A 10.0-g bullet is fired into a 200-g block of wood at rest on a horizontal surface. after impact, the block slides 8.00 m before coming to rest. if the coefficient of friction between the block and the surface is 0.400, what is the speed of the bullet before impact? (a) 106 m/s (b) 166 m/s (c) 226 m/s (d) 286 m/s (e) none of those answers is correct

Step 1 - - determine the acceleration of the 200-g block after bullet hits it a = (coeff of friction) * g g = acceleration due to gravity = 9.8 m/sec^2 (constant) a = 0.400*9.8 a = 3.92 m/sec^2 Step 2 - - determine the speed of the block after the bullet hits it Vf^2 - Vb^2 = 2 (a) (s) where Vf = final velocity = 0 (since it will stop) Vb = velocity of block after bullet hits it a = - 3.92 m/sec^2 s = stopping distance = 8 m (given) Substituting values, 0 - Vb^2 = 2 (-3.92) (8) Vb^2 = 62.72 Vb = 7.92 m/sec. M1V1 + M2V2 = (M1 + M2) Vb where M1 = mass of the bullet = 10 g (given) = 0.010 kg. V1 = velocity of bullet before impact M2 = mass of block = 200 g (given) = 0.2 kg. V2 = initial velocity of block = 0 Vb = 7.92 m/sec Substituting values, 0.010 (V1) + 0.2 (0) = (0.010 + 0.2) (7.92) Solving for V1, V1 = 166.32 m/sec. Therefore the answer is (B) 166 m/s!