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InterviewSolution
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A cannon of mass 1000 kg located at the base of an inclined plane fires a shell of mass 100 kg in ahorizontal direction with a velocity 180 km/hr. The angle of inclination of inclined plane with thehorizontal is 45^@. The coefficient of friction between the cannon and inclined plane is 0.5. The height inmeter to which the cannon ascends the inclined plane as a result of recoil is : (g = 10 m//s^2): |
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Answer» SOLUTION :Momentum of cannon after recoil = Momentum of SHELL `=100xx180xx5/18` =5000 kg-m/s Velocity of recoil of cannon =`5000/1000`=5 m/s LET the cannon ascends a distance s on the rough inclined surface. The effective deceleration is  `a=(muR+Mgsin45^@)/M` `a=(muMgcos 45^@ + Mgsin 45^@)/M` `a=g/sqrt2[mu+1]` `v^2=u^2-2as` `0=(5)^2-2xxg/sqrt2(mu+1)s` or `s=25/(sqrt2xx10(0.5+1))=25/(sqrt2xx15) = 5/(3sqrt2)` m Now, `H = s.sin45^@=5/(3sqrt2)xx1/sqrt2=5/6`m |
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