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.

Explanation:

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

which is the mathematical way of saying that the momentum after the collision is the same as the momentum before it. Filling in:

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)

increase masses, decrease separation