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ksle82
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I was able to get the answer to Conservation of Momentum but couln't get the same answer using conservation of energy? Steps are shown below. Where did i go wrong using the conserv. of engery method? Assume frictionless surface and inelastic collision.
Q: A 10 gram bullet with a velocity of 1000 m/sec strikes a 100 gram block of wood initially at rest. What is their combined velocity? Can you work the problem using the principle of Conservation of Momentum? Conservation of Energy?
using Momentum:
m1v1+m2v2= (m1+m2)vf, v2=0, m1=mass of bullet, m2=mass of block
v1=velocity of bullet, v2=velocity of block
vf=combined velocity
so vf= (m1v1)/(m1+m2) = (10*1000)/(110)~91 m/s
Using Cons. of Energy:
KE1 + PE1 = KE2 + PE2, KE=kinetic energy, PE=potential E
(1/2)m1(v1)^2 = (1/2)(m1+v2)vf PE1=PE2=0
-if solved for vf, the aswer would be different from the momentum method
Q: A 10 gram bullet with a velocity of 1000 m/sec strikes a 100 gram block of wood initially at rest. What is their combined velocity? Can you work the problem using the principle of Conservation of Momentum? Conservation of Energy?
using Momentum:
m1v1+m2v2= (m1+m2)vf, v2=0, m1=mass of bullet, m2=mass of block
v1=velocity of bullet, v2=velocity of block
vf=combined velocity
so vf= (m1v1)/(m1+m2) = (10*1000)/(110)~91 m/s
Using Cons. of Energy:
KE1 + PE1 = KE2 + PE2, KE=kinetic energy, PE=potential E
(1/2)m1(v1)^2 = (1/2)(m1+v2)vf PE1=PE2=0
-if solved for vf, the aswer would be different from the momentum method