A bullet strikes a wooden block

[Arooj]

Homework Statement

A 25 g bullet is fired at 150 m/s into a 2.0 kg block of wood hanging from a string 3.0 m long. The bullet becomes embedded in the block. To what height do the bullet and block swing?

Homework Equations

m1v1 = (m1+m2)vf
1/2 mv^2 = mgh

The Attempt at a Solution

0.025 (150) = (0.025 + 2)vf
vf = 1.85 m/s

1/2 2.025 1.85^2 = 2.025 9.8 h
h = 0.175 m

I don't know what to do from here.

 

[frogjg2003]

First things first, try not to plug in numbers until after you get the final result. This reduces rounding error and can give you let you see how to do some of the algebra without having to deal with specific numbers.
That being said, you seem to have done everything right. What do you mean "I don't know that to do from here." You were told to find the height and you found the height.

 

[Arooj]

So the length of the string doesn't matter?

 

[frogjg2003]

Yes. It would only matter if the height is more than twice the string length. If it then wants you to calculate the maximum angle or the oscillation frequency, then the string length would also matter.

 

[Nickie]

I would first like to clarify that there are a few assumptions being made in this scenario, such as neglecting air resistance and assuming an inelastic collision between the bullet and the block.

Moving on to the solution, your calculations for the final velocity and the height reached by the bullet and block are correct. However, to fully answer the question, we would also need to consider the conservation of angular momentum. Since the bullet becomes embedded in the block, the total angular momentum of the system remains the same before and after the collision. This means that the initial angular momentum of the bullet (mvL) is equal to the final angular momentum of the bullet and the block together ((m1+m2)v' L'), where v' is the final velocity of the combined system and L' is the new distance from the pivot point of the string to the center of mass of the bullet and block.

Using this information, we can set up the following equation:

mvL = (m1+m2)v'L'

Substituting in the values we know, we get:

0.025 kg x 150 m/s x 3 m = (0.025 kg + 2 kg) x 1.85 m/s x L'

Solving for L', we get L' = 0.00625 m.

Therefore, the height reached by the bullet and block would be 3 m + 0.00625 m = 3.00625 m.

In conclusion, the bullet and block would swing to a height of 3.00625 m after the collision.