If a 1000 kg cannon at initialy at rest fires a 10 kg cannonball at 200 m/s, what is the final velocity of the cannon?

Since the whole cannon AND the lead cannonball inside it are starting out motionless, there's no momentum before the shot.  

We know that momentum is conserved, so the total momentum after the shot must also be zero.  This tells us that whatever momentum the ball carries away in THAT ===> direction, the cannon itself will have the same amount of momentum in THIS <=== direction.  That way, they'll add up to zero like we need them to.

The physics we need:  Momentum = (mass) x (velocity)

OK. All students hearing protection ready now.  Ready on the left.  Ready on the right.  Ready on the firing line.   Cannon discharge in three.  Two.  One.  ZERO.  BOOM !

-- 10 kg ball flying THAT way ==> at 200 m/s.

Momentum= (10kg)x(200 m/s THAT==>way = 2,000 kg-m/s THAT==>way

-- Cannon momentum = (1000 kg) x (velocity THIS <== way)

Cannon momentum = 2,000 kg-m/s THIS<==way

(1000 kg) x (velocity) = 2,000 kg-m/s this<==way

Divide each side by (1,000 kg):

Velocity = (2,000 kg-m/s) / (1,000 kg)  this<==way

Velocity of the cannon = 2 m/s this<==way

i guess it’s a and b

a straight line sloped