2
$\begingroup$

Suppose we have an incline of mass $m_2$ and a ball of mass $m_1$. The incline initially rests on a frictionless surface, where it can move freely, if horizontal force is exerted on it. We drop the ball from a specific height and it collides with the incline with velocity $\vec{u_1}$, which is perpendicular to the surface on which the incline lies. The angle of the incline is $a$.

The problem is to determine the state of the system after the collision, i.e. the velocity $\vec{u_2'}$ of the incline and the velocity of the ball $\vec{u_1'}$.

In my approach, these are 3 variables: $\vec{u_2'}$, $\vec{{u_1}_x'}$ and $\vec{{u_1}_y'}$, if we consider 2 axis of focus: 1 parallel to the ground and one perpendicular to it.

I can use that momentum is conserved in the x axis I described above and that the kinetic energy is also conserved but that doesn't use angle $a$ at all and I'm also missing one equation.

This is a problem I came up myself, so I don't know if any books contain it.

Any help would be greatly appreciated.

$\endgroup$
  • $\begingroup$ So what is your solution? $\endgroup$ – sammy gerbil Sep 6 '16 at 17:27
1
$\begingroup$

In one dimension, the outcome of an elastic collision is fully determined by energy and momentum conservation. In higher dimensions, this isn't true. Instead, you have a scattering problem, and you need the full microscopic dynamics, i.e. the exact form of the force between the two objects. So the answer is indeterminate.

However, in this case, we might say the force is a hard core repulsion (i.e. neither the incline or the ball compress during the collision) that is always normal to the incline. In that case, the third equation is conservation of the ball's component of momentum perpendicular to the incline. This depends on the angle $a$, so it gives you your solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.