Sphere striking 45-45-90 triangle block, collision with restitution

[Bebalo]

Hi, I'm having trouble with an assignment problem. If you happen to have Hibbeler's Tenth Edition of "Dynamics" it is problem 15-79. For those who don't have a copy of the book, the problem is:

The sphere of mass "m" falls and strikes the triangular block with a vertical velocity v. If the block rests on a smooth surface and has a mass 3m, determine its velocity just after the collision. The coefficient of restitution is e.

The block is a 45-45-90 triangle and the sphere is falling and striking the middle of the hypotenuse. The surface the block is resting on is perpendicular to the motion of the falling sphere (ie: surface is like counter on which block is sitting).

I've tried using conservation of momentum in the normal direction (through the center of mass of the triangle and sphere, makes 45 degree angle with surface) but this results in the final velocity of the block being "into" the surface. How do I deal with this.

The solution given in the book is: velocity of block = ((1-e)/7)*initial v of sphere.

Thanks in advace for the help.

 

[Andrew Mason]

The solution given in the book is: velocity of block = ((1-e)/7)*initial v of sphere.

Tough problem. Made all the more difficult due to my lack of familiarity working with coefficient of restitution especially at oblique angles.

I assume the sphere/ball deflects at a 45 degree angle to the surface so its motion immediately after the collision is horizontal. Its speed after a headon collision would be given by the coefficient of restitution (e = speed of approach/speed of separation) but what is the effect of the angle? If you can get that, I think you can figure it out using conservation of linear momentum.

It looks to me like in a head on collision, $KE_{ball}+KE_{\Delta} = e^2KE_{initial}$. In a 45 degree collision, the loss of energy would be 1/2 (.707^2) of that of a head on collision. But that is just a guess. And I am too beat right now to wrap my head around it.

AM