Two identical pucks B and C , each of mass m , are connected by a rod of length

Question

Two identical pucks B and C, each of mass m, are connected by a rod of length l and negligible mass, that is free to rotate about its center. Puck A, of mass m/2 comes in and hits B. After the collision, in which no energy is lost find: a) the rotational speed of the dumbbell and b)the velocity and of puck A

I have asked several people and have gotten several answers, which one is correct?

A first answered:

V_Afinal= -(V_Ainitial)/3

=(8/9)((V_Ainitial^2)/(l))

another answered:

After hitting there is no energy loss, it means that the Kinetic energy is conserved and momentom is also conserved

m/2 is the mass of puck a

m is the mass of pucks b and c

Since the kinetic energy is conserved we can write the conservation as,

1/2mv2=1/2Mv22+1/2Iw2

w=mvd/I (WHERE DID THIS COME FROM??)

Substituting w we get

1/2mv2=1/2Mv22+1/2*I*(mvd/I)2

divide by m2v2

1/m=1/M+d2/I

m=1/(1/M+d2/I)

m=12mML2/(1/12)ML2+Md2)

m=ML2/(L2+12d2)

By conservation of momentom,

mv=Mv2

v2=v*(m/M)

So,v2=(v/M)*(ML2/(L2+12d2))

We know that mvd=Iw

w=mvd/I

So angular speed,w=vdML2/(L2+12d2)I

A third answered:

=(4/5)((V_Ainitial)/(l ))

V_Afinal= -(3/5)(V_Ainitial)

A fourth answered:

V_Afinal= V_Ainitial - 2l 2

= -4/(4l 2- 1)

Who gave the correct answer?

Explanation / Answer

Second is absolutely correct, The portion where you have written as where did this come from is from conservation of angular momentum. I hope this answers your doubt.

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