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

`6.00`

Explanation

Momentum of cannon after recoil `xx` Momentum of shell
`=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`