was zero. Unlike a collision (where kinetic energy is at best conserved, and usually decreases) this explosion increases the total kinetic energy of the system. This kinetic energy is not created from nothing: it comes from the chemical potential energy stored in the explosive charge, which is converted into kinetic energy of the two parts as they are pushed apart.

Final answers

  • (a) Momentum before =0 kg m s−1=\boxed{0}\text{ kg m s}^{-1}; momentum after must also be 0 kg m s−1\boxed{0}\text{ kg m s}^{-1} (no external resultant force)
  • (b) vservice≈3.43 m s−1v_{\text{service}} \approx \boxed{3.43}\text{ m s}^{-1}, opposite direction to the lander
  • (c) Total KE after ≈18500 J\approx \boxed{18500}\text{ J}, from chemical energy stored in the explosive charge