- #1
icesalmon
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Homework Statement
part #1: A wooden block with m1 = 2.9kg is sliding across a frictionless surface in the −x direction at 4.5 m/s. A smaller wooden block with m2 = 1.1 kg block is traveling in the +x direction at 2.9 m/s.
part #2: Now they make a head-on elastic collision. (This means that in the COM frame, the velocity of each is reversed.)
Question #1: Transform back into the original frame by adding vcm to the velocity of each block.
What is the velocity of the m1 block in the lab frame after the collision?
Homework Equations
The Attempt at a Solution
V1,i* = V1,F*
where V* = Vobj,lab + Vlab,CM and Vlab,CM = -1(VCM,lab)
I would assume since the collision is Elastic that the velocities of the objects, since there masses don't change and Kinetic Energy and Momentum are Conserved to be the same as their respective initial velocities in the opposite direction they approached each other but this doesn't seem to be the case since Vm1,F != -Vm1,i so I run through the equation I have been given through pre-lectures and the answer still comes out wrong. My conceptual understanding of these relative velocities is drastically skewed and I am desperately trying to understand this. Thanks in advance for any help and I apologize if I haven't included information that is required.
Here's my work
V,1,i* = V,1,F*
V1,*= V,obj,lab - V,CM,lab
V,CM,Lab = 1/(4kg)*[2.9*4.5-1.1*2.9]
V,CM,lab = 2.465 m/s
V,m1,Lab = 4.5 - 2.465
V,m1,Lab = 2.035
wrong
I tried another method where I use the whole equation and all the velocities in the situation
V,1,i* = V,1,F*
V,Obj,Lab,i - V,CM,Lab,i = V,obj,lab,f - V,CM,F
-4.5m/s -(-2.46 m/s) + 2.46m/s = V,obj,lab,f
.42m/s = V,obj,lab,f
wrong
since they say add on the the V,CM in the question I try that and I get that V,m1,i = -V,m1,F
V,m1,F = 4.5 m/s
wrong
Okay so V,m1,F + V,CM,F is wrong
and V,m1,I + V,CM,F is wrong
And V,m1,I + V,CM,I is wrong
V,M1,F + V,CM,I is wrong
I don't think there are any other ways I could approach that question
Should I try using Conservation of Momentum?
ΔPsystem = 0
ΔPsystem,i - Δsystem,f = 0
or how about the conservation of kinetic energy?
ΔKEsytem = 0
ΔKEsystem,i - ΔKEsystem,F = 0
why would this approach be any different to the way in which I'm approaching the problem, both ideas demonstrate that the velocity of the system doesn't change in an elastic collision.
Maybe I'm reading the question wrong, or my interpretation of the question is wrong? Should I be adding on the Velocity of the Center of mass After the collision to the final velocity of the object, wrt the lab, in question? Maybe what I did above isn't the same as this in some way.
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