Law of Conservation of Momentum:

Just as in any Law of Conservation (like the Laws of Conservation of Energy and Mass) the Law of Conservation of Momentum says, "Before = After" and in this case "p_before = p_after", telling us that all the momentum before an event must equal all the momentum after an event. An event can be objects colliding, two objects pushing off of each other, or something being thrown.

Process:

1) Start with your master equation:

p_before = p_after

2) For the "before" side of the equation put in a momentum (p) term for each independent object; do the same for the "after" side. NOTE: if two objects combine (stick together) they are treated as one independent object with a combined mass (m_total = m₁ + m₂)

p_1b + p_2b = p_1a + p_2a (add extra terms as needed)

3) Since p = mv, replace each p with mv for each object.

m_1bv_1b + m_2bv_2b = m_1av_1a + m_2av_2a

4) Put in the mass and velocities that are given and solve algebraicly.

If the situation is of something thrown (someone throws something, a projectile being launched) and these objects are at rest to begin with, the entire "before side" becomes zero (at rest: v = 0; p = 0) Though it may be tempting to just write: m_1av_1a = m_2av_2a this can lead to directions being wrong (one velocity will be negative). Instead, using the above process with ensure all velocity directions being correct.

NOTE: If objects are in two (or even three) dimensions, vectors should be resolved into their x and y (and z) components first. The above procedure is then utilized for each dimension (x, then y, then z) and recombined at the end. Tedious, yes, but not very difficult.

Example: A 40 kg cart going 3 m/s to the right collides with a 30 kg cart going 5 m/s to the left. If the 40 kg cart is going 4 m/s to the left afterward, how fast (and which direction) is the 30 kg cart going afterward?

1) Start with the master equation:
p_before = p_after

2) Since there are two carts moving both before and afterward:
p_1b + p_2b = p_1a + p_2a

3) Putting in mv for each object:
m_1bv_1b + m_2bv_2b = m_1av_1a + m_2av_2a

4) We know this information (I'm chosing right to be positive):
m_1b = 40 kg; v_1b = 3 m/s; v_1a = -4m/s (negative because it is going to the left afterward)
m_2b = 30kg; v_2b = -5 m/s; v_2a = ? m/s;

5) Putting the above into 3) gives:
(40kg)(3m/s) + (30kg) (-5 m/s) = (40kg)(-4m/s) + (30kg)(v_2a)
(Hint: since you know they are all in standard units [kg and m/s] - drop the units until the end.)
120 - 150 = -160 + 30(v_2a)
-30 = -160 + 30(v_2a)
130 = 30(v_2a)
v_2a = 130/30 = 4.33 m/s (positive means cart 2 ends up going to the right)