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Is my solution correct? About momentum

  • Thread starter noppawit
  • Start date
27
0
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]Pi = [tex]\sum[/tex]Pf
------------>>>>> (6)(8) + (4)(2) = (6)(6.4) + 4vf2

Hence, vf2 = 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)(vf2)
------------>>>>> (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

Mentor
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a) [tex]\sum[/tex]Pi = [tex]\sum[/tex]Pf
------------>>>>> (6)(8) + (4)(2) = (6)(6.4) + 4vf2

Hence, vf2 = 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)(vf2)
------------>>>>> (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.
 

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