Ballistics Pendulum Homework: Solving for Velocity After Bullet Emerges

[12boone]

Homework Statement

A 23 g bullet traveling 235 m/s penetrates a 3.8 kg block of wood and emerges cleanly at 195 m/s.

If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

Homework Equations

(1) mv= (m+M)v'

momentum before= momentum after

(2) 1/2(m+M)v'^2 = (m+M)gh

The Attempt at a Solution

I attempted to solvethis using these two equations and I received an answer for 1.14 m/s and 230 m/s. I am having a lot of trouble even determining how to start this problem. I don't want a quick easy answer but I would like help working through it! Thank You!

 

[tiny-tim]

A 23 g bullet traveling 235 m/s penetrates a 3.8 kg block of wood and emerges cleanly at 195 m/s.

If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
…
(1) mv= (m+M)v'

momentum before= momentum after

(2) 1/2(m+M)v'^2 = (m+M)gh
…
I attempted to solvethis using these two equations …

Hi 12boone!

However did you use equation (2)?

Equation (1) should be enough.

Try again!

 

[baba89]

Dear student,

First, let's break down the problem into smaller parts. We can consider the bullet and the block as separate systems and use the conservation of momentum and energy to solve for the final velocity of the block.

1. Conservation of Momentum:

According to the first equation, the momentum before the collision is equal to the momentum after the collision. This means that:

mv = (m+M)v'

Where:
m = mass of the bullet
M = mass of the block
v = initial velocity of the bullet
v' = final velocity of the bullet and block after the collision.

Substituting the values given in the problem, we get:

(0.023 kg)(235 m/s) = (0.023 kg + 3.8 kg)v'

Solving for v', we get:
v' = 0.0059 m/s

2. Conservation of Energy:

According to the second equation, the initial kinetic energy of the bullet and block system is equal to the final potential energy of the system. This means that:

1/2(m+M)v'^2 = (m+M)gh

Where:
m = mass of the bullet
M = mass of the block
v' = final velocity of the bullet and block after the collision.
g = acceleration due to gravity (9.8 m/s^2)
h = height through which the block rises after the collision.

Substituting the values given in the problem, we get:

1/2(0.023 kg + 3.8 kg)(0.0059 m/s)^2 = (0.023 kg + 3.8 kg)(9.8 m/s^2)h

Solving for h, we get:
h = 0.00015 m

This means that the block rises 0.00015 meters after the collision.

Now, we can use the equation for the final velocity of an object after free fall to find the final velocity of the block:

vf = √(vi^2 + 2gh)

Where:
vf = final velocity of the block
vi = initial velocity of the block (0 m/s)
g = acceleration due to gravity (9.8 m/s^2)
h = height through which the block rises (0.00015 m)

Substituting the values, we get:

vf = √(0^2 + 2(9.8 m/s^2)(