Momentum; bullet block of wood; distance; friction.

by General_Sax
Tags: block, bullet, distance, friction, momentum, wood
General_Sax is offline
Jul2-09, 10:20 PM
P: 450
1. The problem statement, all variables and given/known data
A 10g bullet is travelling 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.

2. Relevant equations
p = F(delta)T

d = 0.5*a*t^2

p = mv

3. 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?
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queenofbabes is offline
Jul2-09, 10:25 PM
P: 211
You made an error while calculating the frictional force
General_Sax is offline
Jul2-09, 10:41 PM
P: 450
Oh, I see. Thanks.

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