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March 5, 2015

March 5, 2015

Posted by **Khalifa** on Sunday, February 3, 2013 at 6:55am.

- Physics -
**Khalifa**, Sunday, February 3, 2013 at 8:15amv1'=((m1-m2)/(m1+m2))v1 for the bullet =176ms^-1

And for the sandbag at rest

v2'=((2m1)/(m1+m2))v1 =396ms^-1

The bullet is conserved after the bullet comes to rest in the bag.

Hence Mgh= 1/2mv^2.

I got stuck here what is going to be the value of M, is it sum of the two masses. and value of v as well.

And i hope the above is collect.

Thanks

- Physics -
**Elena**, Sunday, February 3, 2013 at 8:50ammv=(m+M)u

u=mv/(m+M)=0.045•220/(0.045+5)=1.96 m/s

(m+M)u²/2=(m+M)gh

h= u²/2g=1.96²/2•9.8=0.196 m

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