An 8kg ball travelling at 4m/s collides head on with a 3kg ball travelling at 14m/s. Rhe balls bounce off each other and travel back the way they came. The 8kg ball travels away at 2m/s calculate velocity of 3kg ball after the collision.

There are a couple of assumptions I had to make here and also a couple of rules based on what I use in my classroom when I teach the Law of Momentum Conservation. First of all, I am going to call the 8kg ball 1 and say that it is moving to the right (and right is positive), and that means that the 3kg ball is ball 2 and say that it is moving to the left (and left is negative). I had to assume that the 2 balls were moving towards each other; hence, the different signs assigned to their movement. I also added in another significant digit since we have only 1 in most of these values and adding in a .0 is not going to change the value of any number. The Law of Momentum Conservation in this particular instance says

[tex][m_1v_1+m_2v_2]_b=[m_1v_1+m_2v_2]_a[/tex] which is the mathematical way of saying that the momentum after the collision is the same as the momentum before it. Filling in:

[tex][(8.0*4.0)+(3.0*-14)]_b=[(8.0*-2.0)+(3.0*v)]_a[/tex] and doing the math here simplifies to

32 - 42 = -16 + 3.0v and

-10 = -16 + 3.0v and

6.0 = 3.0v so

v = 2.0 (and the positive indicates that ball 2 is now moving to the right)