conana
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Homework Statement
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.)
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.