Posted by **confused** on Sunday, November 21, 2010 at 7:08pm.

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.

(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?

- Physics -
**Damon**, Sunday, November 21, 2010 at 7:28pm
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

- whoops forgot g -
**Damon**, Sunday, November 21, 2010 at 7:31pm
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

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