A 0.0429 kg bullet is fired from a gun at 115m/s horizontally. It immediately st
ID: 1390586 • Letter: A
Question
A 0.0429 kg bullet is fired from a gun at 115m/s horizontally. It immediately strikes a 4.550 kg block of wood that is at rest at time to = 0, colliding with the wood inelasticaily and becoming embedded in the wood. The wood itself is suspended by a thin, massless string of length 1.20 m. W hat is the momentum of the bullet just before it collides with the block of wood? What is the total momentum of the block with the embedded bullet just after the collision? What is the speed of the block of wood (with the embedded bullet) just after the collision? What is the kinetic energy of the block of wood (with the embedded bullet) just after the collision? As the block and bullet swing on the string, what is the highest point they reach, as measured from the initial height of the block?Explanation / Answer
A)
Total momentum of the system (the two objects) before the collision is:
pbefore = m1v1 + m2v2
wood is intially at rest so v2=0
pbefore = m1v1
=0.0429*115
=4.933kgm/s
B)
Total momentum of the system after the collision:
pafter = (m1 + m2)v
momentum is conserved,
pbefore=pafter
C)
V=4.933/(0.0429+4.55)
=1.074m/s
D)
KE=0.5*(m1+m2)V2
=0.5*(0.429+4.55)*1.0742
=2.871J
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