Posted by **Anonymous** on Sunday, December 23, 2012 at 7:40pm.

A(n) 15 g bullet is ﬁred into a(n) 116 g block

of wood at rest on a horizontal surface and

stays inside. After impact, the block slides

15 m before coming to rest.

The acceleration of gravity is 9.8 m/s

2

.

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**, Monday, December 24, 2012 at 7:26pm
The kinetic energy of the (bullet+block) after impact equals the work done against friction at the point where it stops. Solve for the velocity right after impact, V'. Then use conservatiuon of momentum to solve for the bullet velocity, V.

(1/2)(m+M) V'^2 = (m+M)*Mu*g*X

V' = 14.35 m/s

m*V = (m+M)*V'

- Physics -
**Serena**, Sunday, November 2, 2014 at 1:21am
What does the M in Mu stand for in the equation? Momentum?

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