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?