A bullet with a mass of 8 g is fired into a 11 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for the wood is 0.20. What is the speed of the bullet?

609.18m/s

Given

A bullet with a mass of 8 g is fired into a 11 kg wood block that is at rest on a wood table.

ie. ?m1?=8gor0.008kg? , ?m2?=11kg?

Also given,

The block, with the bullet embedded, slides 5.0 cm across the table. The coefficient of kinetic friction for the wood is 0.20.

ie. d=5cm or 0.05m, ?μ=0.20?

Let the velocity of block after bullet embedded be '?v0?? '

negative acceleration due to friction will be ?a=μg⇒0.20×9.8⇒a=1.96m/s2?

According to Newton's third equation of motion: ?v2=v02?+2ad?

Since the block finally stops after moving 5cm so, v=0

putting the values we have: ?0=v02?−2×1.96×0.05?

?⇒v0?=0.196??

?⇒v0?=0.442m/s?

Let initially speed of the bullet be 'v'

According to Law of Conservation of momentum : ?Pinitial?=Pfinal??

?m1?v1?+m2?v2?=MV? {in the case of bullet embedded}

{Where, M is the combined mass of bullet and block and V is the speed of combined mass just after collision.}

M=11+0.008=11.008kg, ?V=v0?=0.442m/s? , ?v2?? is the speed of the block before collision which is zero.

Putting all the available values in this expression we have:

?0.008v+0=11.008×0.442?

?⇒0.008v=4.87?

?⇒v=609.18m/s?

So, the speed of the bullet is 609.18m/s.