Cannon Velocity after firing Cannonball

[HarleyM]

Homework Statement

A stationary 2000 Kg cannon fires a 25 kg cannonball horizontally at 250 m/s. Find the velocity of the cannon after firing the cannonball . You may assume friction is negligible.

Homework Equations

M1V1+M2V2=(M1+M2)V'

The Attempt at a Solution

(25)(250)+ (2000)(0)= (2000+25)V'
V'=(25)(250)/(2025)
V'=3.086 m/s

I just need confirmation as I am not 100% positive with my physics... thanks !

 

[Delphi51]

Why 2025?
I would write:
momentum before shot = momentum after shot
0 = mv + MV
0 = 25*250 + 2000*V

 

[MrB8rPhysics]

The total momentum of the situation before the firing is zero. This means that after the firing, the total must remain at zero. Find the momentum of the cannonball once it has been fired. The cannon must have equal momentum in the opposite direction for momentum to be conserved at zero kgm/s

i.e:

Before: (2000*0)+(25*0) = 0

After: (2000*V)-(25*250) = 0, or 2000V = 6250

Taking the cannons velocity to be positive, and the motion of the cannonball in the other direction to be negative. Should be trivial to solve from there (3.125 m/s).

 

[HarleyM]

Why 2025?
I would write:
momentum before shot = momentum after shot
0 = mv + MV
0 = 25*250 + 2000*V

LOL because my understanding of momentum is limited.. this makes much more sense and gives a velocity backwards which is consistent.

Thanks!

 

[Delphi51]

Most welcome! It pays to write the general principle down, perhaps highly abbreviated, to orient yourself and your readers. I used to be too brief but had to change when I became a school teacher. Students couldn't understand unless I wrote the whole story.