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**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) [tex]\sum[/tex]P**

------------>>>>> (6)(8) + (4)(2) = (6)(6.4) + 4v

_{i}= [tex]\sum[/tex]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)(vi

^{2}) - 1/2ks

^{2}= 1/2(m)(v

_{f}

^{2})

------------>>>>> (6)(8

^{2}) - ks

^{2}= (6)(6.4

^{2}) [Because I cancel out 1/2 all equation]

------------>>>>> ks

^{2}= 138.24

Hence, potential energy of spring = (1/2)(ks

^{2}) = 69.12J

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

Thank you