Does anyone know the answer because i got the wrong answer

Question

A 18 g bullet strikes and becomes embedded in a 1.30 kg block of wood placed on a horizontal surface just in front of the gun.

Part A If the coefficient of kinetic friction between the block and the surface is 0.36, and the impact drives the block a distance of 9.5 m before it comes to rest, what was the muzzle speed of the bullet? Express your answer to two significant figures and include the appropriate units.

Answer 1

In this question given mass of bullet 18g and block mass m = 1.3kg , coefficient of surface kinetic friction is s = 0.36 and block traveled before stopped are = 9.5m

First we calculate deceleration then calculate speed of bullet and block then apply conservation of momentum and find initial velocity of bullet which is equal to nozzle speed .

For deceleration we can say that ma = smg

a = sg = a = 0.36 x 10

a = 3.6 m/s²

Hence use kinametics equation v² = u² + 2ax

Where v is final velocity and u is initial velocity.

0² = u² -2 x 3.6 x 9.5 u2

u² = 68.4 = u = 8.27 m/s

This is combined speed of block and bullet hence apply conservation of momentum.

Initial momentum. = Final Momentum

18×10^-3 x v = ( 0.018+ 1.30) x 8.27

v = 605.54 m/s. Answer.