A bullet of mass 5.15 g is fired horizontally into a 2.74 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.185. The bullet comes to rest in the block, which moves 1.84 m.

Question

(a) What is the speed of the block immediately after the bullet comes to rest within it?
(b) At what speed is the bullet fired?

Answer 1

total mass = 5.15 + 2.74 = 2.74515 kg

momentum before
.00515 V
momentum after = .00515 V = 2.74515 Vi

work done on system by friction = mu M d
= .185 * 2.74515 * 1.84
= .934 Joules
that was the Ke of the system at Vi
(1/2)(2.74515)Vi^2 = .934 Joules
Vi = .825 m/s part a
V = .825 ( 2.74515/.00515) = 440 m/s

Answer 2

work done on system by friction = mu M g d
= .185 * 2.74515 * 1.84 * 9.8
= 9.15 Joules
that was the Ke of the system at Vi
(1/2)(2.74515)Vi^2 = 9.15 Joules
Vi =2.58 m/s part a
V = 2.58 ( 2.74515/.00515) = 1376 m/s