Answer to Question #234620 in Classical Mechanics for Vino

Question #234620
A bullet weighing 4.5*10^-3 kg is fixed horizontally into a 1.8 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet comes to rest in the block which moves 1.8m. Find the speed of the bullet.
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Expert's answer
2021-09-08T08:14:11-0400

Given:

m=4.5103kgm=4.5*10^{-3}\:\rm kg

M=1.8kgM=1.8\:\rm kg

μ=0.20\mu=0.20

d=1.8md=1.8\:\rm m


Let the initial speed of the bullet we denoted as uu.Then initial speed of the block is given by the law of conservation of momentum

v=umM+mumMv=u\frac{m}{M+m}\approx u\frac{m}{M}

The energy-work theorem gives

Mv22=Ffd=μMgd\frac{Mv^2}{2}=F_fd=\mu Mgd

So,

v=2μgd=umMv=\sqrt{2\mu g d}=u\frac{m}{M}

u=2μgdMmu=\sqrt{2\mu g d}\frac{M}{m}

u=20.209.81.81.84.5103=1063m/su=\sqrt{2*0.20*9.8*1.8}*\frac{1.8}{4.5*10^{-3}}\\=1063\:\rm m/s


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