Bullet hitting block attached to spring

  • Thread starter needhelp83
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  • #1
needhelp83
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A 0.0125 kg bullet strikes a 0.300-kg block attached to a fixed horizontal spring whose spring constant is 2.25 x 103 N/m and sets it into vibration with an amplitude of 12.4 cm. What was the speed of the bullet if the two objects move together after impact?

E=(1/2)kA^2=(1/2)(2250 N/m)(0.124 m)2=17.30 J

m=0.0125 kg + 0.3 kg=0.3125 kg

K=(1/2)mv^2
17.30 J=(1/2)(0.3125 kg)v^2
v2=0.009
v=0.095 m/s

Conservation of momentum:
(.0125 kg)v=(0.3125 kg)(0.095 m/s)
v=2.375 m/s

Have I done this properly? The velocity doesn't seem right to me. Have I missed anything?
 

Answers and Replies

  • #2
Doc Al
Mentor
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needhelp83 said:
K=(1/2)mv^2
17.30 J=(1/2)(0.3125 kg)v^2
v2=0.009
v=0.095 m/s
Redo this calculation.
 
  • #3
needhelp83
199
0
K=(1/2)mv2
17.30 J=(1/2)(0.3125 kg)v2
v2=110.72
v=10.52 m/s

Conservation of momentum:
(.0125 kg)v=(0.3125 kg)(10.52 m/s)
v=152.91 m/s

Yes, this looks better. Thanks for the help!
 

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