Question #114455

Please answer using grade 11 Conservation of Energy and Momentum please!


Thank-you!


You and a colleague are on a spacewalk, repairing your spacecraft that has stalled in deep space. Your 90.0 kn colleague, initially at rest, asks you to throw her a hammer, which has a mass of 3.0 kg. You throw it to her with a velocity of 4.5 m/s [forward].


a) What is her velocity after catching the hammer?


b) What impulse does the hammer exert on her?


c) What percentage of kinetic energy is lost in the collision?

Expert's answer

(a)initial momentum = final momentum

So


3×4.5=(3+90) vorv=0.1452 m/s3\times4.5= (3+90)\ v\\or\\v=0.1452\ m/s

(b) impulse = change in momentum of hammer=

m(v4.5)=3(0.14524.5)=13.0644 kgm/sm(v-4.5)=3(0.1452-4.5)=13. 0644\ kg-m/s

(C) initial kinetic energy =

12m×4.52=0.5×3×20.25=30.375 J\frac{1}{2}m\times4.5^2=0.5\times 3\times 20.25=30.375\ J

Final kinetic energy


12(90+3)(0.1452)=6.7518 J\frac{1}{2}(90+3)(0.1452)=6.7518\ J

Percentage loss in energy = 30.3756.751830.375×100=77.77 percent\frac{30.375-6.7518}{30.375}\times100= 77 .77\ percent


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