Find the velocity of the slower object after the collision

  • Thread starter Trent2011
  • Start date
  • #1
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Homework Statement


A(n) 12 g object moving to the right at
31 cm/s overtakes and collides elastically with
a 29 g object moving in the same direction at
15 cm/s.

Find the velocity of the slower object after
the collision.

Answer in units of cm/s.


Homework Equations





The Attempt at a Solution



I know it probably looks very confusing. I am not the best at keeping track of the work I am doing. Some steps may have changed or are irrelevant to the answer:

m1*v1+m2*v2=m1*v1f+m2*v2f

v2f=[(m1*v1+m2*v2)-m1*v1f]/m2


.5(m1)(v1)^2 + .5(m2)(v2)^2 = .5(m1)v1f^2 + .5(m2)v2f^2

-----------------------------------------------------------------

v2f=(807-12v1f)/29


18057= 12v1f^2 + 29v2f^2

18057= 12v1f^2 + 29[(807-12v1f)/29]^2

[(807-12v1f)/29][(807-12v1f)/29]

(807-12v1f)(807-12v1f)

651249 - 2(9684v1f) + 144v1f^2

c.................b.................a

-19368 (+-) [19368^2 - 4(144)(651249)]^.5
______________________________________
2(144)


------------------------------------------------------------------------------

v1f = 67.25

v2f=(807-12v1f)/29

v2f=(807-12[67.25])/29

18057= 54270.75 + 29v2f^2

-36213.75 = 29v2f^2
 

Answers and Replies

  • #2
129
0


Ok so your last correct step was this.

18057= 12v1f^2 + 29[(807-12v1f)/29]^2

And then jumped into the quadratic equation.

Yeah, you can't do that in this case. You should expand the 29[(807-12v1f)/29]^2 term which yields

(1/29)[(144 x^2)-(19368 x)+651249]

Substitute this back to your previous equation.

18057= 12v1f^2 + (1/29)[(144 x^2)-(19368 x)+651249]

THEN you can move everything over to right, simplify, and use the quadratic equation.

Best of luck, quite annoying numbers there.
 

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