A prisoner escapee (mass M1) got himself trapped in a light framed closed train

Question

A prisoner escapee (mass M1) got himself trapped in a light framed closed train boxcar (mass M2) of length L. He recalls that shooting his toy pistol may get the boxcar to start rolling down the horizontal tracks (we neglect all forms of friction and drag)

A) Explain Conceptually, using Newtons 2nd Law, why the boxcar will move

B) Using conservation principles, calculate algebraically the speed of the boxcar after the bullet (mass m) is fired with a muzzle speed vb. Calculate that speed for M1=60kg, M2=1000kg , m=50kg, vb=100 m/s , L=10m

C) Calculate by how much the boxcar moves with respect to the track during the flight of the bullet form one end of the boxcar to the other end.

D) Assume the center of mass of the boxcar+man is in the middle of the car. Show that the position of the combined center of mass of the bullet and the boxcar+man does not move with respect to the tracks as the bullet travels.

E) The bullet eventually embeds itself in a block of wood along the far wall of the boxcar. Explain what happends to the boxcar. Evaluate his chances of escape.

Explanation / Answer

the bullet will be exerted a force so that it ll has a velocity. so that it ll exert another force backward to the prison and car system ------ v*m=(M1+M2)*v' so v' =vm/(M1+M2). muzzle speed v0=v+v'. so we have v'*(1+(M1+M2)/m)=v0. so v'=mv0/(m+M1+M2) --------time of flying is L/(v'+v)=L/v0. distance of moving v'*L/v0=L*m/(m+M1+M2). ---------- the center of mass never moved if no exeternal forces applied. ----------- the box car ll stop, because the total momentum must always be zero.

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