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

Answer 1

1st Case Wwood = DK F d          = Kf         – Ki         F(.1)       = 0 - ½(.0067)v2 F              = -0.00335 v2

2^nd Case

mv = (M+m+m)v_f

0.0067 v = 0.9134 v_f

v_f = 0.00734v

W_wood    = DK   

      F      d  =    ½ 0.9134 v_f²       -    ½(.0067)v²

(-0.00335 v²)d = ½ 0.9134 (0.00734v)² -   ½(.0067)v²

                             v’s now cancel

d = 0.993 meters