# Block hits a rod with axle in middle

• call-me-kiko

#### 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

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

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?

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?

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

Last edited:
.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.