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December 21, 2014

December 21, 2014

Posted by **greg** on Sunday, November 4, 2012 at 9:46pm.

- phisics -
**Elena**, Monday, November 5, 2012 at 12:28pmm=7,45 g, m1=1220 g, m2=1603 g,

v=348 m/s, v1 = 0.648 m/s

The law of conservation of linear momentum for the 1st collision

m•v+0=m1•v1+m•u

u=(m•v-m1•v1)/m = v –(m1/m)v1=

348 – (1220/7.45) •0.648 = 241.9 m/s,

The law of conservation of linear momentum for the 2nd collision

mu+0=(m+m1)u1

u1= mu/(m1+m) = 7.45•241.9/(1603+7.45) =1.12 m/s.

Before the 1st collision KE= m•v²/2,

After the 1st collision

KE1=m1•v1²/2+m•u²/2

After the 2nd collision

KE2=(m+m1) •u1²/2

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