Is my solution correct? About momentum

[noppawit]

Block 1 (mass 6 kg) is moving rightward at 8.0m/s and block 2 (mass 4 kg) is moving rightward at 2.0m/s. The surface is frictionless and a spring with a spring constant of 8000N/m is fixed to block 2. Eventually block 1 overtakes block 2. At instant block 1 is moving rightward at 6.4m/s, what are (a) the speed of block 2 and (b) the elastic potential energy of the spring?

a) $$\sum$$P_i = $$\sum$$P_f
------------>>>>> (6)(8) + (4)(2) = (6)(6.4) + 4v_f2
Hence, v_f2 = 4.4m/s

and

b) I think that kinetic energy before collision is equal to kinetic energy after collision
I tried with

1/2(m)(vi²) - 1/2ks² = 1/2(m)(v_f²)
------------>>>>> (6)(8²) - ks² = (6)(6.4²) [Because I cancel out 1/2 all equation]
------------>>>>> ks² = 138.24
Hence, potential energy of spring = (1/2)(ks²) = 69.12J

Am I all correct? For b) problem, do I have to calculate the Block2 kinetic energy?

Thank you

 

[Doc Al]

a) $$\sum$$P_i = $$\sum$$P_f
------------>>>>> (6)(8) + (4)(2) = (6)(6.4) + 4v_f2
Hence, v_f2 = 4.4m/s

This is good, since momentum is conserved.

b) I think that kinetic energy before collision is equal to kinetic energy after collision

No. Total energy remains constant, not kinetic energy.

I tried with

1/2(m)(vi²) - 1/2ks² = 1/2(m)(v_f²)
------------>>>>> (6)(8²) - ks² = (6)(6.4²) [Because I cancel out 1/2 all equation]
------------>>>>> ks² = 138.24
Hence, potential energy of spring = (1/2)(ks²) = 69.12J

Am I all correct? For b) problem, do I have to calculate the Block2 kinetic energy?

Not correct. Yes, you must include the KE of block 2 as well.