A cannon of mass 1000 kg located at the base 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 cannon 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 = 10 m//s^​2​)`:

Explanation

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