A cannonball of mass 50 kg is fired with a velocity of 40 m`cdot s^(-1)` form a cannon of mass 1000 kg. what will be the recoil velocity of the cannon?
If the force of friction between the surface and the wheels of the cannon is `(1)/(10)`th of the weight of the cannon, how far will the cannon move before coming to rest? [ Given, g = 10 `m cdot s^(-2)`]

Explanation

Let the velocity of recoil of the cannon be V. Hence, from the law of conservation of momentum,
0 = mv + MV or, V = - `(mv)/(M)`
Here, m = mass of the cannonball = 50 kg, v = its velocity = 40 `m cdot s^(-1) `and M = mass of the connon = 1000 kg.
From given data,
V =`- (50 xx 40)/(1000) = -2 " m"cdot s^(-1)`
Hence, recoil velocity of the cannon = 2 m`cdot s^(-1)` (backward). Frictional force between the wheels and the land surface
`= (1)/(10) xx 1000 xx 10 = 1000 N `
`therefore` Retardation of the cannon = `(1000N)/(1000kg) = 1 "m" cdot s^(-2)`
`therefore` Distance travelled by the cannon before coming to rest is,
` s = (V^(2))/(2a) = (2^(2))/(2 xx 1)` = 2 m