Kinematics problem - Bullet in block of wood

[paytona]

Homework Statement

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! :)

 

[PhanthomJay]

Homework Statement

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! :)

looks good! Bullets travel mighty fast.

 

[voko]

Apart from too many significant figures in the final answer, I do not see anything wrong in your solution. And 612 m/s is not too high for a bullet.

 

[haruspex]

One cause for concern: it says "topples". That implies a rolling movement, so some of the energy has gone into rotation.

 

[Andrew Mason]

Homework Statement

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)

Why are you multiplying by 3.512kg?

AM

 

[haruspex]

Why are you multiplying by 3.512kg?

AM

Mass of block+bullet?

 

[Andrew Mason]

Mass of block+bullet?

The problem does not say that the bullet stays in the block but perhaps it is implied. I cannot imagine how a 3.5kg block of wood is going to stop a 12g rifle bullet at 2000 feet per second.

AM