Posted by **Jason** on Friday, January 6, 2012 at 9:23pm.

A(n) 11 g bullet is ﬁred into a(n) 111 g block

of wood at rest on a horizontal surface and

stays inside. After impact, the block slides

14 m before coming to rest.

The acceleration of gravity is 9.8 m/s

If the coeﬃcient of friction between the

surface and the block is 0.7, ﬁnd the speed of

the bullet before impact.

Answer in units of m/s

- physics -
**drwls**, Friday, January 6, 2012 at 10:04pm
The kinetic energy of the bullet and block after impact equals the work done against friction.

(1/2)(M+m)V2^2 = (M+m)*g*(0.7)*14m

V2^2 = 2*g*(0.7)(14) = 192 m^2/s^2

V2 = 13.9 m/s

Now use conservation of momentum to determine the velocity of the bullet, V1

m*V1 = (m+M)*V2

V1 = V2*(121/11) = 11 V2 = 152 m/s

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