# Impulse, Momentum, and Force

1. Mar 19, 2012

### chubbyorphan

1. The problem statement, all variables and given/known data

2000kg stationary cannon
25kg cannon ball fired horizontally @ 250m/s
Assume no friction acting on the cannon
***What is the velocity of the cannon after firing the ball?

2. Relevant equations
PT = PT’
^initial momentum = final momentum
m1v1 + m2v2 = m1v1’ + m2v2’

3. The attempt at a solution

m1v1 + m2v2 = m1v1’ + m2v2’
(2000kg)(0m/s) + (25kg)(0m/s) = (2000kg)(v1’) + (25kg)(250m/s)
0 = (2000kg)(v1’) + 6250kg•m/s
v1’ = (–6250kg•m/s)/(2000kg)
v1’ = –3.125m/s
After rounding:
v1’ = –3.1m/s
Therefore, the velocity of the cannon after firing the cannonball is 3.1m/s[backward]

I think I got it.. if someone could confirm my answer for me that would be awesome!
thank you forum!

2. Mar 19, 2012

That's right. Also, I don't know how you're writing the problem on your own paper, but it's good practice never to put the numbers in until the last second. So for example,
$$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'\Rightarrow0=m_1v_1'+m_2v_2'\\ \Rightarrow m_1v_1'=-m_2v_2'\Rightarrow v_1'=-\frac{m_2v_2'}{m_1}$$
Then you finally plug in the numbers and your solution falls out. This way it's easier to keep track of dimensions and to rearrange a result to get another result in a more complicated problem.

3. Mar 19, 2012

### chubbyorphan

yeah my book actually doesn't do it that way but I'm definitely going to write it like that from now on. Thanks again SadScholar, you rock!
if you're bored I seem to always have topics I need help with >.>