Momentum and Energy Question
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Momentum and Energy Question

[From: ] [author: ] [Date: 11-08-17] [Hit: ]
h=0.17367m or 17.......
A bullet of mass 45 g is fired at a speed of 220 m/s into a 5.0-kg 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?

not sure how to do this question

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m1 = 45g
v1 = 220 m/s
m2 = 5kg = 5000g
v2 = 0 m/s

m1*v1 + m2*v2 = (m1 + m2)*v
45*220 = 5045*v
v = 1.96 m/s

In this case your kinetic energy will transform in to potential once it has reached it optimum height.
Q = U
(mv^2)/2 = mgh (masses cancel and solve for h)
I got: h = 0.20 m (considering sigfigs)

Note Bharat's answer is correct in execution, however he used 200 m/s has the bullet's initial velocity

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since the bullet gets embedded in the sand bag,use law of conservation of momentum...
200*0.045(by converting mass of bullet in kg's)+0*5=(5+0.045)V
V=1.845 m/sec
h=v^2/2g (using v^2-u^2=2as)
h=1.845*1.845/2*9.8
h=0.17367m or 17.367cm
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