A(n) 6.7-g bullet is fired from a gun into a 0.90-kg block of wood held in a vis
ID: 1583506 • Letter: A
Question
A(n) 6.7-g bullet is fired from a gun into a 0.90-kg block of wood held in a vise. The bullet penetrates the block to a depth of 9.00 cm. An identical block of wood (with no bullet inside) is next placed on a frictionless horizontal surface, and a second identical bullet is fired from the same gun into the block.
How much smaller is the penetration depth in the second case?
(Hint: The depth in the second case should be smaller because some of its original kinetic energy is retained in the bullet-block system.) mm smaller What is the muzzle velocity of the bullet if it produces 1030 J of heat in the second case? m/s
Explanation / Answer
1st Case
Wwood = DK
F d = Kf – Ki
F(.1) = 0 - ½(.0067)v2
F = -0.00335 v2
2nd Case
mv = (M+m+m)vf
0.0067 v = 0.9134 vf
vf = 0.00734v
Wwood = DK
F d = ½ 0.9134 vf2 - ½(.0067)v2
(-0.00335 v2)d = ½ 0.9134 (0.00734v)2 - ½(.0067)v2
v’s now cancel
d = 0.993 meters
1st Case
Wwood = DK
F d = Kf – Ki
F(.1) = 0 - ½(.0067)v2
F = -0.00335 v2
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