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Block hits a rod with axle in middle

  • #1
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...
 

Answers and Replies

  • #2
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?
 
  • #3
27
0
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?
 
  • #4
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:
  • #5
Doc Al
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.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.
 

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