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A 500.0 kg module is attached to a 440.0 kg shuttle craft, which moves at 1050. m/s relative to the stationary main spaceship. Then a small explosion sends the module backward with speed 100.0 m/s relative to the new speed of the shuttle craft. As measured by someone on the main spaceship, by what fraction did the kinetic energy of the module and shuttle craft, Ki, increase because of the explosion?

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pmdislamabad

Answer:

0.955286  j

Explanation:

A 500.0 kg module is attached to a 440.0 kg shuttle craft, which moves at 1050. m/s relative to the stationary main spaceship. Then a small explosion sends the module backward with speed 100.0 m/s relative to the new speed of the shuttle craft. As measured by someone on the main spaceship, by what fraction did the kinetic energy of the module and shuttle craft, Ki, increase because of the explosion?

M=500 kg,  m=440 kg

V=1000 m/s, v = 100 m/s

Let relative speed =Vs

Momentum rule says

(M+m)V=mVs+M(Vs-v)

 940(1000)=500(Vs-100)+440Vs

 940000=500Vs-50000+440Vs

 940Vs=940000+50000

 940Vs=990000

 Vs= 990000/940=1053.19 m/s

So, the module speed = Vs-v=1053.19-100=953.19 m/s

Fractional increase in KE is given by;

Total KE after explosion / He before explosion

=500(953.19)2+ 400(1053.19)2/ 940(1000)2= 0.955286

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