A 11.8 g bullet is fired horizontally?

Question

A 11.8 g bullet is fired horizontally into a 94 g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact?

i got 1.1 as an answer but that is not right.
i don't know how to do this so any help is appreciated.thanks.

Answer 1

Start with conservation of momentum for the collision of the block and bullet

11.8*v0=105.8*vi
v0=8.966 *vi

now let's solve for vi

The work done by friction is
7.5*m*g*0.650
is equal to the KE of the combined mass after impact
.5*m*vi^2

solve for vi
vi=sqrt(7.5*9.81*2*0.650)
vi=9.78 m/s
the bullet is
87.7 m/s

That's a slow bullet