Momentum; bullet block of wood; distance; friction.

[General_Sax]

Homework Statement

A 10g bullet is traveling at a speed of 1.5*10^2 m/s parallel to a horizontal surface when it strikes a 5.5kg wooden block. If the bullet becomes lodged in the block, how far will the block move along this surface? The coefficient of friction between the block and the surface is 0.25.

Homework Equations

p = F(delta)T

d = 0.5*a*t^2

p = mv

The Attempt at a Solution

p = (0.010kg)*(150m/s) = 1.5 kg*m/s
Ff = (0.25)(5.51) = 1.3775

Ff = Fp // when the force created by momentum is equal to the force of friction.

Ff = p / (delta)t

(delta)t = p / Ff

(delta)t = 1.0889s

a = F/m

a = p / t*m

a = 0.25m/s^2

d = 0.5*0.25*(1.0889)^2

d =0.15m

The answer in the book is: 0.015m. Did I make a mistake somwhere or is the answer a misprint?

 

[queenofbabes]

You made an error while calculating the frictional force

 

[General_Sax]

Oh, I see. Thanks.