Calculating Horizontal Displacement of Ballistic Pendulum

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Homework Help Overview

The problem involves a ballistic pendulum where a bullet embeds itself in a pendulum, and the task is to calculate the horizontal displacement of the pendulum after the collision. The context includes concepts of inelastic collisions, conservation of momentum, and energy transformations between kinetic and potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the inelastic nature of the collision and the conservation of momentum. There are attempts to calculate the initial velocity of the combined bullet and pendulum system and to determine the height reached using energy conservation principles. Questions arise regarding the geometry of the situation, specifically how to find the angle and the horizontal displacement.

Discussion Status

Several participants are actively engaging with the problem, providing insights and hints about using trigonometric relationships to find the angle. There is a recognition of the need to clarify the geometry involved, and some participants express frustration over the calculations and seek confirmation on their reasoning.

Contextual Notes

Participants mention a deadline for submission, which adds urgency to the discussion. There is also a reference to potential penalties for incorrect answers, indicating the stakes involved in the homework assignment.

Fanman22
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A 16 g rifle bullet traveling 170 m/s buries itself in a 3.2 kg pendulum hanging on a 2.5 m long string, which makes the pendulum swing upward in an arc. Determine the horizontal component of the pendulum's displacement.

I'm having trouble starting this problem. I know it's inelastic because kinetic energy is transformed into potential energy. I think I need to use this equation:

KE1 + PE1 = KE2 + PE2

Using that I can get the height that the block reaches. I need to find the horizontal displacement and I'm having trouble working the geometry because I don't know the angle above the x-axis. If I knew the angle, I could just use the equation:

Horizontal Displacement = h/tan(theta)

Any suggestions?
 
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Fanman22 said:
I'm having trouble starting this problem. I know it's inelastic because kinetic energy is transformed into potential energy.
No. You know the collision is perfectly inelastic because the bullet gets buried in the block! But momentum is still conserved.

After the collision the "block + bullet" has some KE. That energy gets transformed into potential as it rises. (After the collision, energy is conserved.)

Work the problem in two stages.
 
ok, I used mv/(m + M) = V...came to 0.8458m/s. So that's my initial velocity of the bullet/block system

The I used conservation of energy, KE=PE: .5(m + M)V^2 = (m + M)gh ...comes out to a height of 0.036m.

Now, I've got a triangle with the opposite side of theta being 0.036m. I need the adjacent.

0.036/tan(theta) = MY ANSWER :cry:

But how do I find the theta?
 
Fanman22 said:
But how do I find the theta?
Use some trig. Hint: the string length is given.
 
I did this: SqRt. of ((2.5^2) + (2.464^2))...I get 3.51m. That answer niether makes sense, nor is it correct. Please take a look at my geometry in the pic I just made...

http://img.photobucket.com/albums/v225/Fanman22/triangle.jpg
 
Last edited by a moderator:
*bump*

Just trying to get this solved before 6:15pm EST, I have to submit the answer and I can't stand getting something wrong (-7points) for what I know is a simple error.
 
If theta is the angle where the block on the right is then:

Hypotenuse: 2.5
Opposite: 2.5-0.036 = 2.464
Adj = x

What trig functions can give you x? there's 2.
 
Last edited:
I don't have theta, that's why I was using pythagorean theorum with the lengths that you can see in the pic I posted.

Also, the hypotenuse is 2.5, not 2.75, and the opposite is 2.464
 
You can find theta with the information I just cited
 
  • #10
Oh jeez, sorry, I've been doing calculus and physics homework since 10am ...I guess it's time for a break. Thanks though, I got it in just in time.
 

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