Saved Bookmarks
| 1. |
A cannon of mass 1000 kg located at the base of an inclined of an inclined plane fires a shell of mass 100 kg in a horizontal direction with a velocity 180 km//hr.The angle of inclination of inclined plane with the horizontal is 45^(@). The coefficient of friction between the canon and inclined plane is 0.5. The height in meter to which the cannon ascends the inclined plane as a result of recoil is (g=10m//s^(2)) |
|
Answer» `=100xx180xx5/18` `=5000`kg-m/s Velocity of recoil of cannon `=5000/1000=5m//s` Let the cannon ascends a distance s on the rough inclined surface.The effective deceleration is  `a=(muR+Mg sin 45^(@))/M` `a=(mjMgcos 45^(@)+Mgsin45^(@))/M` `a=g/(SQRT(2))[mu+1]` `v^(2()=u^(2)=2as` `0=(5)^(2)-2xxg/(sqrt(2))(mu+1)s` or `s=25/(sqrt(2)xx10(0.5+1))=25/(sqrt(2)xx15)=5/(3sqrt(2))m` Now`h=s.sin45^(@)=5/(3sqrt(2))XX1/(sqrt(2))=5/6m` |
|