Posted by Khalifa on Sunday, February 3, 2013 at 6:55am.
A bullet of mass 45 g is fired at a speed of 220 m/s into a 5.0kg sandbag hanging from a string from the ceiling. The sandbag absorbs the bullet and begins to swing. To what maximum vertical height will it rise? Hint: Energy is not conserved as the bullet enters the bag but is conserved after the bullet comes to rest in the bag and the bag is swinging upward

Physics  Khalifa, Sunday, February 3, 2013 at 8:15am
v1'=((m1m2)/(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 8:50am
mv=(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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