1. The problem statement, all variables and given/known data
A spaceship of mass ##m_0## moves in the absence of external forces with a constant velocity ##v_0##. To change the motion direction, a jet engine is switched on. It starts ejecting a gas jet with velocity u which is constant relative to the spaceship and directed at right angles to the spaceship motion The engine is shut down when the mass of the spaceship decreases to ##m##. Through what angle ##\alpha## did the motion direction of the spaceship deviate due to the jet engine operation?
2. Relevant equations
3. The attempt at a solution
Since there are no external forces, the momentum is conserved. Let the initial direction of motion be the x-axis. At time t, assume that spaceship rotates by an angle ##\theta## and its mass is m(t). The momentum in x direction is ##m(t)v_0\cos(\theta)##. When it ejects mass of dm at t+dt, let it rotate by angle ##d\theta##. The momentum in x-direction at t+dt is ##(m(t)-dm)v_0\cos(\theta+d\theta)+dm(u\sin\theta+v_0\cos\theta)##. Equating them and using ##\cos(\theta+d\theta)=\cos\theta-\sin\theta d\theta## gives
$$dm u=m(t)v_0d\theta$$
Solving the D.E (limits for ##\theta##: 0 to ##\alpha## and for m: ##m_0## to ##m##),
$$\alpha=(v_0/u)\ln(m/m_0)$$
But this is wrong. The given answer is ##\alpha=(v_0/u)\ln(m_0/m)##. I believe there is a sign error but I am not able to spot it. I am not sure if I approached the problem correctly.
Any help is appreciated. Thanks!
A spaceship of mass ##m_0## moves in the absence of external forces with a constant velocity ##v_0##. To change the motion direction, a jet engine is switched on. It starts ejecting a gas jet with velocity u which is constant relative to the spaceship and directed at right angles to the spaceship motion The engine is shut down when the mass of the spaceship decreases to ##m##. Through what angle ##\alpha## did the motion direction of the spaceship deviate due to the jet engine operation?
2. Relevant equations
3. The attempt at a solution
Since there are no external forces, the momentum is conserved. Let the initial direction of motion be the x-axis. At time t, assume that spaceship rotates by an angle ##\theta## and its mass is m(t). The momentum in x direction is ##m(t)v_0\cos(\theta)##. When it ejects mass of dm at t+dt, let it rotate by angle ##d\theta##. The momentum in x-direction at t+dt is ##(m(t)-dm)v_0\cos(\theta+d\theta)+dm(u\sin\theta+v_0\cos\theta)##. Equating them and using ##\cos(\theta+d\theta)=\cos\theta-\sin\theta d\theta## gives
$$dm u=m(t)v_0d\theta$$
Solving the D.E (limits for ##\theta##: 0 to ##\alpha## and for m: ##m_0## to ##m##),
$$\alpha=(v_0/u)\ln(m/m_0)$$
But this is wrong. The given answer is ##\alpha=(v_0/u)\ln(m_0/m)##. I believe there is a sign error but I am not able to spot it. I am not sure if I approached the problem correctly.
Any help is appreciated. Thanks!
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