A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is mu_k_ .
Find an expression for the bullet’s speed Vbullet
useing the variables: d mu_k_ m M
Follow up question if this helps:
What is the speed of a 11g bullet that, when fired into a 9.0kg stationary wood block, causes the block to slide 4.0cm across a wood table? Assume that mu_k_ = .20 .
I tried doing K = Friction * distance
1/2 mv ^2 = mu k * Msys * g * d and solved for V, isnt corret tho =/
Nevermind!!
I figured it out
Vi = ((M+m) * Sqr(uk*g*d))/m
2 before the uk in that
3 Answers

(K + U)initial – NC = (K + U)final, NC being nonconservative forces. μ is kinetic friction
initially, K = mv^2/2 and U = 0, NC = (m + M)gμ
finally, K = 0 and U = 0
mv^2/2 + 0 – (m + M)gμ = 0 + 0
mv^2 = 2(m + M)gμ
v^2 = (2(m + M)gμ) / m
v = √[(2(m + M)gμ) / m]

vi = ((M+m)*(sqrt(2*muk*g*d)))/m

Correct answer is here: https://answers.yahoo.com/question/index?qid=20121…