Momentum and friction?

Question

Mitch throws a 50gram lump of clay with a speed of 20m/s at a 500gram target, which is at rest on a horizontal surface. After impact, the target slides 2m before stopping. Find the coefficient of friction between the target and the surface.

Answer 1

♠ initial momentum is L=m1*v1, where m1=50gram = 0.050kg, v1=20m/s;

final momentum is L=(m1+m2)*v2, where m2=500gram=0.5kg;

since momentum conserves then L=L; that is

m1*v1 =(m1+m2)*v2, hence the speed of clinched composition is

v2= m1*v1/(m1+m2); and their kin energy is

♠ E=0.5*(m1+m2)*v2^2;

This E is dissipated by force of friction f=μ*w, where

w=(m1+m2)*g is their common weight;

So work done by friction is E=f*x= μ*w*x, where x=2m;

thus E=E; 0.5*(m1 +m2)*v2^2 = μ*(m1 +m2)*g*x, hence

μ=0.5*v2^2 /(g*x) = 0.5*(m1*v1/(m1 +m2))^2 /(g*x) =

= 0.5*(0.05*20/0.55)^2 /(9.8*2) = 0.084; correct?

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