This is where we have 2 bodies moving before the collision and at an angle to each other.
This may sound complicated, but really it is just an extension of the previous section, just with an extra dimension to factor in.
This other dimension might not be so important, just remember what we did on the questions when the ball fell on the angle, we only need change one variable as the other doesn't change, check back here
So lets follow the algorithm that we had the last time, in collisions 104
We shall follow this algorithm to solve these problems
- State the Initial velocities, Masses and Final Velocities in a table for both bodies
- Calculate the momentum before the collision and allow it to equal the momentum after, usually we will find ourselves with an equation with 2 unknowns
- Now use Newtons experimental law of restitution, to determine another equation featuring the 2 knowns from the table.
- Using simultaneous equations solve for the unknowns
okay lets start using letters
okay so the velocity in j will not change as the collision will only effect the velocities in i.
so c = s & d = t
Momentum in both direction is obviously conserved, but because there is no change in j we only need to worry about the i direction
m1a + m2b = m1p + m2q
and finally there is no change to the coefficient of resitiution formulae as the bounce only effects the i plane
There's one more thing, the size of the angle of deflection
Where 1 is the original path and 2 is the deflected path,
m's are obviously the slopes of the lines,
these are easily found by m = -i/j
It is always good practice to roughly draw the diagram as below then you can tell the approximate angle as if the
Problem solving techniques,
Some problems will state that there are 2 directions at 90o to each other. Therefore these directions are perpendicular. We know the product of the slopes of 2 perpendicular lines is = - 1