Block hits a rod with axle in middle

[call-me-kiko]

A cube of mass m slides without friction at speed v₀ and undergoes a perfectly elastic collision with the bottom tip of a rod of length d and mass 2m. The rod is pinned through its center through a frictionless axle. And initially hangs straight down at rest. What is the cubes velocity both speed and direction after?

Im kinda lost on this one on where to start.

So i tried momentum equations
mv₀ =mv₁+2mv₁

but the answer is v₁=(1/5)v₀

Im guessing both objects do not receive the same speed so this would be where my equation fails...

 

[call-me-kiko]

So this just popped in my head but i'd like to check if my idea is correct
M/V is a constant. Thus the velocity for the rod must be 2v₁

Go back to the momentum equation
mv₀= mv₁ +2m*2v₁
solve for v₁
v₁=1/5 v₀

Thoughts?

 

[DorianG]

Why are you using a common value for the velocities after the collision? (v1)
That's generally for an inelastic collision, where both objects 'stick together' and move off with a common v.
Here the collision is perfectly elastic.
What do you know about perfectly elastic collisions? What quantities are preserved?

 

[call-me-kiko]

I tried pairing momentum and energy equations together and solving in terms of v₀ and v₁. but the answer was incorrect
Kinetics:
.5m v₀² = .5mv₁²+.5Iw²
where I=(1/12)MR^2
M=2m
and w= v₂/R
thus mv₀²=mv₁²+(1/6)mv₂²
v₀²=v₁²+(1/6)v₂²
Momentum:
mv₀=mv₁+Iw
v₀=v₁+(1/6)v₂R
With the R left i was unsure of what i could do here

 

[Doc Al]

.5m v₀² = .5mv₁²+.5Iw²
where I=(1/12)MR^2
M=2m

OK.

and w= v₂/R
thus mv₀²=mv₁²+(1/6)mv₂²
v₀²=v₁²+(1/6)v₂²

No need to introduce v₂; stick with ω.

Momentum:
mv₀=mv₁+Iw

You want angular momentum, not just momentum. Fix two of those terms.