# Can someone please check this modified pendulum velocity problem

• yamugushi
In summary, The conversation is about finding the velocity of a marble shot from a simple machine (ballistic pendulum) using the equation 1/2mv^2=mgh and trigonometry. The person is concerned about getting a velocity of less than 1m/s when the manufacturer's velocity is 5.5m/s. They figure out their mistake and correct their derivation to get v=rad(GH)/.5.
yamugushi
[solved] Can someone please check this modified pendulum velocity problem

## Homework Statement

I'm looking for the ballistic pendulum velocity for a simple machine (pull back lever, it shoots a marble into a block that raises to a certain degree measured)

I have the angle that the block is raised
length of the string
mass of block

## Homework Equations

Ke=Pe (1/2MV^2=MGH)
cos@=length of sting - height/length of string

## The Attempt at a Solution

.20cos25.3 = .20 - h

which gives me .02cm, which in turn gives me .6m/s, which seems a bit slow, I feel as if I'm messing up my trig but I don't know where...

I figured out I was using an incorrect derivation, I was just so fixed on thinking I had messed up my trig work

Last edited:
I'm confused in what you want us to check? What specific quantity are you concerned about?

The intial velocity of the object after its pulled back a certain distance?

jegues said:
I'm confused in what you want us to check? What specific quantity are you concerned about?

The intial velocity of the object after its pulled back a certain distance?
I've been given the manufacturers velocity, which is 5.5m/s, I'm getting less than 1m/s and honestly I don't see what I'm doing wrong.
BTW here is the pendulum I used:

What exactly is this:

?

jegues said:
What exactly is this:

?

Ke=Pe
1/2mv^2=mgh
masses cancel, divide both sides by .5
take the square root of both sides to get v=rad(GH)/.5

Just your notation--rad is confused with radians. $$\sqrt{}$$ is available on the latex option on advanced setting with the icon:$$\Sigma$$ Or simply denote as sqrt(gh/.5)

## 1. What is a modified pendulum velocity problem?

A modified pendulum velocity problem is a physics problem that involves calculating the velocity of a pendulum with a different set of conditions or parameters compared to a typical pendulum. This can include changes in the length of the pendulum, the mass of the pendulum, or the force acting on the pendulum.

## 2. How do I solve a modified pendulum velocity problem?

To solve a modified pendulum velocity problem, you can use the equation v = √(2gh), where v is the velocity, g is the acceleration due to gravity, and h is the height of the pendulum. You will need to adjust the values of g and h based on the conditions given in the problem, and then calculate the square root to find the velocity.

## 3. Are there any other equations that can be used to solve a modified pendulum velocity problem?

Yes, there are other equations that can be used to solve a modified pendulum velocity problem. These include the equations for kinetic and potential energy, as well as equations that take into account air resistance or friction. The specific equation to use will depend on the conditions given in the problem.

## 4. What are some common mistakes when solving a modified pendulum velocity problem?

One common mistake is forgetting to adjust the values of g and h based on the conditions in the problem. Another mistake is using the wrong equation or forgetting to include all of the relevant variables. It is important to carefully read and understand the given conditions before attempting to solve the problem.

## 5. Can you provide an example of a modified pendulum velocity problem?

Sure, an example of a modified pendulum velocity problem could be: A pendulum with a length of 1 meter and a mass of 2 kg is released from a height of 1.5 meters. What is the velocity of the pendulum at the bottom of its swing? In this case, g would need to be adjusted to 9.8 m/s² and h would be 1.5 meters. Plugging these values into the equation v = √(2gh), we get a velocity of approximately 7.67 m/s.

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