posted by Anonymous .
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
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
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'
What does the M in Mu stand for in the equation? Momentum?