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## Homework Statement

A hockey-disk with mass m, radius "r" collides with a wall in an angle "alpha" with velocity "v". There is friction, but the place where the ball touches the wall doesn't slide. How does the velocity component that is parallel to the wall change? The collision is completely elastic

## Homework Equations

Fdt = mdx

Frdt = (theta)*w

## The Attempt at a Solution

I actually have a problem after the solution, with the no sliding part. First, here's my solution:

Let's the parallel be the x axis, the other one y. Since the collision is elastic, the y component will only become the opposite of what it was.

Let the collision take place during a dt amount of time, in which the average friction force upon the ball is F. The x component of the velocity was v*sin(alpha) before the collision, and is changed by dvx. So we can write:

F*dt = mdvx

Also, during this time a torque causes the hockey-disk to rotate, and if the angular momentum will be w, we can write:

F*r*dt = (1/2)mr^2 *w

From the above two equations, we get that dvx = r*w/2

Since no other forces except F are applied in the x direction, the energy is conserved:

m(vsinalpha)^2 = m(vsinalpha - dvx)^2 + (theta)*w^2

m*dvx^2 + 1/2*m*r^2*w^2 = 2mvsin(alpha)*dvx ,substitute in dvx:

m*(1/4)*r^2*w^2 + (1/2)*m*r^2*w^2 = mvrwsin(alpha) , divide by mrw:

rw/4 + rw/2 = vsin(alpha)

3/4*r*w = vsin(alpha)

w=4vsin(alpha)/(3r)

and so dvx = 2vsin(alpha)/3.

That would be great and the problem is supposed to be done, but here's my real problem. Since the point where the ball and the wall touch isn't supposed to slide, it's velocity should be zero during and at the end of the collision process. Let this velocity be ux, we know that at the end:

ux = vsin(alpha) - dvx - r*w = vsin(alpha) -2vsin(alpha)/3 - 4vsin(alpha)/3 = -vsin(alpha)

Now that isn't zero! What did I do wrong?