# Kinematics problem – Bullet in block of wood

**1. The problem statement, all variables and given/known data**

A 3.50kg block of wood is at rest on a 1.75-m high fencepost. When a 12.0g bullet is fired horizontally into the block, the block topples off the post and lands 1.25m away. What was the speed of the bullet immediately before the collision?

**The attempt at a solution**

d=1/2at^2

t=*sqrt(2d-vertical-)/a

t=*sqrt[(2)(1.75m)/(9.81m/s^2)

t= 0.597309633s

d= vt

vf = d(horizontal)/t

vf= 1.25m/0.597309633m/s^2

vf= 2.09216961m/s

Conservation of momentum (P1+P2)before = (Psystem) after

(m of bullet * vi of bullet) + (m of block * vi of block) = (m of system * vf of system)

(0.012kg*vi) + (0) = (3.512kg*2.09216961m/s)

(0.012kg)vi = 7.349621967kg*m/s

vi = 7.349621967kg/.012kg

vi = 612.4684m/s

The initial velocity seems a bit high and I’m not sure how else to approach the question. Help please!! ðŸ™‚

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