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
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
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
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