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November 28, 2014

November 28, 2014

Posted by **Jessica** on Monday, October 29, 2012 at 3:35pm.

(a) Determine the velocity of each block just before the collision.

(b) Determine the velocity of each block immediately after the collision.

(c) Determine the maximum heights to which m1 and m2 rise after the collision.

- Physics -
**Elena**, Monday, October 29, 2012 at 4:10pm(a) The law of conservation of energy:

PE=KE

mgh =mv²/2

v=sqrt(2gh) = sqrt(2•9.8•5) = 9.9 m/s

m1==> v1 =+9.9 m/s

m2 ==> v2 = - 9.8 m/s

(b) the law of conservation of linear momentum:

After elastic collision

u1 = [(m1 - m2)/(m1 + m2)]•v1 + [ 2 •m2/(m1 + m2) ]•v2

u1 = [(2 kg – 3.7)/(2 + 3.7)]•(9.9) + [ 2•3.7 /(2 + 3.7)]•( - 9.9)=...

u2 = [2• m1 /(m1 + m2)]•v1 + [(m2 - m1) / (m1 + m2) ]•v2

u2= [2 •2 ) / (2 + 3.7)] •9.9 + [ (3.7- 2 ) / (2 + 3.7)]•( - 9.9 )=...

(c) law of conservation of energy:

KE=PE

mu²/2= mgh

h= u²/2g,

=>

h1= u1²/2g

h2= u2²/2g

- Physics -
**Elena**, Monday, October 29, 2012 at 4:12pmmgh =mv²/2

v=sqrt(2gh) = sqrt(2•9.8•5.6) = 10.5 m/s

Correct further calculations...

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