Find the height the block rises to after an elastic collision

[Neon32]

Homework Statement

Two blocks are free to slide along a frictionless wooden track ABC as shown below. The block of mass m1 = 5.09 kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2 = 9.50 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

Answer in m

figure:

Homework Equations

We'll use law of energy conservation: KEi+P.Ei=K.Ef+P.E
+
Vf=(m1-m2/m1+m2)Vi

The Attempt at a Solution

K.Ei+P.Ei=K.Ef+P.Ef
m1gh=1/2 m1 V1i²[/B]
So we get V1i=9.90 m/s.

Substituting in V1f=(m1-m2/m1+m2)Vi, w get:
V1f=-3.3m/s

1)I want to know if what I did above is correct.

2) My second question is

I'll use the law of energy conservation again so we can find the maximum height:
The second body will acquire velocity so there is kinetic energy after collision:

1/2m1 v1²+1/2 m2 V2²=m1gh(new height)
Is this one correct?

 

[Neon32]

Updated.

 

[Buffu]

m1gh=1/2 m1 V1i2

Did not get this part.
it is given that $V_{1i} = 0$.

 

[Neon32]

Did not get this part.
it is given that $V_{1i} = 0$.

This is velocity of mass 1 right before it collides with body 2.

 

[Buffu]

Substituting in V1f=(m1-m2/m1+m2)Vi, w get:
V1f=-3.3m/s

1)I want to know if what I did above is correct.

Yes that is correct. (if the calculations are correct).

1/2m1 v12+1/2 m2 V22=m1gh(new height)

You did not clarify what v12 and v22 means here so I assume :-
v12 - velocity of m1 after collision.
v22 - velocity of m2 after collision.

You will get the same height this way (because 1/2m1 v12+1/2 m2 V22 = initial energy of the body m1). Which is incorrect since some energy is transferred this way.
You should not consider the kinetic energy of second body in your equation, i.e 1/2 m2 V22 should not come.

Please use subscript and superscript in writing variables, it is hard to understand.

 

[Buffu]

Which is incorrect since some energy is transferred this way.

Sorry for this ambiguous statement. I meant to say that some energy is transferred to m2, so height of m1 will be less than initial.

 

[haruspex]

The question looks a bit flawed to me. If the magnets are strong enough to prevent collision, m₂ will start to move when m₁ is still some way short of it. From the diagram, it will still be on the arc. This means momentum will not be conserved, since the normal force has a horizontal component and it will be more than merely enough to support m₁.
Glossing over that...

Substituting in V1f=(m1-m2/m1+m2)Vi, w get:

Please use parentheses correctly. That cannot be what you mean.

V1f=-3.3m/s

Please check that. I make it a bit less.

 

[PeroK]

Homework Equations

Vf=(m1-m2/m1+m2)Vi

If you know that equation, can see you a good way to solve the problem without calculating any velocities?