# Collision of a mass with a stick

1. Jan 18, 2010

### conana

1. The problem statement, all variables and given/known data

A frictionless stick of mass m and length l lies at rest on a frictionless horizontal table. A mass km (where k is some number) moves with speed v0 at a 45 degree angle to the stick and collides elastically with it very close to an end. What should k be so that the mass ends up moving in the y-direction? (HINT: Remember that the stick is frictionless.)

3. The attempt at a solution

$$W=\int_{v_0}^{v_f}km\;dv+\int_0^{v_1}mv\;dv+\int_0^\omega I_0\omega\;d\omega=0$$

$$\Rightarrow \dfrac{1}{2}km(v_f^2-v_0^2)+\dfrac{1}{2}mv_1^2+\dfrac{1}{24}ml^2\omega^2=0.\hspace{.1 in}(1)$$

$$p_x=\dfrac{kmv_0}{\sqrt{2}}=mv_{1x}.\hspace{.1 in}(2)$$

$$p_y=\dfrac{kmv_0}{\sqrt{2}}=kmv_f+mv_{1y}.\hspace{.1 in}(3)$$

$$v1=\sqrt{v_{1x}^2+v_{1y}^2}}.\hspace{.1 in}(4)$$

$$L_A=\dfrac{\sqrt{2}}{4}kmv_0l=\dfrac{1}{12}ml^2\omega.\hspace{.1 in}(5)$$

Where vf is the final velocity of the mass, v1 is the final velocity of the CM of the stick and omega is the angular frequency with which the stick rotates about its CM.

So I am stuck with 6 unknowns and 5 equations unless there is some other relationship I am overlooking (e.g. something to do with the lack of friction from the hint?). Other people in my class seemed to think it was ok to assume that the stick moved strictly horizontally, eliminating the v1x and v1y. If I do this I end up with k=1/4. However, no one could give me an explanation as to why this assumption was ok to make. It doesn't make sense to me, but maybe I am overlooking something. Thanks in advance for any help.

2. Jan 19, 2010

### ehild

Re: Collision!

"What should k be so that the mass ends up moving in the y-direction"

If this means that the mass moves in the x direction after collision, and the road lies along the x axis, you have enough equations.

The road can exert only normal force on the arriving mass, so only the y components of the velocities will change.

You have an equation for the angular momentum: do not forget to include the angular momentum of the mass both before and after collision.

ehild

3. Jan 19, 2010

### conana

Re: Collision!

Rereading my initial post, the problem is not quite as clear without the diagram (however, I have no means of scanning the diagram). The x and y axes both lie within the plane of the table and after the collision the mass km is moving ONLY in the y direction. I also neglected to mention that my angular momentum is taken about a point that coincides with the center of the stick (before collision) so that the angular momentum of the mass after the collision is zero.