Question

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.

Answer 1

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

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

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

$$U=10 m/esc$$

Now, equation of motion after the strike.

$$F_r=M a\\ -\mu R=Ma\\ -\mu (M\times g)=Ma\\ -\mu g=a......Eq[1]$$

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

$$V^2=U^2+2as\\ 0^2=10^2-2\mu g\times15\\ \mu=\cfrac{10}{3g}\\$$

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

$$\mu=\cfrac{1}{3}$$

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