A shell of mass 8.0-kg leaves the muzzle of a cannon with a horizontal velocity of 600 m/s. Find the recoil velocity of the cannon, if its mass is 500kg.

velocity of recoil velocity of cannon  is -9.6 m/sec

Step-by-step explanation:

according to law of conservation of momentum

total momentum of isolated system of body remains constant.

momentum  = mass of body* velocity of body.

in the problem the system is

shell + cannon

momentum of shell = 8*600 = 4800 Kg-m/sec

let the velocity of cannon be x m/sec

momentum of cannon = 500*x = 500x Kg-m/sec

initially the system of body is in rest (before the shell is fired) hence, total momentum of the system i is 0

applying  conservation of momentum

total momentum before shell fired = total momentum after the shell is fired

0 = momentum of shell + momentum of cannon

4800 + 500x = 0

x = -4800/500 = -9.6

Thus, velocity of recoil velocity of cannon  is -9.6 m/sec

here negative sign implies that direction of velocity of cannon is opposite to that of velocity of shell.

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)