Answer to Question #142717 in Mechanics | Relativity for Desmond

Question #142717
A bullet of mass 10 g traveling horizontally with a velocity of 300 ms- strikes a block of wood of mass 290 g which rests on a rough horizontal floor. After impact, the block and bullet move together and come to rest when the block has traveled a distance of 15 m. Calculate the coefficient of sliding friction between the block and the floor.
1
Expert's answer
2020-11-30T14:59:28-0500

Initial speed of bullet before strike =300 m/sec and Block is at rest.Conservation of momentum before and after collision:

(10×300+290×0)×103=(10+290)×103×U(10\times300 + 290\times 0)\times 10^{-3}=(10+290)\times 10^{-3}\times U

Where U is the block-bullet system just after the strike.

U=10m/escU=10 m/esc

Now, equation of motion after the strike.

Fr=MaμR=Maμ(M×g)=Maμg=a......Eq[1]F_r=M a\\ -\mu R=Ma\\ -\mu (M\times g)=Ma\\ -\mu g=a......Eq[1]

Finally system get stopped so final speed is zero and before it travelled S=15 meter.

V2=U2+2as02=1022μg×15μ=103gV^2=U^2+2as\\ 0^2=10^2-2\mu g\times15\\ \mu=\cfrac{10}{3g}\\

Using g=10m/sec2g=10 m/sec^2 ,

μ=13\mu=\cfrac{1}{3}

Hence Coefficient of friction is 13\cfrac{1}{3} .

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