A1200 kg cannon fires a 100.0 kg cannonball at 52 m/s. what is the recoil velocity of the cannon? assume that frictional forces are negligible and the cannon is fired horizontally. a) 4.3 m/s b) 5.2 m/s c) 52 m/s d) 4.7 m/s

A) 4.3 m/s

Explanation:

M = mass of the cannon = 1200 kg

m = mass of the cannonball = 100 kg

v = speed of the cannonball after the fire = 52 m/s

V = speed of cannon after fire

Since the cannon and cannonball were at rest before fire, total initial momentum will be taken as 0

Using conservation of momentum

0 = m v - M V

M V = m v

inserting the values

(1200) V = (100) (52)

V = 4.3 m/s

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer