Assume that the collision is absolutely inelastic. Applying momentum conservation, we have
mv=(m+M)u,u=m+Mmv1. Applying Newton's second law, fond the deceleration:
ma=μ(M+m)g,a=mμg(M+m).
The distance covered upon such deceleration from initial velocity to rest will be
d=2au2=2μg(m+M)3v12m3=0.38 m.
Comments
Leave a comment