Question

a body of mass 1.5kg, traveling along the positive x axis with speed 4.5m/s,collides with another body B of mass 3.2kg which,initially is at rest. A is deflected and moves with a speed of 2.1m/s in a direction which is 30 degrees below the x axis. B Is set in motion at angle b above the x axis. calculate the velocity of B after collision.

Answer 1

This is a conservation of momentum question where we need to consider the momentum in two dimensions. This means that the horizontal momentum after the collision is the same as it was before the collision, and that the vertical momentum after the collision is also the same as it was before the collision. I'll define the variables as follows:

m = smaller mass (the one initially moving)
M = bigger mass (the one initially at rest)
v = the speed of the smaller mass
V = the speed of the bigger mass
x = movement in the horizontal direction
y = movement in the vertical direction
a = angle at which smaller mass is deflected from the x axis
b = angle at which bigger mass is deflected from the x axis
' = denotes final value (i.e. after collision)

We can look at the horizontal momentum first:

\[p_{x(before)}=p_{x(after)}\]\[mv_x+MV_x=mv_x'+MV_x'\]
The larger mass was at rest initially, so Vx is 0. Also, since the final speeds of the masses are diagonal, we need to use trigonometry so that we're just considering the horizontal components of their speed. That means we can rewrite the equation above like this:

\[mv_x=mv' \cos a+MV' \cos b\]
The question gives you most of the information, but you're missing 2 variables: V' and b. We have 2 unknowns, so we need one more equation to be able to solve for them. We can get this equation from considering the vertical momentum. As before, vertical momentum is the same before and after the collision:

\[p_{y(before)}=p_{y(after)}\]\[mv_y+MV_y=mv_y'+MV_y'\]
Here though, the initial vertical momentum was 0 since there was only horizontal movement to begin with! This means the final vertical momenta of both masses must add to equal 0 as well. Since they're moving in opposite directions, the speed of one is negative relative to the speed of the other:

\[0=mv_y'-MV_y'\]\[mv_y'=MV_y'\]
Now same as before, we need to use trigonometry to isolate the vertical component of the final speed:

\[mv' \sin a=MV' \sin b\]
Just like before, you can substitute in all the values except for V' and b. Two equations, two unknowns - now you can solve for both to find your answer!

I know this was a lot of information so let me know if you need me to clarify anything!

Answer 2

Thanks a lot Matt. I finally got an answer. V=1.35m/ s and b=21.4.