Further mechanics collisions help

If I assume that after the collision P doesn't change direction and Q does change direction then by conservation of momentum (taking the initial direction of P as positive):

9mu - 8mu = 3mv_p + 4mv_q
u=3v_p + 4v_q

And then using the coefficient of restitution formula:

(v_q - v_p)/5u = e

And solving those gives v_q = (u/7)(15e+1).

This is the same as the solution given in the textbook but would you not need to check that you get the same v_q if the final velocity directions were different? Or is there a reason why the speed doesn't depend on the direction assumptions?

Reply 1

mqb2766

Original post by 0-)

When you write the equations down, there isnt really signs assumed for v_p and v_q. That should drop out when you do the maths. If you got the wrong direction at the start, you should get a negative answer for the final value/expression. As they ask for the speed, youd give a positive answer, irrespective of the direction.

Also note that you should really have |.| around the velocity difference in the restitution equation. The main argument that should be made is vq > vp after the collision otherwise P would pass through Q. This enables you to remove the |.| and solve as youve done.