1. The problem statement, all variables and given/known data Hey everyone! I'm new to these forums so I apologize in advance if I do something wrong! I just had a quick question about the momentum in one direction equation m_1v_1o +/- m_2v_2o = m_1v_1f +/- m_2v_2f; I'm just not sure when I'm supposed to add these values and when I'm supposed to subtract them. I thought it had to do with the objects vector, but there is conflicting questions in my lesson (I think at least). I'm doing an online course so I don't really have a teacher to ask lol For example: Football player 1 with a mass of 110kg is moving at 8.0m/s when he crashes into football player 2 with a mass of 105kg, moving at 12 m/s [N]. During the tackle football player 1 holds onto football player 2, giving them the same velocity after 0.30s. a) Find the final velocity of each player after collision In a Physics experiment, a 1.2kg cart1 is moving at 3.2m/s [R], when it collides with a 1.8kg cart 2 which is moving at 5.8m/s[L]. After the collision cart1 is moving at 7.6m/s[L].Find the velocity for cart2 after the collision. 2. Relevant equations m_1v_1o + m_2v_2o= (m_1+m_2)v_f m_1v_1o - m_2v_2o = m_1v_1f + m_2v_2f 3. The attempt at a solution So for the football one my book shows; Let [N] be positive P_to=P_tf m_1v_1o + m_2v_2o= (m_1+m_2)v_f (110kg)(-8.0m/s)+(105kg)(12m/s)= (110kg +105kg)v_f v_f=1.77m/s[N] For the Physics experiment one: P_1o-P_2o=P_1f+P_2f Let [R] be positive; m_1v_1o - m_2v_2o = m_1v_1f + m_2v_2f (1.2kg)(3.2m/s)-(1.8kg)(5.8m/s)=-(1.2kg)(7.6m/s)+(1.8kg)v_2f -6.6kgm/s= -9.12kgm/s + (1.8kg)v_2f v_2f= 1.4m/s [R] So Is it because the two forces are combining in the first that you add? and in the second one they bounce away so you subtract? Sorry if this is a really simple question, I haven't done physics in almost 8 years! Thank you in advance!